Question 1

In which one of the following cases is the displacement of the object directly proportional to the elapsed time?

Accepted Answer

When a ball rolls with constant velocity, its displacement is directly proportional to the elapsed time. This is because the velocity is constant, so the displacement increases at a constant rate over time.

Question 2

A car is initially traveling at 50.0 km/h. The brakes are applied and the car stops over a distance of 35 m. What was magnitude of the car's acceleration while it was braking?

Accepted Answer

To find the acceleration, use the formula $ v^2 = u^2 + 2as $. Here, final velocity $ v = 0 $, initial velocity $ u = 50 $ km/h (convert to m/s by multiplying by $ \frac{5}{18} $), and distance $ s = 35 $ m. Solving for $ a $ gives approximately $ 2.8 $ m/s².

Question 3

A car, starting from rest, accelerates in a straight-line path at a constant rate of 2.0 m/s². How far will the car travel in 12 seconds?

Accepted Answer

We can use the equation:
d = 1/2 * a * t^2 + v_i * t
where d is the distance, a is the acceleration, t is the time, and v_i is the initial velocity (which is 0 in this case).
Plugging in the given values, we get:
d = 1/2 * 2.0 m/s^2 * (12 s)^2 + 0
d = 144 m
Therefore, the answer is B) 144 m.

Question 4

A truck accelerates from rest at point A with constant acceleration of magnitude a and, subsequently, passes points B and C as shown in the figure.  The distance between points B and C is x, and the time required for the truck to travel from B to C is t. Which expression determines the average speed of the truck between the points B and C?

Accepted Answer

The answer of A truck accelerates from rest at point...

Question 5

The minimum takeoff speed for a certain airplane is 75 m/s. What minimum acceleration is required if the plane must leave a runway of length 1050 m? Assume the plane starts from rest at one end of the runway.

Accepted Answer

The answer of The minimum takeoff speed for a certain...

Question 6

Which one of the following statements must be true if the expression  is to be used?

Accepted Answer

In the given expression, the only constant element is 'a', as it is not affected by any hexadecimal position changes. Therefore, option D is the only statement that must be true for the given expression to be used.

Question 7

A train with a constant velocity of +28.6 m/s approaches a small town . The operator applies the brake, reducing the train's velocity to +11.4 m/s. If the average acceleration of the train during braking is -1.35 m/s², for what elapsed time does the operator apply the brake?

Accepted Answer

The answer of A train with a constant velocity of...

Question 8

A car is moving at a constant velocity when it is involved in a collision. The car comes to rest after 0.450 s with an average acceleration of 60.0 m/s² in the direction opposite that of the car's velocity. What was the speed, in km/h, of the car before the collision?

Accepted Answer

The answer of A car is moving at a constant...

Question 9

A pitcher delivers a fast ball with a velocity of 43 m/s to the south. The batter hits the ball and gives it a velocity of 51 m/s to the north. What was the average acceleration of the ball during the 1.0 ms when it was in contact with the bat?

Accepted Answer

The change in velocity is 51 m/s (north) - (-43 m/s (south)) = 94 m/s (north). The time is 1.0 ms = 0.001 s. Average acceleration = change in velocity / time = 94 m/s / 0.001 s = 9.4 × 10^4 m/s², north.

Question 10

A landing airplane makes contact with the runway with a speed of 75.0 m/s and moves toward the south. After 18.5 seconds, the airplane comes to rest. What is the average acceleration of the airplane during the landing?

Accepted Answer

First, we need to find the distance that the airplane traveled before coming to rest. We can use the equation:

d = vi * t + 0.5 * a * t^2

where d is the distance, vi is the initial velocity, t is the time, and a is the acceleration.

Since the airplane comes to rest, its final velocity is 0. So we have:

0 = 75.0 m/s + a * 18.5 s

Solving for a, we get:

a = -4.05 m/s^2

The negative sign indicates that the acceleration is in the opposite direction of the airplane's initial velocity, which is north. Therefore, the correct answer is (C) 4.05 m/s^2, north.

Question 11

Starting from rest, a particle confined to move along a straight line is accelerated at a rate of 5.0 m/s². Which one of the following statements accurately describes the motion of this particle?

