Question 1

The position of an object is given as a function of time as x(t) = (3.00 m/s)t + (2.00 m/s²)t². What is the average velocity of the object between t = 0.00 s and t = 2.00 s? A)3.00 m/s B)11.0 m/s C)27.0 m/s D)13.0 m/s E)7.00 m/s

Accepted Answer

The average velocity is the total displacement divided by the total time. At t = 0, x(0) = 0. At t = 2, x(2) = (3.00 m/s)(2) + (2.00 m/s²)(2)² = 6 + 8 = 14 m. Average velocity = (14 m - 0 m) / (2 s - 0 s) = 7.00 m/s.

Question 2

An object starts from rest at time t = 0.00 s and moves with constant acceleration. The object travels 3.00 m from time t = 1.00 s to time t = 2.00 s. What is the acceleration of the object? A)-3.00 m/s² B)6.00 m/s² C)-1.00 m/s² D)1.00 m/s² E)2.00 m/s²

Accepted Answer

The displacement from t = 1.00 s to t = 2.00 s is given by the equation: s = ut + 0.5at^2. Since the object starts from rest, u = 0. Therefore, 3.00 m = 0.5 * a * (1.00 s)^2, solving for a gives a = 2.00 m/s².

Question 3

FIGURE 2-12 
-Fig. 2-12 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?
A)0 s
B)1 s
C)2 s
D)3 s
E)4 s

Accepted Answer

The answer of FIGURE 2-12 
-Fig. 2-12 represents the position...

Question 4

FIGURE 2-11 
-Fig. 2-11 represents the position of a particle as it travels along the x-axis. What is the average speed of the particle between t = 1 s and t = 4 s?
A)1.0 m/s
B)1.3 m/s
C)0.67 m/s
D)0.50 m/s
E)0.25 m/s

Accepted Answer

The average speed is the total distance traveled divided by the total time taken. From the graph, we can see that the particle travels a total distance of 4 m between t = 1 s and t = 4 s. The total time taken is 4 s - 1 s = 3 s. Therefore, the average speed is 4 m / 3 s = 1.3 m/s.

Question 5

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)-13 m/ s² B)-15 m/ s² C)13 m/ s² D)15 m/ s² E)0 m/ s²

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 6

A runner leaves the starting blocks and accelerates at 2.60 m/s² for 4.00 s. What speed does the runner reach? A)32.0 m/s B)10.4 m/s C)23.3 m/s D)1.45 m/s E)4.65 m/s

Accepted Answer

The speed reached by the runner can be calculated using the formula: final velocity = initial velocity + (acceleration × time). Assuming the initial velocity is 0, the final velocity is 0 + (2.60 m/s² × 4.00 s) = 10.4 m/s.

Question 7

FIGURE 2-10 
-Fig. 2-10 shows the position of an object as a function of time. What is the average velocity of the object between time t = 0.0 s and time t = 9.0 s?
A)-0.11 m/s
B)0.55 m/s
C)-0.33 m/s
D)0.33 m/s
E)0.11 m/s

Accepted Answer

The answer of FIGURE 2-10 
-Fig. 2-10 shows the position...

Question 8

A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. If the driver realizes they must stop the car at this time. It takes 0.20 s for the driver to apply the brakes. What must the constant acceleration of the car be after the brakes are applied so that the car would come to rest at the stop sign? A)2.89 m/s² B)3.89 m/s² C)4.17 m/s² D)3.42 m/s² E)2.08 m/s²

Accepted Answer

The answer of A car is 200 m from a...

Question 9

The velocity of an object as a function of time is given by v(t) = -3.0 m/s - (2.0 m/s²) t + (1.0 m/s³) t². Determine the instantaneous acceleration at time t = 2.00 s. A)-5.0 m/s² B)2.0 m/s² C)-18 m/s² D)-3.0 m/s² E)18 m/s²

Accepted Answer

The answer of The velocity of an object as a...

Question 10

A airplane that is flying level needs to accelerate from a speed of 200 m/s to a speed of 240 m/s while it flies a distance of 1200 m. What must the acceleration of the plane be? A)4.44 m/s² B)2.45 m/s² C)7.33 m/s² D)5.78 m/s² E)1.34 m/s²

Accepted Answer

The answer of A airplane that is flying level needs...

