Question 1

A ball is pushed with an initial velocity of 4.0 m/s.The ball rolls down a hill with a constant acceleration of 1.4 m/s².The ball reaches the bottom of the hill in 4.0 s.What is the ball's velocity at the bottom of the hill? A) 5 m/s B) 7 m/s C) 8 m/s D) 10 m/s E) 13 m/s

Accepted Answer

The ball's velocity at the bottom of the hill can be found using the equation:v = u + atwhere:v is the final velocityu is the initial velocitya is the accelerationt is the timeSubstituting the given values into the equation, we get:v = 4.0 m/s + (1.4 m/s^2) * 4.0 sv = 4.0 m/s + 5.6 m/sv = 9.6 m/sTherefore, the ball's velocity at the bottom of the hill is 9.6 m/s, which is closest to choice D) 10 m/s.

Question 2

A European sports car dealer claims that his product will accelerate at a constant rate from rest to a speed of 100 km/hr in 6 s.What distance will the sports car travel during the 6 s acceleration period? (Hint: First convert speed to m/s. )
A) 41.5 m
B) 58.1 m
C) 83 m
D) 166 m
E) ​124.5 m

Accepted Answer

First, we need to convert the speed to m/s.

100 km/hr = 27.78 m/s

To find the distance traveled during the 6 s acceleration period, we can use the following formula:

distance = (1/2) x acceleration x time^2

The acceleration is not given directly, but since the problem states that the acceleration is constant, we can use the following formula to find it:

acceleration = change in velocity / time

The change in velocity is simply the final velocity (100 km/hr or 27.78 m/s) minus the initial velocity (0 m/s):

acceleration = (27.78 m/s - 0 m/s) / 6 s = 4.63 m/s^2

Now we can plug in the values we have:

distance = (1/2) x 4.63 m/s^2 x (6 s)^2 = 83.2 m

So the distance traveled during the 6 s acceleration period is approximately 83 m.

Therefore, the best choice is C.

Question 3

A vehicle designed to operate on a drag strip accelerates from zero to 28 m/s while undergoing a straight line path displacement of 48 m.What is the vehicle's acceleration if its value may be assumed to be constant? A) 6 m/s² B) 4 m/s² C) 8 m/s² D) 12 m/s² E) 16 m/s²

Accepted Answer

We can use the following formula to calculate acceleration:

a = (v_f - v_i)/t

Where v_f is the final velocity, v_i is the initial velocity (which is zero in this case), and t is the time taken to reach the final velocity. We can use the following formula to calculate the time taken:

d = (v_f + v_i)/2 * t

Where d is the displacement (which is 48 m in this case). Rearranging this formula to solve for t, we get:

t = d/(v_f + v_i)/2

Substituting the given values, we get:

t = 48/(28 + 0)/2 = 48/14 = 3.43 s (rounded to two decimal places)

Now, we can substitute the values of v_f, v_i, and t into the first formula to calculate the acceleration:

a = (v_f - v_i)/t = (28 - 0)/3.43 = 8.16 m/s^2 (rounded to two decimal places)

Therefore, the best choice is C, 8 m/s^2.

Question 4

A strobe photograph of a car moving along a straight road shows the interval between each successive image to be diminishing.If the direction of motion of the car is taken as positive,which of the following are negative?
A) the speed of the car
B) the average velocity of the car
C) the average acceleration of the car
D) all of the above

Accepted Answer

When the interval between each successive image is diminishing, it means that the car is slowing down. Therefore, its speed and average velocity are decreasing, which makes them negative. However, its acceleration is positive because it is slowing down, which means that its velocity is changing in the negative direction. Therefore, the correct choice is C, the average acceleration of the car.

Question 5

A 50-g ball traveling at 24.0 m/s is bounced off a brick wall and rebounds at 19.0 m/s.A high-speed camera records this event.If the ball is in contact with the wall for 4.00 ms,what is the average acceleration of the ball during this time interval? A) 10,750 m/s² B) 21,500 m/s² C) 5,375 m/s² D) 10.75 m/s² E) 43 m/s²

Accepted Answer

The average acceleration of the ball is given by the change in velocity divided by the time interval:$$a = \frac{\Delta v}{\Delta t} = \frac{19.0 m/s - 24.0 m/s}{4.00 ms} = -10,750 m/s^2$$The negative sign indicates that the acceleration is in the opposite direction of the ball's initial velocity.

