4.6

Velocity and Acceleration To understand the world around us, we must understand as much about motion as possible. The study of motion is fundamental to the principles of physics, and it is applied to a wide range of topics, such as automotive engineering (crash tests, racing, performance tests), satellite and rocket launches, and improvement of athletic performance (kinesiology). We have already defined and computed velocities in Section 3.4. Now we can compute them more easily with the aid of the differentiation formulas developed in this chapter. Suppose an object moves along a straight line. (Think of a ball being thrown vertically upward or a car being driven along a road or a stone being dropped from a cliff.) Suppose the position function of the object is s(t). Another way to say this is that s(t) is the displacement (directed distance) of the object from the origin at time t. Recall that the instantaneous velocity of the object at time t is defined as the limit of the average velocities over shorter and shorter time intervals. s(t + h) − s(t ) v = lim h→0 h The velocity function is the derivative of the position function, v = s¢(t ) =

ds . dt

Example 1 TV Advertisement for a New Football Helmet To dramatically advertise the effectiveness of a new football helmet, a melon is taped inside it and the helmet is dropped from a tower 78.4 m high. The height of the helmet, h, in metres, after t seconds is given by h = 78.4 - 4.9t2, until it hits the ground. a) Find the velocity of the helmet after 1 s and 2 s. b) When will the helmet hit the ground? c) The manufacturer’s demonstration will be effective if the melon is undamaged after impact. The melon will remain intact if the impact speed is less than 30 m/s. Is the demonstration effective?

Solution 2 a) The position function is h = 78.4 - 4.9t , so the velocity at time t is dh = −9.8t dt The velocity after 1 s is given by dh ù = -9.8(1) dt ûú t = 1 = -9 .8 The velocity after 1 s is -9.8 m/s. The velocity after 2 s is given by dh ù = -9.8(2) dt ûú t = 2 = -19.6 m/s The velocity after 2 s is -19.6 m/s. 4.6 Velocity and Acceleration MHR

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The negative signs for the velocities indicate that the helmet is moving in the negative direction (down). b) The helmet will hit the ground when the height is 0. 2

h = 78.4 - 4.9t =0 t2 = 78.4/4.9 = 16 Since t > 0, t = 4. So, the helmet hits the ground after 4 s. c) Substitute t = 4 into the velocity function.

v(4) = h¢(4) = -9.8(4) = -39.2 Since the speed (that is, the absolute value of the velocity) of 39.2 m/s is greater than 30 m/s, the melon is damaged on impact. Thus, the demonstration fails.

Example 2 Analysis of a Cheetah’s Motion The position function of a cheetah moving across level ground in a straight line chasing after prey is given by the equation s(t) = t3 - 15t2 + 63t where t is measured in seconds and s in metres. a) What is the cheetah’s velocity after 1 s? 4 s? 8 s? b) When is the cheetah momentarily stopped? c) What are the positions of the cheetah in part b)? d) When is the cheetah moving in the positive direction? When is it moving in the negative direction? e) Find the position of the cheetah after 10 s. f) Find the total distance travelled by the cheetah during the first 10 s. g) Compare the results of parts e) and f).

Solution a) The velocity after t seconds is v = s¢(t) = 3t2 - 30t + 63 After 1 s, s¢(1) = 3(1)2 - 30(1) + 63 = 36 The velocity after 1 s is 36 m/s. After 4 s, s¢(4) = 3(4)2 - 30(4) + 63 = -9 The velocity after 4 s is -9 m/s.

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After 8 s, s¢(8) = 3(8)2 - 30(8) + 63 = 15 The velocity after 8 s is 15 m/s. b) The cheetah stops momentarily when v(t) = 0.

3t 2 - 30t + 63 = 0 t 2 - 10t + 21 = 0 (t - 3)(t - 7) = 0 t = 3 or t = 7 Thus, the cheetah is momentarily stopped after 3 s and 7 s. c) At t = 3,

s(3) = (3)3 - 15(3)2 + 63(3) = 81 At t = 7, s(7) = (7)3 - 15(7)2 + 63(7) = 49 The cheetah momentarily stops when it is 81 m and 49 m from its starting position. d) The cheetah moves in the positive direction when v(t) > 0, that is,

t 2 - 10t + 21 = (t - 3)(t - 7) > 0

This inequality is satisfied when both factors are positive, that is, t Î (7, ¥), or when both factors are negative, that is, t Î [0, 3). Thus, the cheetah moves in the positive direction in the time intervals t Î [0, 3) and t Î (7, ¥). It moves in the negative direction when t Î (3, 7). We can demonstrate this using a number line with the factors listed underneath. The +, 0, and - symbols show where the quantities are positive, zero, and negative. 3

7

(t – 3)

–

0

+

(t – 7)

–

–

–

0

+

(t – 3)(t – 7)

+

0

–

0

+

+

The graph of the displacement, s(t) = t3 - 15t2 + 63t, is shown below.

