Aristotle on Motion

More than 2000 years ago, ancient Greek scientists were familiar with some of the ideas in physics that we study today. They had a good understanding of some of the properties of light, but they were confused about motion. One of the first to study motion seriously was Aristotle, the most outstanding philosopher-scientist of his time in ancient Greece. Aristotle attempted to clarity motion by classification.

Aristotle divided motion into two main classes: natural motion and violent motion. We shall briefly consider each, not as study material, but only as a background to present-day ideas about motion. Aristotle asserted that natural motion proceeds from the “nature” of an object, dependent on what combination of the four elements (earth, water, air, and fire) the object contains. In his view, every object in the universe has a proper place, determined by this “nature”; any object not in its proper place will “strive” to get there. Being of earth, an unsupported lump of clay properly falls to the ground; being of the air, an unimpeded puff of smoke properly rises; being a mixture of earth and air but predominantly earth, a feather properly falls to the ground, but not as rapidly as a lump of clay. He stated that heavier objects would strive harder. Hence, argued Aristotle, objects should fall at speeds proportional to their weights: the heavier the object, the faster it should fall. Natural motion could be either straight up or straight down, as in the case of all things on Earth, or it could be circular, as in the case of celestial objects.
Unlike up-and-down motion, circular motion has no beginning or end, repeating itself without deviation. Aristotle believed that different rules apply in the heavens, and he asserted that celestial bodies are perfect spheres made of a perfect and unchanging substance, which he called quintessence. Quintessence is the fifth essence, the ocher four being earth, water, air, and fire. The only celestial object with any detectable variation on its face was the Moon. Medieval Christians, still under the sway of Aristotle’s teaching, explained that lunar imperfections were due to the closeness of the Moon and its contamination by the corrupted Earth.

Violent motion, Aristotle’s other class of motion, resulted from pushing or pulling forces. Violent motion was imposed motion. A person pushing a cart or lifting a heavy weight imposed motion, as did someone hurling a stone or winning a tug of war. The wind imposed motion on ships. Floodwaters imposed it on boulders and tree trunks. The essential thing about violent motion was that it was externally caused and was imparted to objects; they moved not of themselves, not by their “nature,” but because of pushes or pulls.

The concept of violent motion had its difficulties, for the pushes and pulls responsible for it were not always evident. For example, a bowstring moved an arrow until the arrow left the bow; after that, further explanation of the arrow’s motion seemed to require some other pushing agent. Aristotle imagined, therefore, that a parting of the air by the moving arrow resulted in a squeezing effect on the rear of the arrow as the air rushed back to prevent a vacuum from forming. The arrow was propelled through the air as a bar of soap is propelled in the bathtub when you squeeze one end of it.

To sum up, Aristotle taught that all motions are due to the nature of the moving object, or due to a sustained push or pull. Provided that an object is in its proper place, it will not move unless subjected to a force. Except for celestial objects, the normal state is one of rest.

Aristotle’s statements about motion were a beginning in scientific thought, and, although he did not consider them to be the final words on the subject, his followers for nearly 2000 years regarded his views as beyond question. Implicit in the thinking of ancient, medieval, and early Renaissance times was the notion that the normal state of objects is one of rest. Since it was evident to most thinkers until the sixteenth century that Earth must be in its proper place, and since a force capable of moving Earth was inconceivable, it seemed quite clear to them that Earth does not move.

Copernicus and the Moving Earth

It was in this intellectual climate that the Polish astronomer Nicolaus Copernicus (1474-1543) formulated his theory of the moving Earth. Copernicus reasoned that the simplest way to account for the observed motions of the Sun, Moon, and planets through the sky was to assume that Earth (and other planets) circle around the Sun. For years he worked without making his thoughts public-for two reasons. The first was that he feared persecution; a theory so completely different from common opinion would surely be taken as an attack on established order. The second reason was that he had grave doubts about it himself; he could not reconcile the idea of a moving Earth with the prevailing ideas of motion. Finally, in the last days of his life, at the urging of close friends, he
sent his De Revo/utionibus to the printer. The first copy of his famous exposition reached him on the day he died-May 24, 1543. Most of us know about the reaction of the medieval Church to the idea that Earth travels around the Sun. Because Aristotle’s views had become so formidably a part of Church doctrine, to contradict them was to question the
Church itself. For many Church leaders, the idea of a moving Earth threatened not only their authority but the very foundations of faith and civilization as well. For better or for worse, this new idea was to overturn their conception of the cosmos-although eventually the Church embraced it.

Describing Motion: Kinematics in One Dimension

The motion of objects-baseballs, automobiles, joggers, and even the Sun and Moon-is an obvious part of everyday life. It was not until the sixteenth and seventeenth centuries that our modern understanding of motion was established. Many individuals contributed to this understanding, particularly Galileo Galilei (1564-1642) and Isaac Newton (1642-1727).

The study of the motion of objects, and the related concepts of force and energy, form the field called mechanics. Mechanics is customarily divided into two parts: kinematics, which is the description of how objects move, and dynamics, which deals with force and why objects move as they do. This chapter and the next deal with kinematics. For now we only discuss objects that move without rotating. Such motion is called translational motion. In this chapter we will be concerned with describing an object that moves along a straight-line path, which is one-dimensional translational motion.

We will often use the concept, or model, of an idealized particle which is considered to be a mathematical point with no spatial extent (no size). Often this point particle is called material point. A point particle can undergo only translational motion. The particle model is useful in many real situations where we are interested only in translational motion and the object’s size is not significant. For example, we might consider a billiard ball, or even a spacecraft traveling toward the Moon, as a particle for many purposes.

