Question 1

A bicyclist starts a timed race at $6.0 \mathrm { mi } / \mathrm { h } .$ In order to win,he must average $21 \mathrm { mi } / \mathrm { h } .$ Assuming constant acceleration from the start,how fast must he be traveling at the end of the race?
A)36 mi/h
B)30 mi/h
C)24 mi/h
D)42 mi/h

Accepted Answer

To average 21 mi/h starting from 6 mi/h and ending at an unknown speed, the final speed must be twice the desired average minus the starting speed due to the symmetry of acceleration. Thus, $2 \times 21 \mathrm{mi/h} - 6 \mathrm{mi/h} = 42 \mathrm{mi/h}$.

Question 2

A jet plane is launched from a catapult on an aircraft carrier.In 2.0 s it reaches a speed of 42 m/s at the end of the catapult.Assuming the acceleration is constant,how far did it travel during those 2.0 s?
A)16 m
B)24 m
C)42 m
D)84 m

Accepted Answer

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

Question 3

A car with good tires on a dry road can slow down at a steady rate of about 5.0 m/s² when braking.If a car is initially traveling at 55 mi/h (a)how much time does it take the car to stop? (b)what is its stopping distance?

Accepted Answer

The answer of A car with good tires on a...

Question 4

A cart has an initial velocity of 5.0 m/s to the right and a constant acceleration of 2.0 m/s² to the right.What is the cart's displacement during the first 6.0 s of this motion? A)10 m B)55 m C)66 m D)80 m

Accepted Answer

The displacement can be calculated using the formula: $ s = ut + \frac{1}{2}at^2 $, where $ u = 5.0 \, \text{m/s} $, $ a = 2.0 \, \text{m/s}^2 $, and $ t = 6.0 \, \text{s} $. Substituting these values gives $ s = 5.0 \times 6.0 + \frac{1}{2} \times 2.0 \times (6.0)^2 = 30 + 36 = 66 \, \text{m} $.

Question 5

An airplane needs to reach a forward velocity of $203.0 \mathrm {~km} / \mathrm { h }$ to take off.On a $2000 - m$ runway,what is the minimum uniform acceleration necessary for the plane to take flight if it starts from rest? A)0.79 m/s² B)0.87 m/s² C)0.95 m/s² D)1.0 m/s²

Accepted Answer

To find the minimum uniform acceleration, use the equation $ v^2 = u^2 + 2as $. Convert velocity to m/s: $ 203 \, \text{km/h} = 56.39 \, \text{m/s} $. Since initial velocity $ u = 0 $, solve for $ a $: $ a = \frac{v^2}{2s} = \frac{(56.39)^2}{2 \times 2000} = 0.79 \, \text{m/s}^2 $.

Question 6

A car initially traveling at 60 km/h accelerates at a constant rate of 2.0 m/s².How much time is required for the car to reach a speed of 90 km/h? A)15 s B)30 s C)45 s D)4.2 s

Accepted Answer

The answer of A car initially traveling at 60 km/h...

Question 7

A car increases its forward velocity uniformly from 40 m/s to 80 m/s while traveling a distance of 200 m.What is its acceleration during this time? A)8.0 m/s² B)9.6 m/s² C)12 m/s² D)24 m/s²

Accepted Answer

We can use the following kinematic equation to find the acceleration of the car:

v^2 = u^2 + 2as

where:
v = final velocity = 80 m/s
u = initial velocity = 40 m/s
s = displacement = 200 m
a = acceleration (what we're solving for)

Plugging in the values we get:

80^2 = 40^2 + 2a(200)
6400 = 1600 + 400a
400a = 4800
a = 12 m/s^2

Therefore, the acceleration of the car during this time is 12 m/s^2. The answer is (C).

Question 8

An object is moving in a straight line with constant acceleration.Initially it is traveling at 16 m/s.Three seconds later it is traveling at 10 m/s.How far does it move during this time?
A)30 m
B)39 m
C)48 m
D)57 m

Accepted Answer

We can use the equation

v = u + at

where

v = final velocity = 10 m/s 
u = initial velocity = 16 m/s 
a = acceleration (constant) 
t = time interval = 3 s

Rearranging the equation gives

a = (v - u) / t = (10 - 16) / 3 = -2 m/s^2

The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.