Window variables: x Î [0, 10], y Î [0, 10] e) At t = 10 s,

s(10) = (10)3 - 15(10)2 + 63(10) = 130 The cheetah is 130 m from where it started after 10 s. 4.6 Velocity and Acceleration MHR

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The path of the cheetah is modelled in the figure. (Remember that the cheetah goes back and forth along a straight line) f)

t = 0, s = 0 t = 3 s, s = 81 m

t = 7 s, s = 49 m t = 10 s, s = 130 m

To find the total distance travelled, first determine the change in position over each interval for which the cheetah does not change direction. The total distance travelled is the sum of these quantities. From part d), the cheetah changes direction at t = 3 and t = 7. The distance travelled in the time interval [0, 3] is |s(3) - s(0)| = |81 - 0| = 81 The distance travelled in the time interval [3, 7] is |s(7) - s(3)| = | 49 - 81| = 32 The distance travelled in the time interval [7, 10] is | s(10) - s(7)| = |130 - 49| = 81 The total distance travelled by the cheetah is (81 + 32 + 81) m or 194 m. g) The difference between the total distance travelled by the cheetah and its final

position is (194 - 130) m or 64 m. The cheetah’s displacement from its starting point is 64 m less than its total distance run.

Acceleration The acceleration of an object is the rate of change of its velocity with respect to time. Therefore, the acceleration function a(t) at time t is the derivative of the velocity function: a(t ) = v¢(t ) =

dv dt

Since the velocity is the derivative of the position function s(t), it follows that: The acceleration is the second derivative of the position function. Thus, a(t) = v¢(t) = s²(t) or dv d 2 s = a= dt dt 2 If s is measured in metres and t in seconds, the units for acceleration are metres per second squared, or m/s2.

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MHR Chapter 4

Investigate & Inquire: Motion With Variable Acceleration 1. Use a motion detector (CBR) connected to a graphing calculator to record the motion of

a lab cart acted on by a non-constant force, such as a stretched spring or a long chain hanging over the edge of a table. The mass of the chain hanging over the

cart

edge causes both cart and chain to accelerate. The mass hanging over the edge is not constant.

2. Compare the slopes of the tangents to the velocity-time (v-t) graph to the acceleration

from the acceleration-time (a-t) graph at the same time. Record your observations in a table similar to the one shown below. Time (s)

Slope of v-t graph

Acceleration from a-t graph

3. How is the slope of the v-t graph related to the acceleration? 4. Given the velocity of an object as a function of time, how would you find the

acceleration? 5. What is the geometrical interpretation of acceleration?

Example 3 Motion With Non-Constant Acceleration The position function of a person on a bicycle pedalling down a steep hill with steadily 1 1 increasing effort in pedalling is s(t ) = t 3 + t 2 + t , where s is measured in metres and t in 6 2 seconds, t Î [0, 4]. a) Find the velocity and acceleration as functions of time. b) Find the acceleration at 2 s.

Solution a) The velocity is ds v= dt 1 1 = (3t 2 ) + (2t ) + 1 6 2 1 2 = t +t +1 2 4.6 Velocity and Acceleration MHR

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and the acceleration is dv a= dt = t +1 b) We substitute t = 2 in the acceleration equation.

a=2+1 =3 After 2 s, the acceleration is 3 m/s2.

Example 4 Analysis of a Position-Time Graph The graph shows the position function of a bicycle. Determine the sign of the velocity and acceleration in each interval, and relate them to the slope of the graph and whether the slope is increasing or decreasing (that is, how the graph bends). s D C

A

E F

t

B

Solution Interval A to B B to C C to D D D to E E to F F to G

G

Description of graph Slope = 0, horizonal segment Positive slope, slope increasing Positive slope, slope decreasing Slope = 0, momentarily horizontal Negative slope, slope decreasing Negative slope, slope increasing Slope = 0, horizontal segment

Velocity 0 positive positive 0 negative negative 0

Acceleration 0 positive negative negative negative positive 0

Note that the table in Example 4 reflects our understanding that the velocity of the bicycle is the slope of the position-time graph. Furthermore, the acceleration is the rate of change of velocity, which corresponds to the rate of change of the slope of the position-time graph.