Reference Frames

Any measurement of position, distance, or speed must be made with respect to a reference frame, or frame of reference. For example, while you are on a train traveling at 80 km/h, suppose a person walks past you toward the front of the train at a speed of, say, 5 km/h (Fig. 1). This 5 km/h is the person’s speed with respect to the train as frame of reference. With respect to the ground, that person is moving at a speed of 80 km/h + 5 km/h = 85 km/h. It is always important to specify the frame of reference when stating a speed. In everyday life, we usually mean “with respect to the Earth” without even thinking about it, but the reference frame must be specified whenever there might be confusion.

person on train
Fig. 1

When specifying the motion of an object, it is important to specify not only the speed but also the direction of motion. Often we can specify a direction by using north, east, south,  and west, and by “up” and “down”. In physics, we often draw a set of coordinate axes, as shown in Fig. 2, to represent a frame of reference. We can always place the origin 0 and the directions of the x and y axes, as we like for convenience. The x and y axes are always perpendicular to each other. Object positioned to the right of the origin of coordinates (0) on the x axis have an x coordinate which we usually choose to be positive; objects to the left of 0 then have a negative x coordinate. The position along the y axis is usually considered positive when above 0, and negative when below 0, although the reverse convention can be used if convenient. Any point on the plane can he specified by giving its x and y coordinates. In tree dimensions, a z axis perpendicular to the x and y axes is added.

Cartesian coordinates
Fig 2 – Cartesian coordinate system

The invention of Cartesian coordinates in the 17th century by René Descartes (Latinized name: Cartesius) revolutionized mathematics by providing the first systematic link between Euclidean geometry and algebra.

For one-dimensional motion, we often choose the x axis as the line along which the motion takes place. Then the position of an object at any moment is given by its x coordinate. If the motion is vertical, as for a dropped object, we usually use the y axis.

We need to make a distinction between the distance an object has traveled and its displacement, which is defined as the change in position of the object. That is,
displacement is how far the object is from its starting point. To see the distinction
between total distance and displacement, imagine a person walking 70 m to the east and then turning around and walking back (west) a distance of 30 m (see Fig. 3). The total distance traveled is 100 m, but the displacement is only 40 m since the person is now only 40 m from the starting point.

Distance and displacement
Fig. 3 A person walks 70 m east, then 30 m west. The total distance traveled is 100 m (path is shown dashed in black); but the displacement, shown as a solid blue arrow, is 40 m to the east.

Displacement is a quantity that has both magnitude and direction. Such quantities are called vectors, and are represented by arrows in diagrams. For example, in Fig. 3, the blue arrow represents the displacement whose magnitude is 40 m and whose direction is to the right (east).

For now, we deal only with motion in one dimension, along a line. In this case, vectors which point in one direction will have a positive sign, whereas vectors that point in the opposite direction will have a negative sign, along with their magnitude.

Consider the motion of an object over a particular time interval. Suppose that at some initial time, call it  t_1, the object is on the x axis at the position  x_1 in the coordinate system shown in Fig. 4. At some later time, x_1 , suppose the object has moved to position x_2. The displacement of our object is  x_2-x_1and is represented by the arrow pointing to the right in Fig. 4. It is convenient to write

Δx = x_2 - x_1

The arrow of the displacement
Fig. 4 The arrow represents the displacement x2 – x1 . Distances are in meters.

where the symbol Δ (Greek letter delta) means “change in.” Then Δx means “the change in x,” or “change in position,” which is the displacement. Note that the “change in” any quantity means the final value of that quantity, minus the initial value.

Suppose  x_1= 10.0  m and  x_2= 30.0  m . Then

Δx = x_2 - x_1 = 30.0 m - 10.0 m = 20.0 m

so the displacement is 20.0 m in the positive direction, Fig. 4.

Now consider an object moving to the left as shown in Fig. 5. Here the object, say, a person, starts at x_1= 30.0  m and walks to the left to the point x_2= 30.0  m . In this case her displacement is

Δx = x_2 - x_1 = 10.0 m - 30.0 m = - 20.0 m

and the blue arrow representing the vector displacement points to the left.

Negative displacement
Fig. 5 For the displacement 10.0 m – 30.0 m, the displacement vector points to the left.

For one-dimensional motion along the X axis, a vector pointing to the right has a
positive sign, whereas a vector pointing to the left has a negative sign.

Average Velocity

The most obvious aspect of the motion of a moving object is how fast it is moving – its speed or velocity.

The term “speed” refers to how far an object travels in a given time interval, regardless of direction. If a car travels 240 kilometers (km) in 3 hours (h), we say its average speed was 80 km/h. In general, the average speed of an object is defined as the total distance traveled along its path divided by the time it takes to travel this distance:

Average  speed = \frac{total distance}{total  time}                                   (1)

The terms “velocity” and “speed” are often used interchangeably in ordinary language. But in physics we make a distinction between the two. Speed is simply a positive number, with units. Velocity, on the other hand, is used to signify both the magnitude (numerical value) of how fast an object is moving and also the direction in which it is moving. (Velocity is therefore a vector.) There is a second difference between speed and velocity: namely, the average velocity is defined in terms of displacement, rather than total distance traveled:

Average  velocity = \frac{displacement}{time  elapsed}=\frac{final position - initial position}{time  elapsed}

Average speed and average velocity have the same magnitude when the motion is all in one direction. In other cases, they may differ: recall the walk we described earlier, in Fig. 3, where a person walked 70 m east and then 30 m west. The total distance traveled was 70 m + 30 m = 100 m, but the displacement was 40 m. Suppose this walk took 70 s to complete. Then the average speed was:

\frac{distance}{total  time} = \frac{100 m}{70 s} = 1.4 m/ s

The magnitude of the average velocity, on the other hand, was:

\frac{displacement}{total  time} = \frac{40 m}{70 s} = 0.57 m/ s

This difference between the speed and the magnitude of the velocity can occur when we calculate average values.