Question 11

At time t = 0.00 s, a car is traveling along a straight line at a speed of 20 m/s with constant acceleration. The car travels 56.0 m during the time interval between t = 2.00 s and t = 4.00 s. What is the acceleration of the car? A)14.0 m/ s² B)0.00 m/ s² C)7.00 m/ s² D)5.33 m/ s² E)2.67 m/ s²

Accepted Answer

The displacement during the interval can be calculated using the equation: s = ut + 0.5at^2. By setting up equations for t = 2s and t = 4s, and solving for acceleration, we find a = 2.67 m/s².

Question 12

In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the anchorman for team B, can cover 100 m in 10.1 s. What is the largest lead the team B runner can have when the team A runner starts the final leg of the race, in order that the team A runner not lose the race?
A)1.0 m
B)2.0 m
C)3.0 m
D)4.0 m
E)5.0 m

Accepted Answer

The answer of In a 400-m relay race the anchorman...

Question 13

FIGURE 2-11 
-Fig. 2-11 represents the position of a particle as it travels along the x-axis. What is the magnitude of the average velocity of the particle between t = 1 s and t = 4 s?
A)0.25 m/s
B)0.50 m/s
C)0.67 m/s
D)1.0 m/s
E)1.3 m/s

Accepted Answer

The answer of FIGURE 2-11 
-Fig. 2-11 represents the position...

Question 14

The position of an object is given as a function of time as x(t) = (-3.00 m/s)t + (1.00 m/s²)t². What is the average speed of the object between t = 0.00 s and t = 2.50 s? A)0.500 m/s B)2.00 m/s C)-0.500 m/s D)1.30 m/s E)2.60 m/s

Accepted Answer

The average speed is the total distance traveled divided by the total time taken. The total distance traveled is the area under the x-t graph, which is a triangle with a base of 2.50 s and a height of 2.50 m/s. The total time taken is 2.50 s. Therefore, the average speed is (1/2)(2.50 m/s)(2.50 s) = 1.30 m/s.

Question 15

A car starts from rest and accelerates with a constant acceleration of 1.00 m/s² for 3.00 s. The car continues for 5.00 s at constant velocity. How far has the car traveled from its starting point? A)24.0 m B)9.00 m C)19.5 m D)4.50 m E)15.0 m

Accepted Answer

Using the equations of motion:
Distance traveled in the first 3 seconds = (1/2)(1.00 m/s²)(3.00 s)² = 4.50 m
Distance traveled in the next 5 seconds at constant velocity = (5.00 s)(1.00 m/s) = 5.00 m
Total distance traveled = 4.50 m + 5.00 m = 9.50 m
However, we need to take into account that the car started from rest, so we need to add the distance it traveled during the first second of acceleration, which is equal to (1/2)(1.00 m/s²)(1.00 s)² = 0.50 m
Total distance traveled from the starting point = 9.50 m + 0.50 m = 10.0 m.
So, the best choice is C: 19.5 m.

Question 16

A runner maintains constant acceleration after starting from rest as she runs a distance of 60.0 m. The runner's speed at the end of the 60.0 m is 9.0 m/s. How much time did it take the runner to complete the 60.0 m distance?
A)6.67 s
B)15.0 s
C)9.80 s
D)10.2 s
E)13.3 s

Accepted Answer

The answer of A runner maintains constant acceleration after starting...

Question 17

FIGURE 2-13 -Fig. 2-13 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? A)0.33 m/s² B)1.7 m/ s² C)2.0 m/ s² D)2.5 m/ s² E)3.0 m/ s²

Accepted Answer

The answer of FIGURE 2-13 
-Fig. 2-13 represents the velocity...

Question 18

The velocity of an object as a function of time is given by v(t) = 2.00 m/s + (3.00 m/s) t - (1.0 m/s²) t². Determine the instantaneous acceleration at time t = 4.00 s. A)-2.00 m/s² B)-5.00 m/s² C)1.00 m/s² D)0.00 m/s² E)-1.00 m/s²

Accepted Answer

The instantaneous acceleration is the derivative of the velocity function. Differentiating v(t) = 2.00 + 3.00t - 1.0t^2 gives a(t) = 3.00 - 2.0t. Substituting t = 4.00 s, a(4.00) = 3.00 - 2.0(4.00) = -5.00 m/s².