Question 6

An x vs.t graph is drawn for a ball moving in one direction.The graph starts at the origin and at t = 5 s the velocity of the ball is zero.We can be positive that at t = 5 s,
A) the slope of the curve is non-zero.
B) the ball has stopped.
C) the acceleration is constant.
D) the curve is at x = 0,t = 0.

Accepted Answer

At t = 5 s, the velocity of the ball is zero, which means it has stopped momentarily. Therefore, option B is the correct choice. Options A, C, and D are not necessarily true based on the information provided.

Question 7

The value of an object's acceleration may be characterized in equivalent words by which of the following?
A) displacement
B) rate of change of displacement
C) velocity
D) rate of change of velocity

Accepted Answer

The value of an object's acceleration is the rate of change of its velocity. Option D, rate of change of velocity, is the best choice to characterize the acceleration of an object.

Question 8

A Cessna aircraft has a lift-off speed of 120 km/hr.What minimum constant acceleration does this require if the aircraft is to be airborne after a take-off run of 270 m? A) 2.06 m/s² B) 1.03 m/s² C) 2.58 m/s² D) 4.12 m/s² E) 1.55 m/s²

Accepted Answer

To find the minimum constant acceleration, use the equation $ v^2 = u^2 + 2as $, where $ v = 120 $ km/hr = 33.33 m/s (converted), $ u = 0 $ m/s (initial speed), and $ s = 270 $ m. Solving for $ a $, we get $ a = \frac{v^2}{2s} = \frac{(33.33)^2}{2 \times 270} \approx 2.06 $ m/s².

Question 9

A cart is given an initial velocity of 7.0 m/s and experiences a constant acceleration of 2.5 m/s².What is the magnitude of the cart's displacement during the first 6.0 s of its motion? A) 52 m B) 132 m C) 87 m D) 15 m E) 22 m

Accepted Answer

The answer of A cart is given an initial velocity...

Question 10

A strobe photograph shows equally spaced images of a car moving along a straight road.If the time intervals between images is constant,which of the following cannot be positive?
A) the speed of the car
B) the average velocity of the car
C) the acceleration of the car
D) the direction of motion of the car

Accepted Answer

The answer of A strobe photograph shows equally spaced images...

Question 11

A v vs.t graph is drawn for a ball moving in one direction.The graph starts at the origin and at t = 5 s the acceleration of the ball is zero.We know that at t = 5 s,
A) the slope of the curve is non-zero.
B) the velocity of the ball is not changing.
C) the curve is not crossing the time axis.
D) the curve is at v = 0,t = 0.

Accepted Answer

If the acceleration is zero at t = 5 s, then the velocity of the ball is not changing at that moment. Therefore, option B is correct. The slope of the curve may or may not be zero at t = 5 s, so option A is not necessarily true. The curve may or may not cross the time axis, so option C is not necessarily true. Finally, we don't have enough information to know if the curve is at v = 0, t = 0, so option D is not necessarily true.

Question 12

A drag racer starts from rest and accelerates at 11 m/s² for the entire distance of 400 m.What is the velocity of the race car at the end of the run? A) 113 m/s B) 94 m/s C) 71 m/s D) 47 m/s E) 80 m/s

Accepted Answer

Using the equation v² = u² + 2as, where u = 0 m/s, a = 11 m/s², and s = 400 m, we find v = √(0 + 2*11*400) = 94 m/s.

Question 13

A bird,accelerating from rest at a constant rate,experiences a displacement of 25 m in 11 s.What is its acceleration? A) 0.21 m/s² B) 0.41 m/s² C) 0.51 m/s² D) 0.31 m/s² E) 0.62 m/s²

Accepted Answer

The answer of A bird,accelerating from rest at a constant...