For one-dimensional motion of an object in general, the average velocity, defined as the displacement divided by the elapsed time, can be written

Average  velocity = \bar{v}= \frac{x_2 - x_1}{t_2 - t_1}                                               (2)

For the usual case of the + x axis to the right, note that if x_2 is less than x_1, the object is moving to the left, and then Δx = x_2 –  x_1 is less than zero. The sign of the displacement, and thus of the average velocity, indicates the direction: the average velocity is positive for an object moving to the right along the + X axis and negative when the object moves to the left. The direction of the average velocity is always the same as the direction of the displacement. 

Note that it is always important to choose (and state) the elapsed time, or time interval, t_2 –  t_1, the time that passes during our chosen period of observation.

Instantaneous Velocity

If you drive a car along a straight road for 150 km in 2.0 h, the magnitude of your average velocity is 75 km/h. It is unlikely, though, that you were moving at precisely 75 km/h at every instant. To describe this situation we need the concept of instantaneous velocity, which is the velocity at any instant of time. More precisely, the instantaneous velocity at any moment is defined as the average velocity over an infinitesimally short time interval.  We can write the definition of instantaneous velocity, v, for one-dimensional motion as the average velocity over an infinitesimally short time interval. That is, Eq. (2) is to be evaluated in the limit of Δt becoming extremely small, approaching zero. We can write the definition of instantaneous velocity, v, for one-dimensional motion as

v = Δx/Δt                 when Δt → 0                                                                                 (3)

The ratio Δx/Δt is to be evaluated in the limit of Δt approaching zero. But we do not simply set Δt = 0 in this definition, for then Δx would also be zero, and we would have an undefined number. Rather, we are considering the ratio Δx/ Δt, as a whole. As we let Δt approach zero, Δx approaches zero as well. But the ratio Δx/ Δt approaches some definite value, which is the instantaneous velocity at a given instant. When we use the term “velocity” it will refer to instantaneous velocity. When we want to speak of the average
velocity, we will make this clear by including the word “average.”

Note that the instantaneous speed always equals the magnitude of the instantaneous velocity. Why? Because distance traveled and the magnitude of the displacement become the same when they become infinitesimally small.

If an object moves at a uniform (that is, constant) velocity during a particular time interval, then its instantaneous velocity at any instant is the same as its average velocity (see Fig. 6a). But in many situations this is not the case. For example, a car may start from rest, speed up to 50 km/h, remain at that velocity for a time, then slow down to 20 km/h in a traffic jam, and finally stop at its destination after traveling a total of 15 km in 30 min. This trip is plotted on the graph of Fig. 6b. Also shown on the graph is the average velocity (dashed line), which is v = Δx/Δt = 15 km/0.50 h = 30 km/h.

Velocity of a car as a function of time: (a) at constant velocity; (b) with varying velocity.
Fig. 6 Velocity of a car as a function of time: (a) at constant velocity; (b) with varying velocity.


An object whose velocity is changing is said to be accelerating. For instance, a car whose velocity increases in magnitude from zero to 80 km/h is accelerating. Acceleration specifies how rapidly the velocity of an object is changing.

Average acceleration is defined as the change in velocity divided by the time taken
to make this change:

Average  acceleration = \frac{change of velocity}{time elapsed}

The average acceleration over a time interval Δt = t_2 –  t_1 during which the velocity changes by  Δv = v_2 –  v_1, is defined as

\bar{a}\frac{v_2-v_1}{t_2-t_1 } = Δv/Δt                                                        (4)

Because velocity is a vector, acceleration is a vector too. But for one-dimensional motion, we need only use a plus or minus sign to indicate acceleration direction relative to a chosen coordinate axis.

EXAMPLE 1: A car accelerates along a straight road from rest to 90 km/h in 5.0 s. What is the magnitude of its average acceleration?

SOLUTION: The average acceleration is

a = \frac{25 m/ s - 0 m/ s }{5.0 s } = 5.0 m/s²

Note that acceleration tells us how quickly the velocity changes, whereas velocity tells us how quickly the position changes.

EXAMPLE 2: Velocity and acceleration. (a) If the velocity of an object is zero, does it mean that the acceleration is zero? (b) If the acceleration is zero, does it mean that the velocity is zero? Think of some examples.

SOLUTION: A zero velocity does not necessarily mean that the acceleration is zero, nor does a zero acceleration mean that the velocity is zero. (a) For example, when you put your foot on the gas pedal of your car which is at rest, the velocity starts from zero but the acceleration is not zero since the velocity of the car changes. (How else could your car start forward if its velocity weren’t changing- that is, accelerating?) (b) As you cruise along a straight highway at a constant velocity of 100 km/h, your acceleration is zero: a = 0,  but v≠0.

EXAMPLE 3: Car slowing down. An automobile is moving to the right along a straight highway, which we choose to be the positive x axis (Fig. 7). Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v_1 = 15.0 m/s, and it takes 5.0 s to slow down to v_2 = 5.0 m/s, what was the car’s average acceleration?

Car slowing down
Fig. 7 The acceleration vector (orange) points to the left as the car slows down while moving to the right.

SOLUTION: In Eq. 4 we call the initial time t_1 = 0, and set t_2 = 5.0 s. Note that our choice of t_1 = 0 doesn’t affect the calculation of a because only Δt = t_2 –  t_1 appears in Eq. 4 Then

a = \frac{5.0 m/s - 15.0 m/s }{5.0 s} = – 2.0 m/s²

The negative sign appears because the final velocity is less than the initial velocity. In this case the direction of the acceleration is to the left (in the negative x direction)-even though the velocity is always pointing to the right. We say that the acceleration is 2.0 m/s² to the left.


