When air resistance is neglected all objects fall toward the ground the same acceleration called?

By the end of this section, you will be able to:

Describe the effects of gravity on objects in motion.
Describe the motion of objects that are in free fall.
Calculate the position and velocity of objects in free fall.

Índice Show

One-Dimensional Motion Involving Gravity
Section Summary
Slopes and General Relationships
Graph of Displacement vs. Time (a = 0, so v is constant)
Graphs of Motion when a is constant but a≠0
Graphs of Motion Where Acceleration is Not Constant

Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.

The most remarkable and unexpected fact about falling objects is that, if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration, independent of their mass. This experimentally determined fact is unexpected, because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones.

In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball will reach the ground after a hard baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, while friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them. For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free-fall.

The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called the acceleration due to gravity. The acceleration due to gravity is constant, which means we can apply the kinematics equations to any falling object where air resistance and friction are negligible. This opens a broad class of interesting situations to us. The acceleration due to gravity is so important that its magnitude is given its own symbol, g. It is constant at any given location on Earth and has the average value g = 9.80 m/s2.

Although g varies from 9.78 m/s2 to 9.83 m/s2, depending on latitude, altitude, underlying geological formations, and local topography, the average value of 9.80 m/s2 will be used in this text unless otherwise specified. The direction of the acceleration due to gravity is downward (towards the center of Earth). In fact, its direction defines what we call vertical. Note that whether the acceleration a in the kinematic equations has the value +g or −g depends on how we define our coordinate system. If we define the upward direction as positive, then a = −g = −9.80 m/s2, and if we define the downward direction as positive, then a = g = 9.80 m/s2.

One-Dimensional Motion Involving Gravity

The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So we start by considering straight up and down motion with no air resistance or friction. These assumptions mean that the velocity (if there is any) is vertical. If the object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g. We will also represent vertical displacement with the symbol y and use x for horizontal displacement.

A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s. The rock misses the edge of the cliff as it falls back to earth. Calculate the position and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance. Draw a sketch.

We are asked to determine the position y at various times. It is reasonable to take the initial position y0 to be zero. This problem involves one-dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity is downward, so a is negative. It is crucial that the initial velocity and the acceleration due to gravity have opposite signs. Opposite signs indicate that the acceleration due to gravity opposes the initial motion and will slow and eventually reverse it.

Since we are asked for values of position and velocity at three times, we will refer to these as y1 and v1; y2 and v2; and y3 and v3.

1. Identify the knowns. We know that y0 = 0; v0 = 13.0 m/s; a = −g = −9.80 m/s2; and t = 1.00 s.

2. Identify the best equation to use. We will use

$y=y0+v0t+12at2y={y}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}\\$

because it includes only one unknown, y (or y1, here), which is the value we want to find.

3. Plug in the known values and solve for y1.

$s)2=8.10my{}_{1}\text{}=0+\left(\text{13}\text{.}\text{0 m/s}\right)\left(1\text{.}\text{00 s}\right)+\frac{1}{2}\left(-9\text{.}\text{80}{\text{m/s}}^{2}\right){\left(1\text{.}\text{00 s}\right)}^{2}=8\text{.}\text{10}\text{m}\\$

Discussion

The rock is 8.10 m above its starting point at t = 1.00 s, since y1 > y0. It could be moving up or down; the only way to tell is to calculate v1 and find out if it is positive or negative. 1. Identify the knowns. We know that y0 = 0; v0 = 13.0 m/s; a = −g = −9.80 m/s2; and t = 1.00 s. We also know from the solution above that y1 = 8.10 m.

2. Identify the best equation to use. The most straightforward is

$v=v0−gtv={v}_{0}-\text{gt}\\$

(from

$v={v}_{0}+{at}\\$

where a = gravitational acceleration = −g). 3. Plug in the knowns and solve.

$m/s{v}_{1}={v}_{0}-\text{gt}=\text{13}\text{.}\text{0 m/s}-\left(9\text{.}{\text{80 m/s}}^{2}\right)\left(1\text{.}\text{00 s}\right)=3\text{.}\text{20 m/s}\\$

Discussion

The positive value for v1 means that the rock is still heading upward at t = 1.00 s. However, it has slowed from its original 13.0 m/s, as expected. The procedures for calculating the position and velocity at t = 2.00 s and 3.00 s are the same as those above. The results are summarized in Table 1 and illustrated in Figure 3.