Example 5 Motion Along a Vertical Line In a circus stunt, a person is projected straight up into the air, and then falls straight back down into a safety net. The position function is s(t) = 30t - 5t2 for the upward part of the motion and s(t) = -4.8(t - 3)2 + 45 for the downward part of the motion, where s is in metres and t is in seconds. a) Determine the acceleration for the upward part of the motion. b) Determine the acceleration for the downward part of the motion. c) Compare the results of parts a) and b) and explain the difference.

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MHR Chapter 4

Solution 2 a) s(t) = 30t - 5t ds = 30 − 10t dt d2s a= 2 dt = −10 The acceleration for the upward part of the motion is 10 m/s2, directed downward. 2 b) s(t) = -4.8(t - 3) + 45 ds = −9.6(t − 3) dt d2s a= 2 dt = −9 . 6 The acceleration for the downward part of the motion is 9.6 m/s2, directed downward. 2 c) The acceleration due to gravity near Earth’s surface is about -9.8 m/s . When the person is moving upward, gravity acts to slow the person. Since air resistance opposes the motion, it also acts to slow the person down. The effect of gravity and air resistance together produce the acceleration of 10 m/s2. When the person is moving downward, gravity acts to make the person go faster. In this case, air resistance still opposes the motion, which means it acts to slow the person down. Thus, air resistance opposes gravity, so that the absolute value of the total acceleration is a little less than what gravity by itself would produce¾only 9.6 m/s2 instead of 9.8 m/s2. Negative acceleration, dv a= 0 a= dt means that velocity is increasing (as at B).

v

A

B t

0

This can be a little confusing, since there is a distinction between velocity and speed. Velocity has both magnitude and direction, whereas speed is simply the magnitude of the velocity, with no direction indicated. If the velocity is negative and the acceleration is negative, then the velocity is decreasing but the speed is increasing (because speed is the absolute value of the velocity). A summary of the effect of acceleration on speed (how fast the object is moving without regard for the direction) is as follows: If a(t)v(t) > 0, then the object is speeding up at time t. If a(t)v(t) < 0, then the object is slowing down at time t. 4.6 Velocity and Acceleration MHR

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In words, we can say that, if the acceleration and the velocity have the same sign at a particular time (they are in the same direction), then the object is speeding up at that time. This occurs because the object is being pushed in the direction of motion. If the acceleration and the velocity have opposite signs at a particular time (they are in opposite directions), then the object is slowing down at that time, since the object is being pushed in a direction opposite to its motion.

y 4

a

v s

5

6

2

0

1

2

3

4

t

–2 –4

A

C

B

D

–6

We can use similar reasoning to determine if an A: slows down C: slows down object is moving away from the origin or toward B: speeds up D: speeds up the origin. The origin is defined as the position equal to 0. If an object has a positive position and positive velocity, it is moving away from the origin. Similarly, if an object has a negative position and a negative velocity, it is moving away from the origin. In both cases, the product of the position and the velocity is positive. v

Object v

Object

0 s < 0, v < 0, sv > 0 Object is moving away from origin.

0 s > 0, v > 0, sv > 0 Object is moving away from origin.

If an object has a positive position and negative velocity, it is moving toward the origin. Similarly, if an object has a negative position and a positive velocity, it is moving toward the origin. In both cases, the product of the position and the velocity is negative, indicating that the object is moving toward the origin. Object

v

v

0 s < 0, v > 0, sv < 0 Object is moving toward the origin.

Object

0

s > 0, v < 0, sv < 0 Object is moving toward the origin.

To summarize: If s(t)v(t) > 0, then the object is moving away from the origin at time t. If s(t)v(t) < 0, then the object is moving toward the origin at time t.

Example 6 Speeding Up or Slowing Down? In a science-fiction movie, a computer-generated robot appears to be moving about the scene. The actors and the robot are not on the set at the same time. The robot is added to the shot later. For the movie to look realistic, the actors must be given exact instructions about the robot’s motion so they can respond at exactly the right time. The position of the robot is defined by the function s(t) = t3 - 12t2 + 36t, where t is measured in seconds, t Î [0, 10], and s in metres. a) Find the velocity and the acceleration at time t. b) Find the velocity and the acceleration after 3 s and 5 s. Is the robot speeding up or slowing down at these times? c) Graph the position, velocity, and acceleration functions for the first 10 s. d) When is the robot speeding up? When is it slowing down?