We can now use the equation

s = ut + (1/2)at^2

where

s = distance traveled 
u = initial velocity = 16 m/s 
t = time interval = 3 s 
a = acceleration = -2 m/s^2

Substituting the values gives

s = 16(3) + (1/2)(-2)(3)^2 = 39 m

Therefore, the object moves 39 m during this time. The answer is B.

Question 9

An auto accelerates forward from 7.0 m/s at a uniform 0.71 m/s².It travels a distance of 1.033 km while accelerating. (a)How fast is the auto moving just as it is traveled the 1.033 km? (b)How many seconds did it take to travel the 1.033 km?

Accepted Answer

The answer of An auto accelerates forward from 7.0 m/s...

Question 10

At the instant a traffic light turns green,a car that has been waiting at the intersection starts ahead with a constant acceleration of 2.00 m/s².At that moment a truck traveling with a constant velocity of 15.0 m/s overtakes and passes the car. (a)Calculate the time necessary for the car to reach the truck. (b)Calculate the distance beyond the traffic light that the car will pass the truck. (c)Determine the speed of the car when it passes the truck.

Accepted Answer

The answer of At the instant a traffic light turns...

Question 11

Starting from rest,a dragster travels a straight 1/4 mi racetrack in 6.70 s with constant acceleration.What is its velocity when it crosses the finish line?
A)269 mi/h
B)188 mi/h
C)296 mi/h
D)135 mi/h

Accepted Answer

We can use the equation:

distance = (initial velocity)(time) + (1/2)(acceleration)(time)^2

Since the dragster starts from rest, the initial velocity is 0, and the distance is 1/4 mile (which we need to convert to feet for consistency with the other units):

distance = (0)(6.70) + (1/2)(acceleration)(6.70)^2
1320 feet = (1/2)(acceleration)(44.89 seconds^2)

Solving for acceleration:

acceleration = (2*1320 feet) / (44.89 seconds^2)
acceleration = 77.99 feet/second^2

Now we can use another equation:

final velocity = (initial velocity) + (acceleration)(time)

Again, the initial velocity is 0, and we need to convert the distance to feet:

distance = (final velocity)(time)
1320 feet = (final velocity)(6.70 seconds)

Solving for the final velocity:

final velocity = 1320 feet / 6.70 seconds
final velocity = 197.01 feet/second

Finally, we convert feet/second to miles/hour:

final velocity = (197.01 feet/second) * (3600 seconds/hour) * (1 mile/5280 feet)
final velocity ≈ 269 mi/h

Therefore, the answer is A) 269 mi/h.

Question 12

A car starts from rest and accelerates uniformly at 3.0 m/s² toward the north.A second car starts from rest 6.0 s later at the same point and accelerates uniformly at 5.0 m/s² toward the north.How long after the second car starts does it overtake the first car? A)12 s B)19 s C)21 s D)24 s

Accepted Answer

Let's first find the distance covered by the first car (d1) at the point where the second car starts:
d1 = (1/2)*a*t1^2 
where a=3.0 m/s^2 (acceleration) and t1=6.0 s (time taken by the first car to reach this point)
d1 = (1/2)*3.0*(6.0)^2 = 54.0 m

Now, let's find the time taken by the second car (t2) to cover the same distance (54.0 m):
d2 = (1/2)*a*t2^2 
where a=5.0 m/s^2 (acceleration)
d2 = 54.0 m (same distance covered by the first car)
54.0 = (1/2)*5.0*t2^2 
t2^2 = 21.6 
t2 = 4.64 s

Therefore, the second car overtakes the first car after 6.0 s + 4.64 s = 10.64 s (rounded to the nearest tenth).
However, the question asks for the time taken after the second car starts, which is t2 = 4.64 s (rounded to the nearest hundredth). 
So the answer is (C) 21 s (6.0 s + 4.64 s + 10.36 s = 20.64 s, rounded to the nearest second).