Question 14

If the displacement of an object is given in SI units by Δx = −3 t + 4 t ²,at t = 2 s its velocity and acceleration are,respectively: A) positive,positive. B) positive,negative. C) negative,negative. D) negative,positive.

Accepted Answer

The answer of If the displacement of an object is...

Question 15

An automobile driver puts on the brakes and decelerates from 35.0 m/s to zero in 11.0 s.What distance does the car travel?
A) 193 m
B) 97 m
C) 241 m
D) 290 m
E) ​145 m

Accepted Answer

The distance traveled by the car can be found using the formula for distance under uniform acceleration, $d = v_i t + \frac{1}{2} a t^2$, where $v_i$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. Here, $v_i = 35.0 \, \text{m/s}$, $t = 11.0 \, \text{s}$, and the final velocity $v_f = 0 \, \text{m/s}$. First, find the acceleration: $a = \frac{v_f - v_i}{t} = \frac{0 - 35.0}{11.0} = -3.18 \, \text{m/s}^2$. Then, calculate the distance: $d = 35.0 \times 11.0 + \frac{1}{2} \times (-3.18) \times (11.0)^2 = 385 - 192 = 193 \, \text{m}$.

Question 16

When a drag strip vehicle reaches a velocity of 63 m/s,it begins a negative acceleration by releasing a drag chute and applying its brakes.While reducing its velocity back to zero,its acceleration along a straight line path is a constant −7.2 m/s².What displacement does it undergo during this deceleration period? A) 552 m B) 4 m C) 1,985 m D) 276 m E) 9 m

Accepted Answer

Using the equation $ v^2 = u^2 + 2as $, where $ v = 0 $, $ u = 63 \, \text{m/s} $, and $ a = -7.2 \, \text{m/s}^2 $, we find $ s = \frac{63^2}{2 \times 7.2} = 276 \, \text{m} $.

Question 17

The slope of the acceleration vs.time curve represents:
A) the velocity.
B) the rate of change of acceleration.
C) the rate of change of displacement.
D) the area under the position vs.time curve.

Accepted Answer

The slope of the acceleration vs. time curve represents the rate of change of acceleration. This means that it shows how quickly the acceleration is changing over time. It does not represent velocity, displacement, or the area under the position vs. time curve.

Question 18

In the case of constant acceleration,the average velocity equals the instantaneous velocity:
A) at the beginning of the time interval.
B) at the end of the time interval.
C) half-way through the time interval.
D) three-fourths of the way through the time interval.

Accepted Answer

When acceleration is constant, the average velocity is equal to the instantaneous velocity at the midpoint of the time interval. This is because, in uniform acceleration, the velocity changes linearly over time, making the average velocity (initial velocity + final velocity)/2 equal to the instantaneous velocity at the midpoint.

Question 19

Norma releases a bowling ball from rest;it rolls down a ramp with constant acceleration.After half a second it has traveled 0.55 m.How far has it traveled after two seconds?
​
A) 4.4 m
B) 11 m
C) 6.6 m
D) 8.8 m
E) ​17.6 m

Accepted Answer

The distance traveled under constant acceleration is proportional to the square of the time. If it travels 0.55 m in 0.5 seconds, it will travel 0.55 * (2/0.5)^2 = 8.8 m in 2 seconds.

Question 20

A bird,accelerating from rest at a constant rate,experiences a displacement of 25 m in 11 s.What is the final velocity after 11 s?
​
A) 3.4 m/s
B) 2.3 m/s
C) 4.5 m/s
D) 5.1 m/s
E) ​zero

Accepted Answer

The displacement (s) is 25 m, time (t) is 11 s, and initial velocity (u) is 0 m/s. Using the equation s = ut + 0.5at^2, we find acceleration (a) = 0.413 m/s^2. The final velocity (v) is given by v = u + at, which calculates to approximately 4.5 m/s.

Quiz 2: Motion in One Dimension

A ball is pushed with an initial velocity of 4.0 m/s.The ball rolls down a hill with a constant acceleration of 1.4 m/s2.The ball reaches the bottom of the hill in 4.0 s.What is the ball's velocity at the bottom of the hill?

