Relationship Between Velocity and Acceleration

Velocity and acceleration are in the same direction

Acceleration is uniform (blue arrows maintain the same length)

Velocity is increasing (red arrows are getting longer)

Relationship Between Acceleration and Velocity

Uniform velocity (shown by red arrows maintaining the same size)

Acceleration equals zero

Relationship Between Velocity and Acceleration

Acceleration and velocity are in opposite directions

Acceleration is uniform (blue arrows maintain the same length)

Velocity is decreasing (red arrows are getting shorter)

Kinematic Equations Used in situations with uniform

acceleration

2

2 2

2

12

2

oaverage

f o

o

o

v vx v t t

v v at

x v t at

v v a x

Notes on the equations

2o

averagev vx v t t

Gives displacement as a function of velocity and time

Notes on the equations

Shows velocity as a function of acceleration and time

ov v at

Graphical Interpretation of the Equation

Notes on the equations

Gives displacement as a function of time, velocity and acceleration

2o at

21tvx

Notes on the equations

Gives velocity as a function of acceleration and displacement

xa2vv 2o

2f

Problem-Solving Hints Be sure all the units are consistent

Convert if necessary Choose a coordinate system Sketch the situation, labeling initial and

final points, indicating a positive direction

Choose the appropriate kinematic equation

Check your results

EXAMPLE 2.4 The Daytona 500

Goal Apply the basic kinematic equations. Problem (a) 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 1.00 102 ft? (b) How much time has elapsed? Strategy (a) We've read the problem, drawn the diagram in the figure, and chosen a coordinate system (steps 1 and 2). We'd like to find the velocity v after a certain known displacement Δx. The acceleration a is also known, as is the initial velocity v0 (step 3, labeling, is complete), so the third equation in this table looks most useful for solving part (a). Given the velocity, the first equation in the table can then be used to find the time in part (b).

SOLUTION

(a) Convert units of Δx to SI, using the conversion table. Write the kinematics equation for v2 (step 4). v2 = v0

2 + 2aΔx Solve for v, taking the positive square root because the car moves to the right (step 5). v = √v0

2 + 2aΔx

Substitute v0 = 0, a = 5.00 m/s2, and Δx = 30.5 m. v = √v0

2 + 2aΔx = √ (0)2 + 2(5.00 m/s2)(30.5 m) = 17.5 m/s

(b) How much time has elapsed? Apply the equation to the right. v = at + v0 Substitute values and solve for time t. 17.5 m/s = (5.00 m/s2)t

t = 17.5 m/s

= 3.50 s 5.0 m/s2

LEARN MORE

Remarks The answers are easy to check. An alternate technique is to use Δx = v0t + 1/2at2 to find t and then use the equation v = v0 + at to find v. Question What is the final speed if the displacement is increased by a factor of 4? 34.95 m/s

EXAMPLE 2.5 Car Chase

A speeding car passes a hidden trooper. When does the trooper catch up to the car? Goal Solve a problem involving two objects, one moving at constant acceleration and the other at constant velocity. Problem A car traveling at a constant speed of 24.0 m/s passes a trooper hidden behind a billboard, as in the figure. One second after the speeding car passes the billboard, the trooper sets off in chase with a constant acceleration of 3.00 m/s2. (a) How long does it take the trooper to overtake the speeding car? (b) How fast is the trooper going at that time?

Strategy Solving this problem involves two simultaneous kinematics equations of position, one for the trooper and the other for the car. Choose t = 0 to correspond to the time the trooper takes up the chase, when the car is at xcar = 24.0 m because of its head start (24.0 m/s 1.00 s). The trooper catches up with the car when their positions are the same, which suggests setting xtrooper = xcar and solving for time, which can then be used to find the trooper's speed in part (b).