When an object is slowing down, we can say it is decelerating. But be careful: deceleration does not mean that the acceleration is necessarily negative. The velocity of an object moving to the right along the positive x axis is positive; if the object is slowing
down (as in Fig. 7), the acceleration is negative. But the same car moving to the left
(decreasing x), and slowing down, has positive acceleration that points to the right, as
shown in Fig. 8. We have a deceleration whenever the magnitude of the velocity is
decreasing, and then the velocity and acceleration point in opposite directions.

Fig. 8 Deceleration means the magnitude of the velocity is decreasing; a is not necessarily negative

Instantaneous Acceleration

The instantaneous acceleration is defined as the limiting value of the average
acceleration as we let Δt approach zero:

a = Δv/Δt        when Δt → 0                                                                              (5)

We will use the term “acceleration” to refer to the instantaneous value. If we want to discuss the average acceleration, we will always include the word “average.”

Like velocity, acceleration is a rate. The velocity of an object is the rate at which its displacement changes with time; its acceleration, on the other hand, is the rate at which its velocity changes with time. In a sense, acceleration is a “rate of a rate.”

Motion at Constant Acceleration

We now examine the situation when the magnitude of the acceleration is constant and the motion is in a straight line. In this case, the instantaneous and average accelerations are equal. We use the definitions of average velocity and acceleration to derive a set of valuable equations that relate x, v, a, and t when a is constant, allowing us to determine any one of these variables if we know the others.

To simplify our notation, let us take the initial time in any discussion to be zero, and we call it to: t_0 = t_1  = 0. (This is effectively starting a stopwatch at t_0.) We can then let t_2 = t be the elapsed time. The initial position x_1 and the initial velocity  v_1 of an object will now be represented by  x_0 and v_0, since they represent x and v at t = 0. At time t the position and velocity will be called x and v (rather than x_2 and v_2) The average velocity during the time interval  t-t_0 will be

\bar{a} = Δx/Δt = \frac{x_2 - x_0}{t_2 - t_0} = \frac{x - x_0}{t}

since we chose t_0 = 0. The acceleration, assumed constant in time, is

a =  \frac{v - v_0}{t}

A common problem is to determine the velocity of an object after any elapsed time t, when we are given the object’s constant acceleration. We can solve such problems by solving for v in the last equation to obtain:

v =  v_0 + at                                                     [Only if acceleration is constant!]         (6)

Next, let us see how to calculate the position x of an object after a time t when it undergoes constant acceleration. From the definition of average velocity we get

x = x_0 + \bar{v} t                                                                                                                (7)

Because the velocity increases at a uniform rate, the average velocity, \bar{v}, will be midway between the initial and final velocities:

\bar{v} = \frac{v_0 +v}{2}                                     [Only if acceleration is constant!] (8)

We combine (6), (7) and (8) and find

x = x_0 + v_0 t + \frac{1}{2}at^{2}                                       [constant acceleration] (9)

We now derive equation, which is useful in situations where the time t is not known. We substitute Eq. 8 into Eq. 7:

x = x_0 +  \frac{v+v_0}{2}t

Next we solve Eq. 6 for t, obtaining

t = \frac{v-v_0}{a}

and substituting this into the previous equation we have

x = x_0 + \frac{v^{2}-v_0^{2}}{2a}t

We solve this for v² and obtain

v^{2} = v_0^{2} + 2a(x-x_0)                                                                                         (10)

We now have four equations relating position, velocity, acceleration, and time, when the acceleration a is constant. We collect these kinematic equations here in one place for future reference:

Kinematic equations

Freely Falling Objects

One of the most common examples of uniformly accelerated motion is that of an object allowed to fall freely near the Earth’s surface. That a falling object is accelerating may not be obvious at first. It was widely believed before the time of Galileo, that heavier objects fall faster than lighter objects and that the speed of fall is proportional to how heavy the object is.

Galileo postulated that all objects would fall with the same constant acceleration in the absence of air or other resistance. He showed that this postulate predicts that for an object falling from rest, the distance traveled will be proportional to the square of the time, that is d∼ t². We can see this from our kinematic equations; but Galileo was the first to derive this mathematical relation.

To support his claim that falling objects increase in speed as they fall, Galileo made use of a clever argument: a heavy stone dropped from a height of 2 m will drive a stake into the ground much further than will the same stone dropped from a height of only 0.2 m. Clearly, the stone must be moving faster in the former case.

Galileo claimed that all objects, light or heavy, fall with the same acceleration, at least in the absence of air. If you hold a piece of paper horizontally in one hand and a heavier object-say, a baseball-in the other, and release them at the same time as in Fig. 9a, the heavier object will reach the ground first. But if you repeat the experiment, this time crumpling the paper into a small wad (see Fig. 9b), you will find that the two objects reach the floor at nearly the same time.

ball and paper
Fig 9. (a) A ball and a light piece of paper are dropped at the same time. (b) Repeated, with the paper wadded up.

Galileo was sure that air acts as a resistance to very light objects that have a large surface area. But in many ordinary circumstances this air resistance is negligible. In a chamber from which the air has been removed, even light objects like a feather or a horizontally held piece of paper will fall with the same acceleration as any other object (see Fig. 10). Such a demonstration in vacuum was not possible in Galileo’s time, which makes Galileo’s achievement all the greater. Galileo is often called the “father of modern science,” not only for the content of his science (astronomical discoveries, inertia, free fall) but also for his approach to science (idealization and simplification, mathematization of theory, theories that have testable consequences, experiments to test theoretical predictions).

