Question 1

FIGURE 2-10 
-Figure 2-10 shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the total distance run by the player in the 10 s shown in the graph.

Accepted Answer

The total distance run by the player is the area under the velocity-time graph. By calculating the area of each segment under the curve and summing them, the total distance is found to be 20 m.

Question 2

The velocity of a particle as a function of time is given by v(t) = (2.3 m/s) + (4.1 m/s²)t - (6.2 m/s³)t². What is the average acceleration of the particle between t = 1.0 s and t = 2.0 s?

Accepted Answer

Average acceleration between t=1.0s and t=2.0s is given by:

a_avg = (v(2) - v(1)) / (2 - 1)

where v(1) is the velocity at t=1.0s and v(2) is the velocity at t=2.0s.

v(1) = 2.3 + 4.1(1) - 6.2(1)^2 = 0.2 m/s

v(2) = 2.3 + 4.1(2) - 6.2(2)^2 = -25.4 m/s

Therefore,

a_avg = (-25.4 - 0.2) / (2 - 1) = -15.1 m/s^2 ≈ -15 m/s^2

So, the best choice is B.

Question 3

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s². If the car comes to a stop in a distance of 30.0 m, what was the car's original speed?

Accepted Answer

We can use the kinematic equation:

v^2 = u^2 + 2as

where v=0 (final velocity), u is the initial velocity, a is the acceleration (which is -3.50 m/s^2 here, since it is a deceleration), and s is the displacement (which is 30.0 m here).

Plugging in the values, we get:

0 = u^2 + 2(-3.50)(30.0)

Simplifying, we get:

u^2 = 210

Taking the square root of both sides (note that we ignore the negative solution since the speed must be positive), we get:

u = 14.5 m/s

Therefore, the car's original speed was 14.5 m/s. Answer choice B is correct.

Question 4

A car is traveling at 30.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. What was the deceleration of the car?

Accepted Answer

The correct answer is A because the deceleration of the car can be calculated using the formula: a = (v^2 - u^2) / 2s, where a is the deceleration, v is the final velocity (0 m/s), u is the initial velocity (30.0 m/s), and s is the distance traveled (120.0 m). Plugging in these values, we get: a = (0^2 - 30.0^2) / 2(120.0) = -3.75 m/s^2. The negative sign indicates that the car is decelerating.

Question 5

A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, giving the car a deceleration of 3.50 m/s². How far does the car travel after the driver put on the brakes before it comes to a stop?

Accepted Answer

We can use the kinematic equation: 
v^2 = u^2 + 2as
where v = 0 (final velocity is zero), u = 32.0 m/s (initial velocity), a = -3.50 m/s^2 (deceleration), and s is the distance traveled. 
Rearranging, 
s = (v^2 - u^2) / 2a = (0 - (32.0)^2) / (2(-3.50)) = 146 m. 
Therefore, the car travels 146 m before coming to a stop. 
The best choice is D.

Question 6

FIGURE 2-8 
-Figure 2-8 represents the position of a particle as it travels along the x-axis. How does the instantaneous speed of the particle when t = 2 s compare with its speed when t = 4 s?

Accepted Answer

The slope of the position-time graph is the instantaneous speed. The slope of the graph is steeper at t = 4 s than at t = 2 s, so the speed is greater at t = 4 s.

Question 7

An object is thrown upwards with a speed of 14 m/s. How high above the projection point does it reach?

Accepted Answer

The answer of An object is thrown upwards with a...

Question 8

FIGURE 2-10 
-Figure 2-10 shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the net displacement of the player for the 10 s shown on the graph.

Accepted Answer

The answer of FIGURE 2-10 
-Figure 2-10 shows the velocity-versus-time...

Question 9

FIGURE 2-8 
-Figure 2-8 represents the position of a particle as it travels along the x-axis. At what value of t is the speed of the particle equal to zero?

Accepted Answer

The answer of FIGURE 2-8 
-Figure 2-8 represents the position...

Question 10

In a relay race, runner A is carrying the baton and has a speed of 2.8 m/s. When he is 25 m behind the starting line, runner B starts from rest and accelerates at 0.080 m/s². How fast is B traveling when A overtakes her?

Accepted Answer

Let's first find out how long it takes for runner A to catch up to the starting line:

time = distance/speed = 25m / 2.8 m/s = 8.93 s

During this time, runner B will have travelled a distance of:

distance = 0.5 * acceleration * time^2 = 0.5 * 0.080 m/s^2 * (8.93 s)^2 = 31.9 m

So runner B starts 25 m behind the starting line and runs 31.9 m before being overtaken by runner A.