SOLUTION

(a) How long does it take the trooper to overtake the car? Write the equation for the car's displacement. Δxcar =xcar - x0 = v0t + 1/2acart2

Take x0 = 24.0 m, v0 = 24.0 m/s and acar = 0. Solve for xcar. xcar =x0 + vt = 24.0 m + (24.0 m/s) t

Write the equation for the trooper's position, taking x0 = 0, v0 = 0, and atrooper = 3.00 m/s2. xtrooper = 1/2atroopert2 =1/2(3.00 m/s2)t2 = (1.50 m/s2)t2

Set xtrooper = xcar, and solve the quadratic equation. Only the positive root is meaningful. (1.50 m/s2)t2 =24.0 m + (24.0 m/s) t

(1.50 m/s2)t2 - (24.0 m/s)t - 24.0 m = 0

t = 16.9 s

(b) Find the trooper's speed at this time. Substitute the time into the trooper's velocity equation. vtrooper = v0 + atroopert = 0 + (3.00 m/s2)(16.9 s) = 50.7 m/s

LEARN MORE

Remarks The trooper, traveling about twice as fast as the car, must swerve or apply his brakes strongly to avoid a collision! This problem can also be solved graphically by plotting position versus time for each vehicle on the same graph. The intersection of the two graphs corresponds to the time and position at which the trooper overtakes the car. Question The graphical solution corresponds to finding the intersection of what two types of curves in the xt-plane? (Select all that apply.)

An upward-shaped curve whose slope increases as the

displacement x increases. A straight line sloped downward.

A straight line sloped upward. A straight horizontal line. A downward-shaped curve whose slope decreased and then increases as x increases.

EXAMPLE 2.6 Runway Length

Coasting and braking distances for a landing jetliner. Goal Apply kinematics to horizontal motion with two phases. Problem A typical jetliner lands at a speed of 160 mi/h and decelerates at the rate of (10 mi/h)/s. If the plane travels at a constant speed of 160 mi/h for 1.0 s after landing before applying the brakes, what is the total displacement of the aircraft between touchdown on the runway and coming to rest? Strategy First, convert all quantities to SI units. The problem must be solved in two parts, or phases, corresponding to the initial coast after touchdown, followed by braking. Using the kinematic equations, find the displacement during each part and add the two displacements.

SOLUTION

Convert units of speed and acceleration to SI.

v0 = (160 mi/h) ( 0.447 m/s ) = 71.5 m/s 1.00 mi/ h

a = (-10 (mi/h)/ s) ( 0.447 m/s ) = -4.47 m/s2 1.00 mi/ h

Taking a = 0, v0 = 71.5 m/s, and t = 1.00 s, find the displacement while the plane is coasting. Δxcoasting = v0t + ½at2 = (71.5 m/s)(1.00 s) + 0 = 71.5

Use the time-independent kinematic equation to find the displacement while the plane is braking. v2 = v0

2 + 2aΔxbraking

Take a = -4.47 m/s2 and v0 = 71.5 m/s. The negative sign on a means that the plane is slowing down.

Δxbraking = v2 - v0

2 =

0 - (71.5 m/s)2 = 572 m

2a 2.00 (-4.47 m/s2) Sum the two results to find the total displacement. Δxcoasting + Δxbraking = 72 m +572 m = 644 m

LEARN MORE

Remarks To find the displacement while braking, we could have used the two kinematics equations involving time, namely, Δx = v0t + ½ at2 and v = v0 + at, but because we weren't interested in time, the time-independent equation was easier to use. Question By how much would the answer change if the plane coasted for 2.0 s before the pilot applied the brakes? (71.5m)

EXAMPLE 2.7 The Acela: The Porsche of American Trains

The Acela, 1,250,000 lb of cold steel thundering along at 170 mi/h. Goal Find accelerations and displacements from a velocity vs. time graph. Problem The sleek high-speed electric train known as the Acela (pronounced ahh-sell-ah) is currently in service on the Washington-New York-Boston run. The Acela consists of two power cars and six coaches and can carry 304 passengers at speeds up to 170 mi/h. In order to negotiate curves comfortably at high speeds, the train carriages tilt as much as 6° from the vertical, preventing passengers from being pushed to the side. A velocity vs. time graph for the Acela is shown in Figure a.