Table 1. Results

Time, t	Position, y	Velocity, v	Acceleration, a
1.00 s	8.10 m	3.20 m/s	− 9.80 m/s 2
2.00 s	6.40 m	− 6.60 m/s	− 9.80 m/s 2
3.00 s	− 5.10 m	− 16.4 m/s	− 9.80 m/s 2

Graphing the data helps us understand it more clearly.

A simple experiment can be done to determine your reaction time. Have a friend hold a ruler between your thumb and index finger, separated by about 1 cm. Note the mark on the ruler that is right between your fingers. Have your friend drop the ruler unexpectedly, and try to catch it between your two fingers. Note the new reading on the ruler. Assuming acceleration is that due to gravity, calculate your reaction time. How far would you travel in a car (moving at 30 m/s) if the time it took your foot to go from the gas pedal to the brake was twice this reaction time?

What happens if the person on the cliff throws the rock straight down, instead of straight up? To explore this question, calculate the velocity of the rock when it is 5.10 m below the starting point, and has been thrown downward with an initial speed of 13.0 m/s. Draw a sketch.

Since up is positive, the final position of the rock will be negative because it finishes below the starting point at y0 = 0. Similarly, the initial velocity is downward and therefore negative, as is the acceleration due to gravity. We expect the final velocity to be negative since the rock will continue to move downward. 1. Identify the knowns. y0 = 0; y1 = −5.10 m; v0 = −13.0 m/s; a = −g = −9.80 m/s2.

2. Choose the kinematic equation that makes it easiest to solve the problem. The equation

$v2=v02+2a(y−y0){v}^{2}={v}_{0}^{2}+2a\left(y-{y}_{0}\right)\\$

works well because the only unknown in it is v. (We will plug y1 in for y.)

3. Enter the known values v2 = (−13.0 m/s)2+2(−9.80 m/s2)(−5.10 m−0 m) = 268.96 m2/s2, where we have retained extra significant figures because this is an intermediate result.

Taking the square root, and noting that a square root can be positive or negative, gives v = ±16.4 m/s.

The negative root is chosen to indicate that the rock is still heading down. Thus, v = −16.4 m/s.

Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same initial speed. (See Example 1 and Figure 5(a).) This is not a coincidental result. Because we only consider the acceleration due to gravity in this problem, the speed of a falling object depends only on its initial speed and its vertical position relative to the starting point. For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from Example 1) when the initial velocity is 13.0 m/s straight up, a result of ±3.20 m/s is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction.

Another way to look at it is this: In Example 1, the rock is thrown up with an initial velocity of 13.0 m/s. It rises and then falls back down. When its position is y=0 on its way back down, its velocity is −13.0 m/s. That is, it has the same speed on its way down as on its way up. We would then expect its velocity at a position of y=−5.10 m to be the same whether we have thrown it upwards at +13.0 m/s or thrown it downwards at −13.0 m/s. The velocity of the rock on its way down from y=0 is the same whether we have thrown it up or down to start with, as long as the speed with which it was initially thrown is the same.

The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physics laboratory course. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, Figure 6. Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.

Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location? Draw a sketch.

We need to solve for acceleration a. Note that in this case, displacement is downward and therefore negative, as is acceleration. 1. Identify the knowns. y0 = 0; y = –1.0000 m; t = 0.45173; v0 = 0.

2. Choose the equation that allows you to solve for a using the known values.

$y=y0+v0t+12at2y={y}_{0}+{v}_{0}t+\frac{1}{2}{{at}}^{2}\\$

3. Substitute 0 for v0 and rearrange the equation to solve for a. Substituting 0 for v0 yields

$y=y0+12at2y={y}_{0}+\frac{1}{2}{{at}}^{2}\\$

Solving for a gives

$a=2(y−y0)t2a=\frac{2\left(y-{y}_{0}\right)}{{t}^{2}}\\$

4. Substitute known values yields

$m/s2a=\frac{2(-1.0000\text{ m} - 0)}{(0.45173 \text{ s})^{2}}=-9.8010 \text{ m/s}^{2}\\$

so, because a = −g with the directions we have chosen,

g = 9.8010 m/s2.

Discussion The negative value for a indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of 9.80 m/s2, so 9.8010 m/s2 makes sense. Since the data going into the calculation are relatively precise, this value for g is more precise than the average value of 9.80 m/s2; it represents the local value for the acceleration due to gravity.