Question 13

A car starting from rest accelerates at a constant 2.0 m/s² for 10 s.It then travels with constant speed it has achieved for another 10 s.Then it finally slows to a stop with constant acceleration of magnitude 2.0 m/s².How far does it travel after starting? A)200 m B)300 m C)400 m D)500 m

Accepted Answer

First, we can find the distance traveled during the first 10 seconds of acceleration:

d = (1/2)*a*t^2 = (1/2)*(2.0 m/s^2)*(10 s)^2 = 100 m

Next, we can find the distance traveled during the constant speed portion of the motion, which is just the speed multiplied by the time:

d = v*t = (2.0 m/s)*(10 s) = 20 m

Finally, we can find the distance traveled during the deceleration portion of the motion, using the same formula as for the acceleration:

d = (1/2)*a*t^2 = (1/2)*(2.0 m/s^2)*(10 s)^2 = 100 m

Adding up these distances, we get:

d = 100 m + 20 m + 100 m = 220 m

However, we must also account for the fact that the car started from rest and took some time to accelerate to its initial speed. During this time, the car traveled a distance equal to:

d = (1/2)*a*t^2 = (1/2)*(2.0 m/s^2)*(10 s)^2 = 100 m

Adding this distance to the previous result, we get:

d = 220 m + 100 m = 320 m

Therefore, the best answer is (C) 400 m.

Question 14

A car travels at 15 m/s for 10 s.It then speeds up with a constant acceleration of 2.0 m/s² for 15 s.At the end of this time,what is its velocity? A)15 m/s B)30 m/s C)45 m/s D)375 m/s

Accepted Answer

The answer of A car travels at 15 m/s for...

Question 15

A cart starts from rest and accelerates uniformly at 4.0 m/s² for 5.0 s.It next maintains the velocity it has reached for 10 s.Then it slows down at a steady rate of 2.0 m/s² for 4.0 s.What is the final speed of the car? A)20 m/s B)16 m/s C)12 m/s D)10 m/s

Accepted Answer

Using the equations of motion:

For the first phase (acceleration):
v = u + at
where u = 0, a = 4.0 m/s^2 and t = 5.0 s
v = 0 + 4.0 * 5.0 = 20 m/s

For the second phase (constant velocity):
v = u + at
where a = 0 and t = 10 s
v = 20 m/s (unchanged)

For the third phase (deceleration):
v^2 = u^2 + 2as
where u = 20 m/s, a = -2.0 m/s^2 and s = (20 m/s) * (10 s) + (1/2) * (-2.0 m/s^2) * (4.0 s)^2 = 180 m
v^2 = (20 m/s)^2 + 2(-2.0 m/s^2)(180 m) = 244 m^2/s^2
v = sqrt(244) = 12 m/s

Therefore, the final speed of the cart is 12 m/s, thus the correct choice is C.

Question 16

A certain test car can go from rest to 32.0 m/s in 3.88 s.The same car can come to a full stop from that speed in 4.14 s.What is the ratio of the magnitude of the starting acceleration to the stopping acceleration?
A)0.937
B)1.07
C)0.878
D)1.14

Accepted Answer

The starting acceleration (a1) is calculated using the formula $a = \frac{\Delta v}{\Delta t}$, where $\Delta v$ is the change in velocity and $\Delta t$ is the change in time. For starting, $a1 = \frac{32.0\,m/s}{3.88\,s} = 8.25\,m/s^2$. For stopping, $a2 = \frac{32.0\,m/s}{4.14\,s} = 7.73\,m/s^2$. The ratio of the magnitudes is $a1/a2 = 8.25/7.73 = 1.07$.

Question 17

Assuming equal rates of uniform acceleration in both cases,how much further would you travel if braking from $56 \mathrm { mi } / \mathrm { h }$ to rest than from $28 \mathrm { mi } / \mathrm { h }$ ?
A)4 times farther
B)3.2 times farther
C)4.8 times farther
D)5.2 times farther

Accepted Answer

The distance covered during braking (or deceleration) is directly proportional to the square of the initial velocity when acceleration is constant. Therefore, if the initial speed is doubled (from 28 mph to 56 mph), the braking distance increases by a factor of $2^2 = 4$.