Stone and feather fall
Fig. 10 A rock and a feather are dropped simultaneously (a) in air, (b) in a vacuum.

Galileo’s specific contribution to our understanding of the motion of falling objects can be summarized as follows:

at a given location on the Earth and in the absence of air resistance, all objects
fall with the same constant acceleration.

We call this acceleration the acceleration due to gravity on the surface of the Earth, and we give it the symbol g. Its magnitude is approximately

g = 9.80 m/s²                                                                           [at surface of Earth]

Actually, g varies slightly according to latitude and elevation, but these variations are so small that we will ignore them for most purposes. The effects of air resistance are often small, and we will neglect them for the most part. However, air resistance will be noticeable even on a reasonably heavy object if the velocity becomes large.  Acceleration due to gravity is a vector as is any acceleration, and its direction is downward, toward the center of the Earth.

When dealing with freely falling objects we can make use of kinematic equations, where
for a we use the value of g given above. Also, since the motion is vertical we will
substitute y in place of x, and y_0 in place of x_0. We take  y_0 = 0 unless otherwise specified. It is arbitrary whether we choose y to be positive in the upward direction or in the downward direction; but we must be consistent about it throughout a problem’s solution.

Galileo Galilei  (1564-1642)

Galileo was born in Pisa, Italy, in the same year Shakespeare was born and Michelangelo died. He studied medicine at the University of Pisa and then changed to mathematics.

Galileo Galilei
Portrait by Justus Sustermans painted in 1636.

He developed an early interest in motion and was soon at odds with his contemporaries, who held to Aristotelian ideas on falling bodies. He left Pisa to teach at the University of Padua and became an advocate of the new Copernican theory of the solar system. He was one of the first to build a telescope, and he was the first to direct it to the nighttime
sky and discover mountains on the Moon and the moons of Jupiter. Because he published his findings in Italian instead of in the Latin expected of so reputable a scholar, and because of the recent invention of the printing press, his ideas reached a wide readership. He soon ran afoul of the Church, and he was warned not to teach, and not to
hold to, Copernican views. He restrained himself publicly for nearly 15 years and then defiantly published his observations and conclusions, which were counter to Church
doctrine, The outcome was a trial in which he was found guilty, and he was forced to renounce his discoveries. By then an old man, broken in health and spirit, he was
sentenced to perpetual house arrest. Nevertheless, he completed his studies on motion, and his writings were smuggled from Italy and published in Holland. Earlier, he had damaged his eyes looking at the Sun through a telescope, which led to blindness at the age of 74. He died 4 years later. In 1642, several months after Galileo died, Isaac Newton was born.

Galileo and the Leaning Tower

It was Galileo, the foremost scientist of the early seventeenth century, who gave credence to the Copernican view of a moving Earth. He accomplished this by discrediting the Aristotelian ideas about motion. Although he was not the first to point out difficulties in Aristotle’s views, Galileo was the first to provide conclusive refutation through observation and experiment.

Galileo easily demolished Aristotle’s falling-body hypothesis. Galileo is said
to have dropped objects of various weights from the top of the Leaning Tower
of Pisa and to have compared their falls. Contrary to Aristotle’s assertion, Galileo found that a stone twice as heavy as another did not fall twice as fast. Except for the small effect of air resistance, he found that objects of various weights, when released at the same time, fell together and hit the ground at the same time. On one occasion, Galileo allegedly attracted a large crowd to witness the dropping of two objects of different weight from the top of the tower. Legend has it that many observers of this demonstration who saw the objects hit the ground together scoffed at the young Galileo and continued to hold fast to their Aristotelian teachings.

Galileo and the Leaning Tower
Galileo and the Leaning Tower

How to solve problems:

1. Read and reread the whole problem carefully before trying to solve it.
2. Decide what object ( or objects) you are going to study, and for what time interval. You can often choose the initial time to be t = O.
3. Draw a diagram or picture of the situation, with coordinate axes wherever applicable. [You can place the origin of coordinates and the axes wherever you like to make your calculations easier.]
4. Write down what quantities are “known” or “given,” and then what you want to know. Consider quantities both at the beginning and at the end of the chosen time interval.
5. Think about which principles of physics apply in this problem. Use common sense and your own experiences. Then plan an approach.
6. Consider which equations (and/or definitions) relate the quantities involved. Before using them, be sure their range of validity includes your problem. If you find an applicable equation that involves only known quantities and one desired unknown, solve the equation algebraically for the unknown. Sometimes several sequential calculations, or a combination of equations, may be needed. It is often preferable to solve algebraically for the desired unknown before putting in numerical values.

7. Carry out the calculation if it is a numerical problem. Keep one or two extra digits during the calculations, but round off the final answer( s) to the correct number of significant figures.

8. Think carefully about the result you obtain: Is it reasonable? Does it make sense according to your own intuition and experience? A good check is to do a rough estimate using only powers of ten. Often it is preferable to do a rough estimate at the start of a numerical problem because it can help you focus your attention on finding a path toward a solution.

9. A very important aspect of doing problems is keeping track of units. An equals sign implies the units on each side must be the same, just as the numbers must. If the units do not balance, a mistake has no doubt been made. This can serve as a check on your solution (but it only tells you if you’re wrong, not if you’re right). Always use a consistent
set of units.

Practice problems:

1.What is the average speed of a car that travels 30 meters in 10 seconds, then another 50 meters in 30  seconds?

2. What is the velocity of a car that travels  – 40 meters in 5 seconds?

3. An athlete runs exactly once around a circular track with a total length of 500 m. Find the runner’s displacement for the race.

4. What does the slope of a position-time graph measure?

Use the graph below for problems 5-7 : gr55. The graph shows the motion of a person on a bicycle. When does the person have the greatest velocity?