A chunk of ice breaks off a glacier and falls 30.0 meters before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water?

We know that initial position y0=0, final position y = −30.0 m, and a = −g = −9.80 m/s2. We can then use the equation

$y=y0+v0t+12at2y={y}_{0}+{v}_{0}t+\frac{1}{2}{{at}}^{2}\\$

to solve for t. Inserting a=−g, we obtain

$s\begin{array}{lll}y& =& 0+0-\frac{1}{2}{\text{gt}}^{2}\\ {t}^{2}& =& \frac{2y}{-g}\\ t& =& \pm \sqrt{\frac{2y}{-g}}=\pm \sqrt{\frac{2\left(-\text{30.0 m}\right)}{-9.80 m{\text{/s}}^{2}}}=\pm \sqrt{\text{6.12}{s}^{2}}=\text{2.47 s}\approx \text{2.5 s}\end{array}\\$

where we take the positive value as the physically relevant answer. Thus, it takes about 2.5 seconds for the piece of ice to hit the water.

Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. y = bx) to see how they add to generate the polynomial curve.

Click to download the simulation. Run using Java.

Section Summary

An object in free-fall experiences constant acceleration if air resistance is negligible.
On Earth, all free-falling objects have an acceleration due to gravity g, which averages
g = 9.8 m/s2.
Whether the acceleration a should be taken as +g or -g is determined by your choice of coordinate system. If you choose the upward direction as positive, a = -g = -9.8 m/s2 is negative. In the opposite case, a = g = 9.8 m/s2 is positive. Since acceleration is constant, the kinematic equations above can be applied with the appropriate +g or
-g substituted for a.
For objects in free-fall, up is normally taken as positive for displacement, velocity, and acceleration.

1. What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down? 2. An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration due to gravity have the same sign on the way up as on the way down? 3. Suppose you throw a rock nearly straight up at a coconut in a palm tree, and the rock misses on the way up but hits the coconut on the way down. Neglecting air resistance, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down? Explain. 4. If an object is thrown straight up and air resistance is negligible, then its speed when it returns to the starting point is the same as when it was released. If air resistance were not negligible, how would its speed upon return compare with its initial speed? How would the maximum height to which it rises be affected? 5. The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration due to gravity being the same, how many times higher could a safe fall on the Moon be than on Earth (gravitational acceleration on the Moon is about 1/6 that of the Earth)?

6. How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (gravitational acceleration on the Moon is about 1/6 of g on Earth)?

Assume air resistance is negligible unless otherwise stated.

1. Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be yo = 0.

2. Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

3. A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

4. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.

5. A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.

6. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (a) How long are her feet in the air? (b) What is her highest point above the board? (c) What is her velocity when her feet hit the water?

7. (a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long would it take to reach the ground if it is thrown straight down with the same speed?

8. A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?

9. You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?

10. A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

11. Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

12. An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

13. There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day.

14. A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 1.30 s to go past the window. What was the ball’s initial velocity?

15. Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return. (a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s. (b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.

16. A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms

$(8.00times10−5s)\left(8\text{.}\text{00}times {\text{10}}^{-5}\text{s}\right)\\$

. (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

17. A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

18. A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50 m × 10-3). (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

free-fall:the state of movement that results from gravitational force onlyacceleration due to gravity:acceleration of an object as a result of gravity

1. (a) y1 = 6.28 m; v1 = 10.1 m/s (b) y2 = 10.1 m; v2 = 5.20 m/s (c) y3 = 11.5 m; v3 = 0.300 m/s (d) y4 = 10.4 m; v4 = −4.60 m/s

3. v0 = 4.95 m/s

5. a) a = −9.80 m/s2; v0 = 13.0 m/s; y0 = 0 m (b) v = 0 m/s. Unknown is distance y to top of trajectory, where velocity is zero. Use equation

$v2=v02+2a(y−y0){v}^{2}={v}_{0}^{2}+2a\left(y-{y}_{0}\right)\\$

because it contains all known values except for y, so we can solve for y. Solving for y gives

Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result.