Accepted Answer

The acceleration of 5.0 m/s² means that the speed of the particle increases by 5.0 m/s every second.

Question 12

An elevator is moving upward with a speed of 11 m/s. Three seconds later, the elevator is still moving upward, but its speed has been reduced to 5.0 m/s. What is the average acceleration of the elevator during the 3.0 s interval?

Accepted Answer

The average acceleration is calculated using the formula: (final velocity - initial velocity) / time. Here, (5.0 m/s - 11 m/s) / 3.0 s = -2.0 m/s². The negative sign indicates the acceleration is downward.

Question 13

Two objects A and B accelerate from rest with the same constant acceleration. Object A accelerates for twice as much time as object B, however. Which one of the following statements is true concerning these objects at the end of their respective periods of acceleration?

Accepted Answer

Let's assume that both objects have an initial velocity of 0 and a constant acceleration of a. Let's also assume that object B accelerates for a time t, meaning that object A accelerates for a time 2t (twice as much time as object B).

Using the kinematic equation for displacement, we can find the distance traveled by each object during their respective periods of acceleration:

Distance traveled by object A: 
dA = 1/2 * a * (2t)^2 
dA = 2 * (1/2 * a * t^2)
dA = 2 * dB

Distance traveled by object B: 
dB = 1/2 * a * t^2

Therefore, at the end of their periods of acceleration, object A will have traveled four times as far as object B (dA = 2 * dB). Thus, the correct answer is B.

Question 14

Two cars travel along a level highway. It is observed that the distance between the cars is increasing. Which one of the following statements concerning this situation is necessarily true?

Accepted Answer

Both cars could be accelerating at the same rate, but if they started with different velocities, the distance between them could still increase.

Question 15

Which one of the following situations is not possible?

Accepted Answer

A body traveling with constant velocity implies zero acceleration, so it cannot have a time-varying acceleration.

Question 16

A car starts from rest and accelerates at a constant rate in a straight line. In the first second the car moves a distance of 2.0 meters. How much additional distance will the car move during the second second of its motion?

Accepted Answer

The car accelerates uniformly, so the distance covered in the first second is 2.0 m. Using the formula for distance under constant acceleration, $s = ut + \frac{1}{2}at^2$, where $u = 0$, $t = 1$, and $s = 2$, we find $a = 4 \, \text{m/s}^2$. In the second second, the total distance covered is $s = \frac{1}{2} \times 4 \times 2^2 = 8 \, \text{m}$. The additional distance in the second second is $8 - 2 = 6 \, \text{m}$.

Question 17

An object moving along a straight line is decelerating. Which one of the following statements concerning the object's acceleration is necessarily true?

Accepted Answer

When an object is decelerating, its acceleration is in the opposite direction to its velocity.

Question 18

A car starts from rest and accelerates at a constant rate in a straight line. In the first second the car moves a distance of 2.0 meters. How fast will the car be moving at the end of the second second?

Accepted Answer

The car moves 2.0 meters in the first second, which means the average velocity during the first second is 2.0 m/s. Since it starts from rest, the final velocity at the end of the first second is 4.0 m/s (twice the average velocity). With constant acceleration, the velocity increases linearly, so at the end of the second second, the velocity will be 8.0 m/s.

Question 19

A car traveling along a road begins accelerating with a constant acceleration of 1.5 m/s²in the direction of motion. After traveling 392 m at this acceleration, its speed is 35 m/s. Determine the speed of the car when it began accelerating.

Accepted Answer

We can use the kinematic equation:

v^2 = u^2 + 2as

where v = final velocity = 35 m/s, a = acceleration = 1.5 m/s^2, s = distance traveled = 392 m

Plugging in these values, we get:

35^2 = u^2 + 2(1.5)(392)

Solving for u, we get:

u = 7 m/s

Therefore, the speed of the car when it began accelerating was 7 m/s. So, the correct answer is B).

Question 20

Which one of the following is not a vector quantity?

Accepted Answer

A vector quantity has both magnitude and direction. Acceleration, displacement, average velocity, and instantaneous velocity all have direction in addition to a numerical value. Average speed, on the other hand, only has a numerical value and hence is not a vector quantity.

Quiz 2: Kinematics in One Dimension