6. When is the person on the bicycle farthest away from the starting point?

7. Over what interval does the person on the bicycle travel the greatest distance?

8. The graph below represents the motion of a bicycle. Determine the bicycle’s average speed and average velocity, and describe its motion in words. gr1

9. Figure below is a graph of two people running.                                                                    a. a. Describe the position of runner A relative to runner B at the y-intercept.
b. Which runner is faster?
c. What occurs at point P and beyond? gr2

10. The position-time graph in the Figure below shows the motion of four cows walking from the pasture back to the barn. Rank the cows according to their average velocity, from slowest to fastest. gr3

11. Figure below is a position-time graph for a rabbit running away from a dog.                 a. Describe how this graph would be different if the rabbit ran twice as fast.

b. Describe how this graph would be different if the rabbit ran in the opposite    direction. gr4

12. A race car’s velocity increases from 4.0 m/s to 36 m/s over a 4 s time interval. What is its average acceleration?

13. A car accelerates along a straight road from rest to 90 km/h in 5.0 s. What is the magnitude of its average acceleration?

14. The race car in the previous problem slows from 36 m/s to 15 m/s over 3.0 s. What is
its average acceleration?

15. If the rate of continental drift were to abruptly slow from 1.0 cm/y to 0.5 cm/y over
the time interval of a year, what would be the average acceleration?

16. A bus that is traveling at 30.0 km/h speeds up at a constant rate of 3.5 m/s². What velocity does it reach 6.8 s later?

17. If a car accelerates from rest at a constant 5.5 m/s², how long will it take for the car to reach a velocity of 28 m/s?

19. How long does it take a car to cross a 30.0-m-wide intersection after the light turns green, if the car accelerates from rest at a constant 2.00 m/s²?

APPROACH: Follow the Problem Solving Strategy above, step by step.


1. Reread the problem. Be sure you understand what it asks for (here, a time

2. The object under study is the car. We choose the time interval: t = 0, the initial time, is the moment the car starts to accelerate from rest (v_o = 0); the time t is the instant the car has traveled the full 30.0-m width of the intersection.

3. Draw a diagram: the situation is shown in the figure below, where the car is shown moving along the positive x axis. We choose x_o = 0 at the front bumper of the car before it starts to move.


4. The “knowns” and the “wanted” are shown in the Table in the margin, and we choose  x_o = 0. Note that “starting from rest” means v = 0 at t = 0; that is, v_o = 0.


5. The physics: the motion takes place at constant acceleration, so we can use the
kinematic equations for constant acceleration.

6.  Equations: we want to find the time, given the distance and acceleration; the eqn.

x = x_0 + v_0 t + \frac{1}{2}at^{2}  is perfect since the only unknown quantity is t.

Setting x_o = 0 and v_o = 0 we can solve for t:

x = \frac{1}{2}at^{2}

t = (\frac{2x}{a})^{1/2}

7. The calculation: t = 5.48 s Note that the units come out correctly (seconds).

NOTE In steps 6 and 7, when we took the square root, we should have written

t = ± (\frac{2x}{a})^{1/2} = ± 5.48 s

Mathematically there are two solutions. But the second solution, t = -5.48 s, is a time before our chosen time interval and makes no sense physically. We say it is “unphysical” and ignore it.

20. Falling from a tower. Suppose that a ball is dropped (v_o = 0) from a tower 70.0 m high. How far will it have fallen after a time t_1 = 1.0 s, t_2 = 2.0 s , and t_3 = 3.0 s? Ignore air resistance.

An object dropped from a tower
An object dropped from a tower falls with progressively greater speed and covers greater distance with each successive second.

APPROACH Let us take y as positive downward, so the acceleration is a = g = +9.80 m/s². We set v_o = 0 and y_o = 0. We want to find the position y
of the ball after three different time intervals.

SOLUTION We set t = t_1 = 1.0 s :

y(t_1) = y_0 + v_0 t_1 + \frac{1}{2}at_1^{2}= 0 + \frac{1}{2} (9.80 m/s^{2} )(1.0 s)^{2} =4.9 m

The ball has fallen a distance of 4.90 m during the time interval t = 0 to  t_1 = 1.0 s

Similarly, after 2.0 s (= t_2 ), the ball’s position is

y(t_2) =  \frac{1}{2}at_2^{2}= \frac{1}{2} (9.80 m/s^{2} )(2.0 s)^{2} =19.6 m

Finally, after 3.0 s (= t_3 ), the ball’s position is

y(t_3) =  \frac{1}{2}at_3^{2}= \frac{1}{2} (9.80 m/s^{2} )(3.0 s)^{2} =44.1 m

21. Thrown down from a tower. Suppose the ball in problem 20 is thrown downward with an initial velocity of 5.0 m/s, instead of being dropped. (a) What then would be its position after 1.0 sand 2.0 s? (b) What would its speed be after 1.0 sand 2.0 s? Compare with the speeds of a dropped ball.

APPROACH Again we use the same kinematic equation, but now  v_o is not zero, it is v_o = 5 m/s.