(a) 8.26 m

(b) 0.717

9. 1.91 s 11. (a) 94.0 m (b) 3.13 s 13. (a) -70.0 m/s (downward)(b) 6.10 s 15. (a) 19.6 m (b) 18.5 m 17. (a) 305 m (b) 262 m, -29.2 m/s (c) 8.91 s

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Page 2

By the end of this section, you will be able to:

Describe a straight-line graph in terms of its slope and y-intercept.
Determine average velocity or instantaneous velocity from a graph of position vs. time.
Determine average or instantaneous acceleration from a graph of velocity vs. time.
Derive a graph of velocity vs. time from a graph of position vs. time.
Derive a graph of acceleration vs. time from a graph of velocity vs. time.

A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships between physical quantities. This section uses graphs of displacement, velocity, and acceleration versus time to illustrate one-dimensional kinematics.

Slopes and General Relationships

First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one another in such a graph, the horizontal axis is usually considered to be an independent variable and the vertical axis a dependent variable. If we call the horizontal axis the x-axis and the vertical axis the y-axis, as in Figure 1, a straight-line graph has the general form

Here m is the slope, defined to be the rise divided by the run (as seen in the figure) of the straight line. The letter b is used for the y-intercept, which is the point at which the line crosses the vertical axis.

Graph of Displacement vs. Time (a = 0, so v is constant)

Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement versus time would, thus, have x on the vertical axis and t on the horizontal axis. Figure 2 is just such a straight-line graph. It shows a graph of displacement versus time for a jet-powered car on a very flat dry lake bed in Nevada.

Using the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity

$vˉ\bar{v}\\$

and the intercept is displacement at time zero—that is, x0. Substituting these symbols into

$y=mx+by=\text{mx}+b\\$

gives

Thus a graph of displacement versus time gives a general relationship among displacement, velocity, and time, as well as giving detailed numerical information about a specific situation. From the figure we can see that the car has a displacement of 400 m at time 0.650 m at t = 1.0 s, and so on. Its displacement at times other than those listed in the table can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph.

Find the average velocity of the car whose position is graphed in Figure 2. The slope of a graph of x vs. t is average velocity, since slope equals rise over run. In this case, rise = change in displacement and run = change in time, so that Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is larger.) 1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points.)

2. Substitute the x and t values of the chosen points into the equation. Remember in calculating change (Δ) we always use final value minus initial value.

yielding This is an impressively large land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of 60 mi/h (27 m/s or 96 km/h), but considerably shy of the record of 343 m/s (1234 km/h or 766 mi/h) set in 1997.

Graphs of Motion when a is constant but a≠0

The graphs in Figure 3 below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the displacement and velocity are initially 200 m and 15 m/s, respectively.

The graph of displacement versus time in Figure 3(a) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 3(a). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 3(b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 3(c).

Calculate the velocity of the jet car at a time of 25 s by finding the slope of the x vs. t graph in the graph below.

The slope of an x vs. t graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent at that point.

The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 5, where Q is the point at t=25 s. 1. Find the tangent line to the curve at t = 25 s. 2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.

3. Plug these endpoints into the equation to solve for the slope, v.

Thus,

$m/s.{v}_{Q}=\frac{\text{1820 m}}{\text{13 s}}=\text{140 m/s.}\\$

Discussion

This is the value given in this figure’s table for v at t = 25 s. The value of 140 m/s for vQ is plotted in Figure 5. The entire graph of v vs. t can be obtained in this fashion.

Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a v vs. t graph, rise = change in velocity Δv and run = change in time Δt.

The slope of a graph of velocity v vs. time t is acceleration a.

Since the velocity versus time graph in Figure 3(b) is a straight line, its slope is the same everywhere, implying that acceleration is constant. Acceleration versus time is graphed in Figure 3(c).

Additional general information can be obtained from Figure 5 and the expression for a straight line,

$y=mx+by=\text{mx}+b\\$

In this case, the vertical axis y is V, the intercept b is v0, the slope m is a, and the horizontal axis x is t. Substituting these symbols yields

$v=v0+atv={v}_{0}+\text{at}\\$

A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension.

It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to discover physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in any way. Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships.

Graphs of Motion Where Acceleration is Not Constant

Now consider the motion of the jet car as it goes from 165 m/s to its top velocity of 250 m/s, graphed in Figure 6. Time again starts at zero, and the initial displacement and velocity are 2900 m and 165 m/s, respectively. (These were the final displacement and velocity of the car in the motion graphed in Figure 3.) Acceleration gradually decreases from 5.0 m/s2 to zero when the car hits 250 m/s. The slope of the x vs. t graph increases until t=55 s, after which time the slope is constant. Similarly, velocity increases until 55 s and then becomes constant, since acceleration decreases to zero at 55 s and remains zero afterward.

Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the v vs. t graph in Figure 6(b). The slope of the curve at t = 25 s is equal to the slope of the line tangent at that point, as illustrated in Figure 6(b). Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope, a.

$s−1.0s)\text{slope}=\frac{\Delta v}{\Delta t}=\frac{\left(\text{260 m/s}-\text{210 m/s}\right)}{\left(\text{51 s}-1.0 s\right)}\\$

$s=1.0m/s2a=\frac{\text{50 m/s}}{\text{50 s}}=1\text{.}0 m{\text{/s}}^{2}\\$

Note that this value for a is consistent with the value plotted in Figure 6(c) at t = 25 s.

A graph of displacement versus time can be used to generate a graph of velocity versus time, and a graph of velocity versus time can be used to generate a graph of acceleration versus time. We do this by finding the slope of the graphs at every point. If the graph is linear (i.e., a line with a constant slope), it is easy to find the slope at any point and you have the slope for every point. Graphical analysis of motion can be used to describe both specific and general characteristics of kinematics. Graphs can also be used for other topics in physics. An important aspect of exploring physical relationships is to graph them and look for underlying relationships.

A graph of velocity vs. time of a ship coming into a harbor is shown below. (a) Describe the motion of the ship based on the graph. (b)What would a graph of the ship’s acceleration look like?

(a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving. (b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.

Graphs of motion can be used to analyze motion.
Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.
The slope of a graph of displacement x vs. time t is velocity v.
The slope of a graph of velocity v vs. t graph is acceleration a.
Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.

1. (a) Explain how you can use the graph of position versus time in Figure 9 to describe the change in velocity over time. Identify: (b) the time (ta, tb, tc, td, or te) at which the instantaneous velocity is greatest, (c) the time at which it is zero, and (d) the time at which it is negative.

Figure 9.

2. (a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure 10. (b) Identify the time or times (ta, tb, tc, etc.) at which the instantaneous velocity is greatest. (c) At which times is it zero? (d) At which times is it negative?

3. (a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in Figure 11. (b) Based on the graph, how does acceleration change over time?

4. (a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure 12. (b) Identify the time or times (ta, tb, tc, etc.) at which the acceleration is greatest. (c) At which times is it zero? (d) At which times is it negative?

Figure 12.

5. Consider the velocity vs. time graph of a person in an elevator shown in Figure 13 Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of: (a) position vs. time and (b) acceleration vs. time for this trip.

6. A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.

Note: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you.

1. a) By taking the slope of the curve in Figure 14, verify that the velocity of the jet car is 115 m/s at t=20 s. (b) By taking the slope of the curve at any point in Figure 15 verify that the jet car’s acceleration is 5.0 m/s2.

2. Using approximate values, calculate the slope of the curve in Figure 16 to verify that the velocity at t=10.0 s is 0.208 m/s. Assume all values are known to 3 significant figures.

3. Using approximate values, calculate the slope of the curve in Figure 16 to verify that the velocity at t = 30.0 s is 0.238 m/s. Assume all values are known to 3 significant figures.

4. By taking the slope of the curve in Figure 17, verify that the acceleration is 3.2 m/s2 at t = 10 s.

Figure 17.

5. Construct the displacement graph for the subway shuttle train as shown in Figure 2 from Acceleration (shown again below). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

A subway train in Sao Paulo, Brazil, decelerates as it comes into a station. It is accelerating in a direction opposite to its direction of motion. (credit: Yusuke Kawasaki, Flickr)

6. (a) Take the slope of the curve in Figure 11 to find the jogger’s velocity at t = 2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 18.

7. A graph of v(t) is shown for a world-class track sprinter in a 100-m race. (See Figure 21). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at t = 5 s? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

8. Figure 22 shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs.

independent variable: the variable that the dependent variable is measured with respect to; usually plotted along the x-axis dependent variable: the variable that is being measured; usually plotted along the y-axis slope: the difference in y-value (the rise) divided by the difference in x-value (the run) of two points on a straight line y-intercept: the y-value when x = 0, or when the graph crosses the y-axis

1. (a) 115 m/s (b) 5.0 m/s

$m/sv=\frac{\left(\text{11.7}-6.95\right)\times {\text{10}}^{3}\text{m}}{\left(40\text{.}\text{0 - 20}.0\right)\text{s}}=\text{238 m/s}\\$