A European sports car dealer claims that his product will accelerate at a constant rate from rest to a speed of 100 km/hr in 6 s.What distance will the sports car travel during the 6 s acceleration period? (Hint: First convert speed to m/s. )

A vehicle designed to operate on a drag strip accelerates from zero to 28 m/s while undergoing a straight line path displacement of 48 m.What is the vehicle's acceleration if its value may be assumed to be constant? ​

A strobe photograph of a car moving along a straight road shows the interval between each successive image to be diminishing.If the direction of motion of the car is taken as positive,which of the following are negative?

A 50-g ball traveling at 24.0 m/s is bounced off a brick wall and rebounds at 19.0 m/s.A high-speed camera records this event.If the ball is in contact with the wall for 4.00 ms,what is the average acceleration of the ball during this time interval?

An x vs.t graph is drawn for a ball moving in one direction.The graph starts at the origin and at t = 5 s the velocity of the ball is zero.We can be positive that at t = 5 s,

The value of an object's acceleration may be characterized in equivalent words by which of the following?

A Cessna aircraft has a lift-off speed of 120 km/hr.What minimum constant acceleration does this require if the aircraft is to be airborne after a take-off run of 270 m? ​

A cart is given an initial velocity of 7.0 m/s and experiences a constant acceleration of 2.5 m/s2.What is the magnitude of the cart's displacement during the first 6.0 s of its motion? ​

A strobe photograph shows equally spaced images of a car moving along a straight road.If the time intervals between images is constant,which of the following cannot be positive?

A v vs.t graph is drawn for a ball moving in one direction.The graph starts at the origin and at t = 5 s the acceleration of the ball is zero.We know that at t = 5 s,

A drag racer starts from rest and accelerates at 11 m/s2 for the entire distance of 400 m.What is the velocity of the race car at the end of the run?

A bird,accelerating from rest at a constant rate,experiences a displacement of 25 m in 11 s.What is its acceleration?

If the displacement of an object is given in SI units by Δx = −3 t + 4 t 2,at t = 2 s its velocity and acceleration are,respectively:

An automobile driver puts on the brakes and decelerates from 35.0 m/s to zero in 11.0 s.What distance does the car travel?

The slope of the acceleration vs.time curve represents:

In the case of constant acceleration,the average velocity equals the instantaneous velocity:

Norma releases a bowling ball from rest;it rolls down a ramp with constant acceleration.After half a second it has traveled 0.55 m.How far has it traveled after two seconds? ​

A bird,accelerating from rest at a constant rate,experiences a displacement of 25 m in 11 s.What is the final velocity after 11 s? ​

A ball is pushed with an initial velocity of 4.0 m/s.The ball rolls down a hill with a constant acceleration of 1.4 m/s².The ball reaches the bottom of the hill in 4.0 s.What is the ball's velocity at the bottom of the hill?

A vehicle designed to operate on a drag strip accelerates from zero to 28 m/s while undergoing a straight line path displacement of 48 m.What is the vehicle's acceleration if its value may be assumed to be constant?

A Cessna aircraft has a lift-off speed of 120 km/hr.What minimum constant acceleration does this require if the aircraft is to be airborne after a take-off run of 270 m?

A cart is given an initial velocity of 7.0 m/s and experiences a constant acceleration of 2.5 m/s².What is the magnitude of the cart's displacement during the first 6.0 s of its motion?

A drag racer starts from rest and accelerates at 11 m/s² for the entire distance of 400 m.What is the velocity of the race car at the end of the run?

If the displacement of an object is given in SI units by Δx = −3 t + 4 t ²,at t = 2 s its velocity and acceleration are,respectively:

Norma releases a bowling ball from rest;it rolls down a ramp with constant acceleration.After half a second it has traveled 0.55 m.How far has it traveled after two seconds?

A bird,accelerating from rest at a constant rate,experiences a displacement of 25 m in 11 s.What is the final velocity after 11 s?