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MHR Chapter 4

Solution 3

2

a) s(t) = t - 12t + 36t

ds dt = 3t 2 − 24t + 36 dv a(t ) = dt = 6t − 24

v(t ) =

b) At t = 3,

v = 3(3)2 - 24(3) + 36 = -9 and a = 6(3) - 24 = -6 After 3 s, the velocity is -9 m/s and the acceleration is -6 m/s2. Since a(3)v(3) = 54 > 0, the robot is speeding up at this time. At t = 5, v = 3(5)2 - 24(5) + 36 = -9 and a = 6(5) - 24 =6 After 5 s, the velocity is -9 m/s and the acceleration is 6 m/s2. Since a(5)v(5) = -54 < 0, the robot is slowing down at this time. c) The three graphs are shown below on the same set of axes.

Window variables: x Î [0, 10], y Î [-25, 160] d) Factoring the velocity function gives

v = 3t2 - 24t + 36 = 3(t - 2)(t - 6) and so the velocity is positive for t Î [0, 2) and t Î (6, 10] and negative for t Î (2, 6) (see the graph). For the acceleration function, we have a = 6t - 24 = 6(t - 4) and so the acceleration is negative for t Î [0, 4) and positive for t Î (4, 10] (see the graph). 4.6 Velocity and Acceleration MHR

235

We can summarize this information in the following number line diagrams. Velocity

6

2

(t – 2)

–

(t – 6)

–

–

0

+

(t – 2)(t – 6)

+

–

0

+

0

Acceleration 0

8

4 0

–

t–4

+

+

0

+

The table and the graph below show when the robot is slowing down and when it is speeding up. Slowing Down v > 0, a < 0 v < 0, a > 0 t Î [0, 2) t Î (4, 6)

Speeding Up v < 0, a < 0 v > 0, a > 0 t Î (2, 4) t Î (6, 10]

40 s 20

a

0

1

2

3

4

5v

6

7 t

–20 –40

slows down

speeds up

slows down speeds up

Key Concepts ·

If a particle has position function s(t), then its velocity function is ds v(t ) = = s¢(t ) dt and its acceleration function is dv d 2 s = = v¢(t ) = s² (t ) dt dt 2 a(t)v(t) > 0, then the object is speeding up. a(t)v(t) < 0, then the object is slowing down. s(t)v(t) > 0, then the object is moving away from the origin. s(t)v(t) < 0, then the object is moving toward the origin.

a(t ) = · · · ·

236

If If If If

MHR Chapter 4

Communicate Your Understanding 1. Explain how to find the velocity and acceleration of a particle given its position

function. Given the position function of a particle, explain how to determine when the velocity is zero the acceleration is zero a) Under what conditions on the acceleration is the velocity of a particle increasing? Under what conditions is the velocity of a particle decreasing? b) Under what conditions on the acceleration is a particle speeding up? Under what conditions is a particle slowing down? c) Why are the results of parts a) and b) different? 4. Discuss the validity of each statement. Provide examples to illustrate your answer. a) “If the acceleration is positive, the object is speeding up. If the acceleration is negative, the object is slowing down.” b) “If the velocity is positive, the object is moving away from the origin. If the velocity is negative, the object is moving toward the origin.” 2. a) b) 3.

Apply, Solve, Communicate B 1. Communication The graph shows the position

3. The graph of a position function is shown. s(t)

function of a car.

A

s(t)

E B

B A

C

D C

E

t

0

D F

t

a) For the part of the graph from 0 to A, use

at A, B, and C? d) What happens between C and D? e) What happens at F?

slopes of tangents to decide whether the velocity is increasing or decreasing. Is the acceleration positive or negative? b) State whether the acceleration is positive, zero, or negative i) from A to B ii) from B to C iii) from C to D iv) from D to E

2. The graph of a velocity function is shown.

4. Communication On the graph are shown

0

a) What is the initial velocity of the car? b) Is the car going faster at A or at B? c) Is the car slowing down or speeding up

s(t) C

A

D

y

B 0

the position function, velocity function, and acceleration function of an object. Identify each curve and explain your reasoning. a

E t

State whether the acceleration is positive, zero, or negative a) from 0 to A b) from A to B c) from B to C d) from C to D e) from D to E

b

t

0

c 4.6 Velocity and Acceleration MHR

237

5. The figure shows the graphs of the position

a) Find the velocity of the ball after 1 s, 2 s, 4 s,

function, the velocity function, and the acceleration function of a car that is undergoing a test to determine its performance level. Determine when the car is speeding up and when it is slowing down.

and 5 s. b) When does the ball reach its maximum height? c) What is its maximum height? d) When does it hit the ground? e) With what velocity does it hit the ground?

y

10. The position of a motorcycle moving down

24

1 a straight highway is s = t2 + 15t, where t is 2 measured in seconds and s, in metres. a) When does the motorcycle reach i) 15 m/s? ii) 25 m/s? b) What is the acceleration of the motorcycle?