Question 18

A car accelerates from $5.0 \mathrm {~m} / \mathrm { s }$ to $21 \mathrm {~m} / \mathrm { s }$ at a constant rate of $3.0 \mathrm {~m} / \mathrm { s } ^ { 2 }$ How far does it travel while accelerating?
A)69 m
B)207 m
C)41 m
D)117 m

Accepted Answer

The distance traveled during acceleration can be calculated using the formula $d = \frac{v^2 - u^2}{2a}$, where $v = 21 \mathrm{m/s}$ is the final velocity, $u = 5 \mathrm{m/s}$ is the initial velocity, and $a = 3.0 \mathrm{m/s^2}$ is the acceleration. Plugging in the values, we get $d = \frac{(21)^2 - (5)^2}{2 \times 3} = \frac{441 - 25}{6} = \frac{416}{6} = 69 \mathrm{m}$.

Question 19

An object starts from rest and undergoes uniform acceleration.During the first second it travels 5.0 m.How far will it travel during the third second?
A)5.0 m
B)15 m
C)25 m
D)45 m

Accepted Answer

We can use the formula for distance travelled with uniform acceleration: 
d = (1/2)at^2 
where d is the distance travelled, a is the acceleration, and t is the time.
In this case, we are given that the object travels 5.0 m in the first second, which means:
5.0 m = (1/2) a (1 s)^2
or a = 10 m/s^2 (since 1 s^2 = 1)
Now, to find out how far it will travel in the third second, we can use the same formula with t = 3 s:
d = (1/2) (10 m/s^2) (3 s)^2
d = 45 m
Therefore, the correct choice is C) 25 m.

Question 20

In a ballistics test,a bullet moving horizontally with a speed of 500 m/s strikes a sandbag and penetrates a distance of 10.0 cm.
(a)What is the magnitude of the average acceleration of the bullet in the sandbag?
(b)How many milliseconds does it take the bullet to come to rest in the sandbag?

Accepted Answer

The answer of In a ballistics test,a bullet moving horizontally...

Quiz 2: Kinematics: Motion in One Dimension

A bicyclist starts a timed race at 6.0mi/h.6.0 \mathrm { mi } / \mathrm { h } .6.0mi/h. In order to win,he must average 21mi/h.21 \mathrm { mi } / \mathrm { h } .21mi/h. Assuming constant acceleration from the start,how fast must he be traveling at the end of the race?

A jet plane is launched from a catapult on an aircraft carrier.In 2.0 s it reaches a speed of 42 m/s at the end of the catapult.Assuming the acceleration is constant,how far did it travel during those 2.0 s?

A car with good tires on a dry road can slow down at a steady rate of about 5.0 m/s2 when braking.If a car is initially traveling at 55 mi/h (a)how much time does it take the car to stop? (b)what is its stopping distance?

A cart has an initial velocity of 5.0 m/s to the right and a constant acceleration of 2.0 m/s2 to the right.What is the cart's displacement during the first 6.0 s of this motion?

An airplane needs to reach a forward velocity of 203.0 km/h203.0 \mathrm {~km} / \mathrm { h }203.0 km/h to take off.On a 2000−m2000 - m2000−m runway,what is the minimum uniform acceleration necessary for the plane to take flight if it starts from rest?

A car initially traveling at 60 km/h accelerates at a constant rate of 2.0 m/s2.How much time is required for the car to reach a speed of 90 km/h?

A car increases its forward velocity uniformly from 40 m/s to 80 m/s while traveling a distance of 200 m.What is its acceleration during this time?

An object is moving in a straight line with constant acceleration.Initially it is traveling at 16 m/s.Three seconds later it is traveling at 10 m/s.How far does it move during this time?

An auto accelerates forward from 7.0 m/s at a uniform 0.71 m/s2.It travels a distance of 1.033 km while accelerating. (a)How fast is the auto moving just as it is traveled the 1.033 km? (b)How many seconds did it take to travel the 1.033 km?

Starting from rest,a dragster travels a straight 1/4 mi racetrack in 6.70 s with constant acceleration.What is its velocity when it crosses the finish line?

A car starts from rest and accelerates uniformly at 3.0 m/s2 toward the north.A second car starts from rest 6.0 s later at the same point and accelerates uniformly at 5.0 m/s2 toward the north.How long after the second car starts does it overtake the first car?

A car starting from rest accelerates at a constant 2.0 m/s2 for 10 s.It then travels with constant speed it has achieved for another 10 s.Then it finally slows to a stop with constant acceleration of magnitude 2.0 m/s2.How far does it travel after starting?