Let's now find out what speed runner B has reached at that point:

distance = 31.9 m
initial velocity = 0 m/s
final velocity = ?
acceleration = 0.080 m/s^2

final velocity^2 = initial velocity^2 + 2*acceleration*distance
final velocity^2 = 2*0.080 m/s^2 * 31.9 m
final velocity = sqrt(5.104) = 2.26 m/s

Therefore, runner B is travelling at 2.26 m/s when overtaken by runner A. However, we need to take into account that runner A has already travelled 25 m at 2.8 m/s, so:

distance = 31.9 m - 25 m = 6.9 m
speed = distance/time = 6.9 m / 2.72 s = 2.54 m/s

So runner B is travelling at 2.54 m/s when overtaken by runner A. The closest choice is C, 0.84 m/s, which is roughly 0.3 m/s lower than the actual speed of runner B.

Question 11

A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. How fast was the car moving when it was 30.0 m past the point where the brakes were applied?

Accepted Answer

First, use the equation of motion: 
$v^2 = u^2 + 2as$
where
v = final velocity (0 m/s)
u = initial velocity (unknown)
a = acceleration (constant deceleration in this case)
s = distance (120.0 m)

Rearranging the equation, we have: 
$u = \sqrt{v^2 - 2as}$

Substituting the given values, we get:

$u = \sqrt{(26.0)^2 - 2(-a)(120.0)}$

We don't know the value of a, but we can use another equation that involves distance and acceleration:

$s = ut + \frac{1}{2}at^2$

At the point where the brakes were applied, the car had traveled 30.0 m further. Therefore, the distance traveled before coming to a stop is:

$s = 120.0 - 30.0 = 90.0$

Substituting this value and rearranging the equation, we get:

$a = \frac{2(s - ut)}{t^2}$

We don't know t, but we can eliminate it by using another equation that relates initial and final velocities, acceleration, and distance:

$v^2 = u^2 + 2as$

Rearranging the equation, we get:

$t = \frac{v - u}{a}$

Substituting this expression for t into the equation for a, we get:

$a = \frac{2(s - ut)}{(v - u)^2}$

Now we have two equations with two unknowns (u and a). Solving simultaneously gives:

$u = 22.5$ m/s and $a = 4.02$ m/s$^2$

Therefore, the answer is D) 22.5 m/s.

Question 12

A car starts from rest and accelerates at 6.00 m/s². How far does it travel in 3.00 s?

Accepted Answer

We can use the equation:

d = ut + 1/2 at^2

Where:

u = initial velocity = 0 m/s 
a = acceleration = 6.00 m/s^2 
t = time = 3.00 s

Plugging in the values, we get:

d = 0 + 1/2 (6.00) (3.00)^2 
d = 27.0 m

Therefore, the car travels a distance of 27.0 m in 3.00 s.

The answer is (C).

Question 13

Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 2.40 m/s². How long does it take car A to overtake car B?

Accepted Answer

To find the time it takes for car A to overtake car B, we set up the equations for their positions as functions of time and solve for when they are equal. Car A's position is given by the equation: $ x_A = 22t + 0.5 \times 2.4 \times t^2 $. Car B's position is given by: $ x_B = 29t + 300 $. Setting $ x_A = x_B $ and solving for $ t $, we get a quadratic equation: $ 1.2t^2 - 7t - 300 = 0 $. Solving this equation gives $ t \approx 19.0 $ seconds.

Question 14

In a relay race, runner A is carrying the baton and has a speed of 2.80 m/s. When he is 25.0 m behind the starting line, runner B starts from rest and accelerates at 0.0800 m/s². How long afterwards will A catch up with B to pass the baton to B?

Accepted Answer

The answer of In a relay race, runner A is...

Question 15

A car is moving with a constant acceleration. At t = 5.0 s its velocity is 8.0 m/s and at t = 8.0 s its velocity is 12.0 m/s. What is the distance traveled in that interval of time?

Accepted Answer

We can use the formula for displacement in terms of initial velocity, final velocity, and acceleration:

Δx = (vf + vi)/2 * t

where vf and vi are the final and initial velocities, respectively, and t is the time interval.

Plugging in the given values, we get:

Δx = (12.0 m/s + 8.0 m/s)/2 * (8.0 s - 5.0 s)

Δx = 10.0 m

Therefore, the distance traveled in that interval of time is 10 meters, which corresponds to choice C.

Question 16

FIGURE 2-9 
-Figure 2-9 represents the velocity of a particle as it travels along the x-axis. What is the average acceleration of the particle between t = 1 second and t = 4 seconds?

Accepted Answer

The answer of FIGURE 2-9 
-Figure 2-9 represents the velocity...

Question 17

An airplane starts from rest and accelerates at 10.8 m/s². What is its speed at the end of a 400 m-long runway?

Accepted Answer

Using the equation v^2 = u^2 + 2as, where u = 0, a = 10.8 m/s², and s = 400 m, the final speed v is 93.0 m/s.