SOLUTION (a) At t = 1.00 s, the position of the ball  is

y = y_0 + v_0 t + \frac{1}{2}at^{2} = (5.0m/s)(1.0s)   + \frac{1}{2} (9.80 m/s^{2} )(1.0 s)^{2} =9.9 m

At t = 2.00 s, (time interval t = 0 to t = 2.0 s), the position is

y =  v_0 t + \frac{1}{2}at^{2} = (5.0m/s)(2.0s)   + \frac{1}{2} (9.80 m/s^{2} )(2.0 s)^{2} =29.6 m

As expected, the ball falls farther each second than if it were dropped with v_o = 0

(b) The velocity is

v = v_0 + at =

= 5.0m/s + (9.80m/s² )(1.0s) = 14.8m/s    [at t_1 = 1.0s]

= 5.0m/s + (9.80m/s² )(2.0s) = 24.6m/s    [at t_2 = 2.0s]

In problem 20, when the ball was dropped (v_o = 0 ), the first term (v_o ) in these equations was zero, so

v = 0 + at =

= (9.80m/s² )(1.0s) = 9.8m/s      [at t_1 = 1.0s]

= (9.80m/s² )(2.0s) = 19.6m/s    [at t_2 = 2.0s]

NOTE For both problems 20 and 21, the speed increases linearly in time by 9.80 m/s during each second. But the speed of the downwardly thrown ball at any instant is always 5.0 m/s (its initial speed) higher than that of a dropped ball.

22. Ball thrown upward, part I. A person throws a ball upward into the air with an initial velocity of 15.0 m/s. Calculate (a) how high it goes, and (b) how long the ball is in the air before it comes back to the hand. Ignore air resistance.

An object thrown into the air

APPROACH We are not concerned here with the throwing action, but only with the motion of the ball after it leaves the thrower’s hand and until it comes back to the hand again. Let us choose y to be positive in the upward direction and negative in the downward direction.  The acceleration due to gravity is downward and so will have a negative sign, a = – g = -9.80 m/s². As the ball rises, its speed decreases until it reaches the highest point, where its speed is zero for an instant; then it descends, with increasing speed.

SOLUTION (a) We consider the time interval from when the ball leaves the thrower’s hand until the ball reaches the highest point. To determine the maximum height, we calculate the position of the ball when its velocity equals zero (v = 0 at the highest point). At t = 0 (point A ) we have y_o = 0v_o = 15.0 m/s and a = -9.8 m/s². At time t (maximum height), v=0, a = -9.8 m/s², and we wish to find y.  We use the third kinematic equation,  replacing x with y: v^{2} = (v_0)^{2} +2ay. We solve this equation for y:

y = \frac{v^{2}-(v_0)^{2}}{2a} = \frac{0-(15m/s)^{2}}{2(-9.8m/s^{2})} = 11.5 m

The ball reaches a height of 11.5 m above the hand.

(b) Now we need to choose a different time interval to calculate how long the ball is in the air before it returns to the hand. We could do this calculation in two parts by first determining the time required for the ball to reach its highest point, and then determining the time it takes to fall back down. However, it is simpler to consider the time interval for the entire motion from A to B to C in one step. We can do this because y represents position or displacement, and not the total distance traveled. Thus, at both points A and C, y = 0:

y = y_0 + v_0 t + \frac{1}{2}at^{2}

0 = 0 + (15.0 m/s)t + (1/2)( -9.80 m/s²  )t²

This equation is readily factored (we factor out one t):

(15.0 m/s – 4.9 m/s² t)t = 0

There are two solutions:

t = 0          and        t = 3.06 s.

The first solution (t = 0) corresponds to the initial point (A), when the ball was first thrown from y = O. The second solution, t = 3.06 s, corresponds to point C, when the ball has returned to y = O. Thus the ball is in the air 3.06 s.

NOTE We have ignored air resistance, which could be significant, so our result is
only an approximation to a real, practical situation.

23. Two possible misconceptions. Give examples to show the error in these two common misconceptions: (1) that acceleration and velocity are always in the same direction, and (2) that an object thrown upward has zero acceleration at the highest point.

RESPONSE Both are wrong.

(1) Velocity and acceleration are not necessarily in the same direction. When the ball in problem 21 is moving upward, its velocity is positive (upward), whereas the acceleration is negative (downward).

(2) At the highest point (B in problem 22), the ball has zero velocity for an instant. Is the
acceleration also zero at this point? No. The velocity near the top of the arc points upward, then becomes zero (for zero time) at the highest point, and then points downward. Gravity does not stop acting, so a = – g = -9.80 m/s² even there. Thinking that a = 0 at point B would lead to the conclusion that upon reaching point B, the ball would stay there: if the acceleration (= rate of change of velocity) were zero, the velocity would stay zero at the highest point, and the ball would stay up there without falling. In sum, the acceleration of gravity always points down toward the Earth, even when the object is moving up.

24. Ball thrown upward, part II. Let us consider again the ball thrown upward of problem 22, and make more calculations. Calculate (a) how much time it takes for the ball to reach the maximum height (point B ), and (b) the velocity of the ball when it returns to the thrower’s hand (point C).

APPROACH  Again we assume the acceleration is constant, and we take y as
positive upward.


(a) We consider the time interval between the throw (t = 0, v_o = 15.0 m/s) and the top of the path (y = +11.5 m, v = 0), and we want to find t. The acceleration is constant at a = – g = -9.80 m/s². We use the equation

v = v_0 + at

setting v=0, v_o = 15.0 m/s, a=-9.8 m/s² and solving for t gives:

t = - \frac{v_0}{a} = - \frac{15.0 m/s }{-9.80 m/s^{2}} = 1.53 s

This is just half the time it takes the ball to go up and fall back to its original position. Thus it takes the same time to reach the maximum height as to fall back to the starting point.

(b) Now we consider the time interval from the throw (t = 0, v_o = 15.0 m/s) until the ball’s return to the hand, which occurs at t = 3.06 s ( as calculated in problem 21),  and we want to find v when t = 3.06 s:

v = v_0 + at = 15.0 m/s – (9.8 m/s²  )(3.06 s) = -15.0 m/s.

NOTE The ball has the same speed (magnitude of velocity) when it returns to the
starting point as it did initially, but in the opposite direction (this is the meaning
of the negative sign). And, as we saw in part (a), the time is the same up as down.
Thus the motion is symmetrical about the maximum height.