v s

18 a 12 6

0

11. The position of a person on a skateboard is 1

2

3

4 t

6. Each position function gives s, in metres, as

a function of t, in seconds. Find the velocity and acceleration as functions of time. a) s = 9 + 4t 2 b) s = 3t + 2t - 5 3 2 c) s = t - 4t + 5t - 7 3 t d) s = t +1 2 e) s = (t − 2 t )(2t − t + t ) 6t f) s = 1+ t 7. Each position function gives s, in metres, as

a function of t, in seconds. Find the velocity and acceleration for any t and for t = 4 s. 2 a) s = 6 + 3t b) s = 4t - 7t + 14 3 t 3 2 c) s = t - 6t + 12t d) s = 1+ t 2 + 3t 2 2 e) s = t (t − 2t ) f) s = t+4 8. If a stone is thrown downward with a speed of 10 m/s, from a cliff that is 90 m high, its height, h, in metres, after t seconds is h = 90 - 10t - 4.9t2. Find the velocity after 1 s and after 2 s. 9. Inquiry/Problem Solving If a ball is thrown

directly upward with an initial velocity of 29.4 m/s, then the height, h, in metres, after 2 t seconds, is given by h = 29.4t - 4.9t .

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MHR Chapter 4

given by s = t2 - 8t + 12, where s is measured in metres and t, in seconds. a) Find the velocity after 2 s and 6 s. b) When is the person at rest? c) When is the person moving in the positive direction? d) Draw a diagram to illustrate the motion of the person. 12. Inquiry/Problem Solving A new motion detector

is used to graph the position of a sports car. If the motion detector is not calibrated properly, it might give unrealistic graphs and data. The position of the car is described by the function s = 2t3 - 21t2 + 60t, t Î [0, 6], where s is measured in metres and t, in seconds. It is known that the largest possible acceleration of the car is 12 m/s2. Investigate if the detector is working properly. a) i) What is the initial velocity of the car? ii) When is the car at rest? iii) When is the car moving in the forward direction? b) Draw a diagram to illustrate the motion of the car. c) Find the total distance travelled by the car in the first 6 s. d) Find the maximum acceleration of the car. e) List all the evidence that you can to show the company that the equation of motion is not realistic for a sports car. 13. Application If a weighted and coiled rescue

rope is thrown upward with a velocity of 2 m/s

from the top of a 23 m cliff, then the height, in metres, of the rope above the base of the cliff, after t seconds, is s = 23 + 2t - 4.9t2. a) When does the rope reach its maximum height? b) Use the quadratic formula to find how long it takes for the rope to reach the ground. c) Find the approximate velocity with which the rope strikes the ground. 14. A position function is given by

1 s = s0 + v0 t + gt2, where s0, v0, and g are 2 constants. Find a) the initial position b) the initial velocity c) the acceleration 15. Application A stunt car driver is practising for a movie. The position function of the car is 1 s = t3 - 25t, t ³ 0, where s is measured in 3 metres and t, in seconds. Find the acceleration at the instant the velocity is zero. 16. An unoccupied test rocket meant for orbit is

fired straight up from a launch pad. Something goes wrong and the rocket comes back down. The height of the rocket above the ground is 1 modelled by h = 5t2 - t3, t ³ 0, where h is 6 measured in metres and t, in seconds. a) When is the acceleration positive and when is it negative? b) When is the velocity zero? c) Find the maximum height of the rocket. d) At what velocity does the rocket hit the ground?

e) When is the rocket speeding up? When is the

rocket slowing down? 17. The position of a cougar chasing its

prey is given by the function s = t3 - 6t2 + 9t, where t is measured in seconds and s, in metres. a) Find the velocity and acceleration at time t. b) When is the cougar moving toward the origin? When is it moving away? c) When is the cougar speeding up? When is it slowing down? d) Graph the position, velocity, and acceleration for the first 4 s.

C 18. The shaft of a well is 15 m deep. A stone is

dropped into the well and a splash occurs after 1.6 s. How far down the well is the surface of the water? 19. Application A hawk is flying at 20 km/h

horizontally and drops its prey from a height of 30 m. a) State equations representing the horizontal displacement, velocity, and acceleration of the falling prey. b) If the prey falls at an acceleration due to gravity of -9.8 m/s2, state equations representing the vertical displacement, velocity, and acceleration. c) When is the vertical speed greater than the horizontal speed? d) Develop an equation representing the overall velocity of the prey. e) What is the equation representing the overall acceleration of the prey? f ) What is the acceleration of the prey after 2 s?

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