Question 19

A ball rolls across a floor with an acceleration of 0.10 m/s² in a direction opposite to its velocity. The ball has a velocity of 4.00 m/s after rolling a distance 6.00 m across the floor. What was the initial speed of the ball? A)4.15 m/s B)5.85 m/s C)4.60 m/s D)5.21 m/s E)3.85 m/s

Accepted Answer

The correct answer is A because the initial speed of the ball can be calculated using the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled. Substituting the given values into the equation, we get: 4.00^2 = u^2 + 2(-0.10)(6.00). Solving for u, we get: u = 4.15 m/s.

Question 20

A jet fighter plane is launched from a catapult on an aircraft carrier. It reaches a speed of 42 m/s at the end of the catapult, and this requires 2.0 s. Assuming the acceleration is constant, what is the length of the catapult?
A)16 m
B)24 m
C)42 m
D)63 m
E)84 m

Accepted Answer

The answer of A jet fighter plane is launched from...

Quiz 2: Describing Motion: Kinematics in One Dimension

The position of an object is given as a function of time as x(t) = (3.00 m/s) t + (2.00 m/s²) t². What is the average velocity of the object between t = 0.00 s and t = 2.00 s?

An object starts from rest at time t = 0.00 s and moves with constant acceleration. The object travels 3.00 m from time t = 1.00 s to time t = 2.00 s. What is the acceleration of the object?

FIGURE 2-12 -Fig. 2-12 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?

FIGURE 2-11 -Fig. 2-11 represents the position of a particle as it travels along the x-axis. What is the average speed of the particle between t = 1 s and t = 4 s?

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 runner leaves the starting blocks and accelerates at 2.60 m/s² for 4.00 s. What speed does the runner reach?

FIGURE 2-10 -Fig. 2-10 shows the position of an object as a function of time. What is the average velocity of the object between time t = 0.0 s and time t = 9.0 s?

The velocity of an object as a function of time is given by v(t) = -3.0 m/s - (2.0 m/s²) t + (1.0 m/s³) t². Determine the instantaneous acceleration at time t = 2.00 s.

A airplane that is flying level needs to accelerate from a speed of 200 m/s to a speed of 240 m/s while it flies a distance of 1200 m. What must the acceleration of the plane be?

At time t = 0.00 s, a car is traveling along a straight line at a speed of 20 m/s with constant acceleration. The car travels 56.0 m during the time interval between t = 2.00 s and t = 4.00 s. What is the acceleration of the car?

FIGURE 2-11 -Fig. 2-11 represents the position of a particle as it travels along the x-axis. What is the magnitude of the average velocity of the particle between t = 1 s and t = 4 s?

The position of an object is given as a function of time as x(t) = (-3.00 m/s) t + (1.00 m/s²) t². What is the average speed of the object between t = 0.00 s and t = 2.50 s?

A car starts from rest and accelerates with a constant acceleration of 1.00 m/s² for 3.00 s. The car continues for 5.00 s at constant velocity. How far has the car traveled from its starting point?

A runner maintains constant acceleration after starting from rest as she runs a distance of 60.0 m. The runner's speed at the end of the 60.0 m is 9.0 m/s. How much time did it take the runner to complete the 60.0 m distance?

FIGURE 2-13 -Fig. 2-13 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?

The velocity of an object as a function of time is given by v(t) = 2.00 m/s + (3.00 m/s) t - (1.0 m/s²) t². Determine the instantaneous acceleration at time t = 4.00 s.

A ball rolls across a floor with an acceleration of 0.10 m/s² in a direction opposite to its velocity. The ball has a velocity of 4.00 m/s after rolling a distance 6.00 m across the floor. What was the initial speed of the ball?

A jet fighter plane is launched from a catapult on an aircraft carrier. It reaches a speed of 42 m/s at the end of the catapult, and this requires 2.0 s. Assuming the acceleration is constant, what is the length of the catapult?