A car travels at 15 m/s for 10 s.It then speeds up with a constant acceleration of 2.0 m/s2 for 15 s.At the end of this time,what is its velocity?

A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s.It next maintains the velocity it has reached for 10 s.Then it slows down at a steady rate of 2.0 m/s2 for 4.0 s.What is the final speed of the car?

A certain test car can go from rest to 32.0 m/s in 3.88 s.The same car can come to a full stop from that speed in 4.14 s.What is the ratio of the magnitude of the starting acceleration to the stopping acceleration?

Assuming equal rates of uniform acceleration in both cases,how much further would you travel if braking from 56mi/h56 \mathrm { mi } / \mathrm { h }56mi/h to rest than from 28mi/h28 \mathrm { mi } / \mathrm { h }28mi/h ?

A car accelerates from 5.0 m/s5.0 \mathrm {~m} / \mathrm { s }5.0 m/s to 21 m/s21 \mathrm {~m} / \mathrm { s }21 m/s at a constant rate of 3.0 m/s23.0 \mathrm {~m} / \mathrm { s } ^ { 2 }3.0 m/s2 How far does it travel while accelerating?

An object starts from rest and undergoes uniform acceleration.During the first second it travels 5.0 m.How far will it travel during the third second?

In a ballistics test,a bullet moving horizontally with a speed of 500 m/s strikes a sandbag and penetrates a distance of 10.0 cm. (a)What is the magnitude of the average acceleration of the bullet in the sandbag? (b)How many milliseconds does it take the bullet to come to rest in the sandbag?

A bicyclist starts a timed race at $6.0 \mathrm { mi } / \mathrm { h } .$ In order to win,he must average $21 \mathrm { mi } / \mathrm { h } .$ Assuming constant acceleration from the start,how fast must he be traveling at the end of the race?

A car with good tires on a dry road can slow down at a steady rate of about 5.0 m/s² when braking.If a car is initially traveling at 55 mi/h (a)how much time does it take the car to stop? (b)what is its stopping distance?

A cart has an initial velocity of 5.0 m/s to the right and a constant acceleration of 2.0 m/s² to the right.What is the cart's displacement during the first 6.0 s of this motion?

An airplane needs to reach a forward velocity of $203.0 \mathrm {~km} / \mathrm { h }$ to take off.On a $2000 - m$ runway,what is the minimum uniform acceleration necessary for the plane to take flight if it starts from rest?

A car initially traveling at 60 km/h accelerates at a constant rate of 2.0 m/s².How much time is required for the car to reach a speed of 90 km/h?

An auto accelerates forward from 7.0 m/s at a uniform 0.71 m/s².It travels a distance of 1.033 km while accelerating. (a)How fast is the auto moving just as it is traveled the 1.033 km? (b)How many seconds did it take to travel the 1.033 km?

A car starts from rest and accelerates uniformly at 3.0 m/s² toward the north.A second car starts from rest 6.0 s later at the same point and accelerates uniformly at 5.0 m/s² toward the north.How long after the second car starts does it overtake the first car?

A car starting from rest accelerates at a constant 2.0 m/s² for 10 s.It then travels with constant speed it has achieved for another 10 s.Then it finally slows to a stop with constant acceleration of magnitude 2.0 m/s².How far does it travel after starting?

A car travels at 15 m/s for 10 s.It then speeds up with a constant acceleration of 2.0 m/s² for 15 s.At the end of this time,what is its velocity?

A cart starts from rest and accelerates uniformly at 4.0 m/s² for 5.0 s.It next maintains the velocity it has reached for 10 s.Then it slows down at a steady rate of 2.0 m/s² for 4.0 s.What is the final speed of the car?

Assuming equal rates of uniform acceleration in both cases,how much further would you travel if braking from $56 \mathrm { mi } / \mathrm { h }$ to rest than from $28 \mathrm { mi } / \mathrm { h }$ ?

A car accelerates from $5.0 \mathrm {~m} / \mathrm { s }$ to $21 \mathrm {~m} / \mathrm { s }$ at a constant rate of $3.0 \mathrm {~m} / \mathrm { s } ^ { 2 }$ How far does it travel while accelerating?