Figure a Velocity vs. time graph for the Acela

(a) Describe the motion of the Acela. (b) Find the peak acceleration of the Acela in miles per hour per second ((mi/h)/s) as the train speeds up from 45 mi/h to 170 mi/h. (c) Find the train's displacement in miles between t = 0 and t = 200 s. (d) Find the average acceleration of the Acela and its displacement in miles in the interval from 200 s to 300 s. (The train has regenerative braking, which means that it feeds energy back into the utility lines each time it stops!) (e) Find the total displacement in the interval from 0 to 400 s.

Strategy For part (a), remember that the slope of the tangent line at any point of the velocity vs. time graph gives the acceleration at that time. To find the peak acceleration in part (b), study the graph and locate the point at which the slope is steepest. In parts (c) through (e), estimating the area under the curve gives the displacement during a given period, with areas below the time axis, as in part (e), subtracted from the total. The average acceleration in part (d) can be obtained by substituting numbers taken from the graph into the definition of average acceleration, a =Δv/Δt.

SOLUTION

(a) Describe the motion. From about -50 s to 50 s, the Acela cruises at a constant velocity in the +x-direction. Then the train accelerates in the +x-direction from 50 s to 200 s, reaching a top speed of about 170 mi/h, whereupon it brakes to rest at 350 s and reverses, steadily gaining speed in the -x-direction.

Figure b The slope of the steepest blue tangent line gives the peak acceleration, while the slope of the green line is the average acceleration between 200 s and 300 s.

(b) Find the peak acceleration.

Calculate the slope of the steepest tangent line, which connects the points (50 s, 50 mi/h) and (100 s, 150 mi/h) (the light blue line in Figure b).

Figure c The area under the velocity vs. time graph in some interval gives the displacement of the Acela in that time interval.

a = slope = Δv

= (1.5 102 − 5.0 101) mi/h

Δt (1.0 102 − 5.00 101) s = 2.0 (mi/h)/s

(c) Find the displacement between 0 s and 200 s.

Using triangles and rectangles, approximate the area in figure c.

Δx0 → 200 s = area1 + area2 + area3 + area4 + area5

≈ (5.0 101 mi/h)(5.0 101 s)

+ (5.0 101 mi/h)(5.0 101 s)

+ (1.6 102 mi/h)(1.0 102 s)

+ ½(5.0 101 s)(1.0 102 mi/h)

+ ½(1.0 102 s)(1.7 102 mi/h − 1.6 102 mi/h)

= 2.4 104 (mi/h)s

Convert units to miles by converting hours to seconds.

Δx0 → 200 s ≈ 2.4 104 mi · s (

1h ) = 6.7 mi h 3600 s

(d) Find the average acceleration from 200 s to 300 s, and find the displacement.

The slope of the green line is the average acceleration from 200 s to 300 s (Fig. b).

a = slope = Δv

= (1.0 101 − 1.7 102) mi/h

Δt 1.0 102 = −1.6 (mi/h)/s The displacement from 200 s to 300 s is equal to area6, which is the area of a triangle plus the area of a very narrow rectangle beneath the triangle.

Δx200 → 300 s ≈ ½(1.0 102 s)(1.7 102 − 1.0 101) mi/h

+ (1.0 101 mi/h)(1.0 102 s) = 9.0 103 (mi/h)(s) = 2.5 mi

(e) Find the total displacement from 0 s to 400 s.

The total displacement is the sum of all the individual displacements. We still need to calculate the displacements for the time intervals from 300 s to 350 s and from 350 s to 400 s. The latter is negative because it's below the time axis.

Δx300 → 350 s ≈ ½(5.0 101 s)(1.0 101 mi/h)

= 2.5 102 (mi/h)(s)

Δx350 → 400 s ≈ ½(5.0 101 s)(−5.0 101 mi/h)

= −1.3 103 (mi/h)(s)

Find the total displacement by summing the parts.

Δx0 → 400 s ≈ (2.4 104 + 9.0 103 + 2.5 102 − 1.3 103) (mi/h)(s)

= 8.9 mi