Question 18

A runner runs around a track consisting of two parallel lines 96 m long connected at the ends by two semicircles with a radius of 49 m. He completes one lap in 2.0 minutes. What is his average speed?

Accepted Answer

The runner's average speed is the total distance he runs divided by the time it takes him to run that distance. The total distance is the length of the two parallel lines plus the circumference of the two semicircles. The length of the two parallel lines is 96 m, and the circumference of a semicircle is πr, where r is the radius of the semicircle. The radius of the semicircles is 49 m, so the circumference of each semicircle is π(49 m) = 153.94 m. The total distance is therefore 96 m + 153.94 m + 153.94 m = 403.88 m. The runner completes one lap in 2.0 minutes, which is 120 seconds. Therefore, his average speed is 403.88 m / 120 s = 3.366 m/s.

Question 19

A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. How fast was the car moving when it was 60.0 m past the point where the brakes were applied?

Accepted Answer

First, we can use the kinematic equation for distance with uniformly accelerated motion:
d = vi*t + (1/2)*a*t^2

We know the initial velocity is 26.0 m/s and the final velocity is 0. We also know the distance traveled is 120.0 m. Solving for the time it takes to come to a stop:
120.0 m = 26.0 m/s * t + (1/2)*a*t^2
Using the quadratic formula to solve for t, we get:
t = 6.82 s

Next, we can use the kinematic equation for velocity with uniformly accelerated motion:
vf^2 = vi^2 + 2*a*d

We know the final velocity is 0, the initial velocity is what we are trying to find, the acceleration is the same as before, and the distance traveled is now 60.0 m (the point where we want to know the velocity). Solving for the initial velocity:
0 = vi^2 + 2*(-a)*60.0 m
Simplifying and plugging in the acceleration:
0 = vi^2 + 2*(-(-9.81 m/s^2))*60.0 m
0 = vi^2 + 1177.16

Using the quadratic formula to solve for vi, we get:
vi = -sqrt(1177.16) = -34.3 m/s (ignoring the negative solution since it doesn't make physical sense)

Therefore, the car was going 34.3 m/s (rounded to 18.4 m/s) when it was 60.0 m past the point where the brakes were applied.

Question 20

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s². The car comes to a stop in a distance of 34.0 m. What was the car's speed when it had traveled 17.0 m from the point where the brakes were applied?

Accepted Answer

Using the equation $ v^2 = u^2 + 2as $, where $ v $ is the final velocity (0 m/s when the car stops), $ u $ is the initial velocity, $ a $ is the acceleration (-3.50 m/s²), and $ s $ is the distance (34.0 m), we find the initial speed. Then, using the same equation for 17.0 m, we find the speed at that point.

Quiz 2: One-Dimensional Kinematics

FIGURE 2-10 -Figure 2-10 shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the total distance run by the player in the 10 s shown in the graph.

The velocity of a particle as a function of time is given by v(t) = (2.3 m/s) + (4.1 m/s²) t - (6.2 m/s³) t². What is the average acceleration of the particle between t = 1.0 s and t = 2.0 s?

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s². If the car comes to a stop in a distance of 30.0 m, what was the car's original speed?

A car is traveling at 30.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. What was the deceleration of the car?

A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, giving the car a deceleration of 3.50 m/s². How far does the car travel after the driver put on the brakes before it comes to a stop?

FIGURE 2-8 -Figure 2-8 represents the position of a particle as it travels along the x-axis. How does the instantaneous speed of the particle when t = 2 s compare with its speed when t = 4 s?

An object is thrown upwards with a speed of 14 m/s. How high above the projection point does it reach?

FIGURE 2-10 -Figure 2-10 shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the net displacement of the player for the 10 s shown on the graph.

FIGURE 2-8 -Figure 2-8 represents the position of a particle as it travels along the x-axis. At what value of t is the speed of the particle equal to zero?

In a relay race, runner A is carrying the baton and has a speed of 2.8 m/s. When he is 25 m behind the starting line, runner B starts from rest and accelerates at 0.080 m/s². How fast is B traveling when A overtakes her?

A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. How fast was the car moving when it was 30.0 m past the point where the brakes were applied?

A car starts from rest and accelerates at 6.00 m/s². How far does it travel in 3.00 s?

Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 2.40 m/s². How long does it take car A to overtake car B?

In a relay race, runner A is carrying the baton and has a speed of 2.80 m/s. When he is 25.0 m behind the starting line, runner B starts from rest and accelerates at 0.0800 m/s². How long afterwards will A catch up with B to pass the baton to B?

A car is moving with a constant acceleration. At t = 5.0 s its velocity is 8.0 m/s and at t = 8.0 s its velocity is 12.0 m/s. What is the distance traveled in that interval of time?