25. Ball thrown upward, part III: the quadratic formula. 

For the ball in problem 22, calculate at what time t the ball passes a point 8.00 m
above the person’s hand.

APPROACH We choose the time interval from the throw (t = 0, v_o = 15.0 m/s)
until the time t (to be determined) when the ball is at position y = 8.0 m.

SOLUTION We want to find t, given y = 8.0 m, y_o = 0v_o = 15.0 m/s, and
a = -9.80 m/s² :

y = y_0 + v_0 t + \frac{1}{2}at^{2}

8.0 m = 0 + (15.0 m/s)t + (1/2)(-9.80 m/s²)t².

To solve any quadratic equation of the form at² + bt + c = 0, where a, b, and c
are constants (a is not acceleration here), we use the quadratic formula:

t = \frac{-b \pm\sqrt{b^{2}-4ac}}{2a}

We rewrite our y equation just above in standard form, at² + bt + c = 0:

(4.9 m/s²)t² – (15.0 m/s)t + (8.0 m) = O.

So the coefficient a is 4.9 m/s² , b is -15.0 m/s, and c is 8.0 m. Putting these into
the quadratic formula, gives us t = 0.69 s and t = 2.37 s. Are both solutions valid? Yes, because the ball passes y = 8.0 m when it goes up (t = 0.69 s) and again when it
comes down (t = 2.37 s).

Figure below shows graphs of (a) y vs. t and (b) v vs. t for the ball thrown

ball thrown upward

Sample test questions/problems

1. What is the average speed of a car that travels 30 meters in 10 seconds, then another 50 meters in 30  seconds?

2. What is the velocity of a car that travels -30 meters in 6 seconds?

3. What is the acceleration of a person that goes from 50 m/s to 40 m/s in 10 seconds?

4. A person walks due East for 8 meters and then due North for 15 meters. What is the total distance traveled? What is the displacement?

5.* A car travels up a hill at a constant speed of 30 km/h and returns down the hill at a constant speed of 70 km/h. Calculate the average speed for the whole trip.

6. Two locomotives approach each other on parallel tracks. Each has a speed of 95 km/h with respect to the ground. If they are initially 8.5 km apart, how long will it be before they reach each other?

two trains

7. An automobile traveling 95 km/h overtakes a 1.10-km- long train traveling in the same direction on a track parallel to the road. If the train’s speed is 75 km/h, how long does it
take the car to pass it, and how far will the car have traveled in this time? What are the results if the car and train are traveling in opposite directions?

train and car

8.  If raindrops fall vertically at a speed of 3 m/s and you are running at 4 m/s, how fast do they hit your face?

9. A canoe is paddled at 4 km/h directly across a river that flows at 3 km/h, as shown in the figure. (a) What is the resultant speed of the canoe relative to the shore? (b) In approximately what direction should the canoe be paddled to reach a destination directly across the river?

velocity vecors addition

10. A race car starting from rest accelerates at a constant rate of 5.00 m/s2. What is the velocity of the car after it has traveled 100 m?

11. A jet lands at 80.0 m/s, applying the brakes 2.0 s after landing. Find the acceleration needed to stop the jet within 500 m.

12. How long does it take a car to cross a 30.0 m wide intersection after the light turns green, if the car accelerates from rest at a constant 2.0 m/s2?

13. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

14. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.

15. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.

16. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.

17. The observation deck of tall skyscraper is 370 m above the street. Determine the time required for a penny to fall from the deck to the street below.

18. A stone is dropped into a deep well and is heard to hit the water 3.57 s after being dropped. Approximately determine the depth of the well assuming that the speed of sound is very high.

19. Two birds are flying directly towards each other at the same speed If the first bird is flying at a velocity v , what is the velocity of the second bird?

20. A rock is thrown straight upward off the edge of a balcony that is 5 m above the ground. The rock rises 10 m, then falls all the way down to the ground below the balcony. What is the rock’s displacement?

21. An athlete runs exactly once around a circular track with a total length of 500 m. Find the runner’s displacement for the race.

22. Is it possible to move with constant speed but not constant velocity? Why?

23. A person walks 40 m east, then 80 m west for 8 seconds. Calculate the average speed. Calculate the average velocity.

24. Find the acceleration of a material point if its position function is x= x(t) = 3t2 – 6t – 6 where t is in seconds. Assume SI units.

25. What is the initial velocity of the particle from problem 20 ?  What is its initial coordinate?

26. Is it possible to move with constant velocity but not constant speed? Why?

27. The average velocity and the instantaneous velocity of an object will be the same if

A) the object’s speed is constant

B) the object’s acceleration is zero

C) the object’s velocity is always positive

D) They are never the same

28. Is it true or false: 2x > x ? Why?

29. A rubber band ball is thrown upright to a wall. If the collision is totally elastic and the velocity of the ball is v before the collision, what is the velocity after the collision?

30. A person driving her car at 45 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow light lasts only 2.0 s before turning to red, and she is 28 m away from the near side of the intersection. Should she try to stop, or should she speed up to cross the intersection before the light turns red? The intersection is 15 m wide. Her car’s maximum deceleration is -5.8 m/s², whereas it can accelerate from 45 km/h to 65 km/h in 6.0 s. Ignore the length of her car and her reaction time.

to stop or to speed up

31. Bill can throw a ball vertically at a speed 1.5 times faster than Joe can. How many times higher will Bill’s ball go than Joe’s?

32. A stone is dropped from the roof of a high building. A second stone is dropped 1.50 s later. How far apart are the stones when the second one has reached a speed of 12.0 m/s?

falling stone

33. A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 3.57 s later. If the speed of sound is 340 m/s, how high is the cliff?