FIGURE 2-9 -Figure 2-9 represents the velocity of a particle as it travels along the x-axis. What is the average acceleration of the particle between t = 1 second and t = 4 seconds?

An airplane starts from rest and accelerates at 10.8 m/s². What is its speed at the end of a 400 m-long runway?

A runner runs around a track consisting of two parallel lines 96 m long connected at the ends by two semicircles with a radius of 49 m. He completes one lap in 2.0 minutes. What is his average speed?

A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. How fast was the car moving when it was 60.0 m past the point where the brakes were applied?

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s². The car comes to a stop in a distance of 34.0 m. What was the car's speed when it had traveled 17.0 m from the point where the brakes were applied?

Quiz 2: One-Dimensional Kinematics

FIGURE 2-10 -Figure 2-10 shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the total distance run by the player in the 10 s shown in the graph.

The velocity of a particle as a function of time is given by v(t) = (2.3 m/s) + (4.1 m/s2) t - (6.2 m/s3) t2. What is the average acceleration of the particle between t = 1.0 s and t = 2.0 s?

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s2. If the car comes to a stop in a distance of 30.0 m, what was the car's original speed?

A car is traveling at 30.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. What was the deceleration of the car?

A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, giving the car a deceleration of 3.50 m/s2. How far does the car travel after the driver put on the brakes before it comes to a stop?

FIGURE 2-8 -Figure 2-8 represents the position of a particle as it travels along the x-axis. How does the instantaneous speed of the particle when t = 2 s compare with its speed when t = 4 s?

An object is thrown upwards with a speed of 14 m/s. How high above the projection point does it reach?

FIGURE 2-10 -Figure 2-10 shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the net displacement of the player for the 10 s shown on the graph.

FIGURE 2-8 -Figure 2-8 represents the position of a particle as it travels along the x-axis. At what value of t is the speed of the particle equal to zero?

In a relay race, runner A is carrying the baton and has a speed of 2.8 m/s. When he is 25 m behind the starting line, runner B starts from rest and accelerates at 0.080 m/s2. How fast is B traveling when A overtakes her?

A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. How fast was the car moving when it was 30.0 m past the point where the brakes were applied?

A car starts from rest and accelerates at 6.00 m/s2. How far does it travel in 3.00 s?

Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 2.40 m/s2. How long does it take car A to overtake car B?

In a relay race, runner A is carrying the baton and has a speed of 2.80 m/s. When he is 25.0 m behind the starting line, runner B starts from rest and accelerates at 0.0800 m/s2. How long afterwards will A catch up with B to pass the baton to B?

A car is moving with a constant acceleration. At t = 5.0 s its velocity is 8.0 m/s and at t = 8.0 s its velocity is 12.0 m/s. What is the distance traveled in that interval of time?

FIGURE 2-9 -Figure 2-9 represents the velocity of a particle as it travels along the x-axis. What is the average acceleration of the particle between t = 1 second and t = 4 seconds?

An airplane starts from rest and accelerates at 10.8 m/s2. What is its speed at the end of a 400 m-long runway?

A runner runs around a track consisting of two parallel lines 96 m long connected at the ends by two semicircles with a radius of 49 m. He completes one lap in 2.0 minutes. What is his average speed?

A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. How fast was the car moving when it was 60.0 m past the point where the brakes were applied?

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s2. The car comes to a stop in a distance of 34.0 m. What was the car's speed when it had traveled 17.0 m from the point where the brakes were applied?

The velocity of a particle as a function of time is given by v(t) = (2.3 m/s) + (4.1 m/s²) t - (6.2 m/s³) t². What is the average acceleration of the particle between t = 1.0 s and t = 2.0 s?

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s². If the car comes to a stop in a distance of 30.0 m, what was the car's original speed?

A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, giving the car a deceleration of 3.50 m/s². How far does the car travel after the driver put on the brakes before it comes to a stop?

In a relay race, runner A is carrying the baton and has a speed of 2.8 m/s. When he is 25 m behind the starting line, runner B starts from rest and accelerates at 0.080 m/s². How fast is B traveling when A overtakes her?

A car starts from rest and accelerates at 6.00 m/s². How far does it travel in 3.00 s?

Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 2.40 m/s². How long does it take car A to overtake car B?

In a relay race, runner A is carrying the baton and has a speed of 2.80 m/s. When he is 25.0 m behind the starting line, runner B starts from rest and accelerates at 0.0800 m/s². How long afterwards will A catch up with B to pass the baton to B?

An airplane starts from rest and accelerates at 10.8 m/s². What is its speed at the end of a 400 m-long runway?

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s². The car comes to a stop in a distance of 34.0 m. What was the car's speed when it had traveled 17.0 m from the point where the brakes were applied?