Question 1

The velocity of an object is given as a function of time by v = 4t - 3t², where v is in m/s and t is in seconds.Its average velocity over the interval from t = 0 s to t = 2 s: A)is 0 B)is -2 m/s C)is 2 m/s D)is -4 m/s E)cannot be calculated unless the initial position is given

Accepted Answer

The answer of The velocity of an object is given...

Question 2

How far does a car travel in 6 s if its initial velocity is 2 m/s and its acceleration is 2 m/s² in the same direction as its initial velocity? A)12 m B)14 m C)24 m D)36 m E)48 m

Accepted Answer

The answer of How far does a car travel in...

Question 3

Consider the following five graphs (note the axes carefully).Which of these represent(s)motion at constant speed? 
A)IV only
B)IV and V only
C)I, II, and III only
D)I and II only
E)I and IV only

Accepted Answer

The answer of Consider the following five graphs (note the...

Question 4

A car starts from rest and goes down a slope with a constant acceleration of 5 m/s².After 5 seconds the car reaches the bottom of the hill.What is its speed at the bottom of the hill? A)1 m/s B)12.5 m/s C)25 m/s D)50 m/s E)160 m/s

Accepted Answer

The speed of the car can be calculated using the formula v = u + at, where u is the initial velocity (0 m/s), a is the acceleration (5 m/s²), and t is the time (5 s). Thus, v = 0 + (5 m/s²)(5 s) = 25 m/s.

Question 5

An object starts from rest and has a constant acceleration of 3 m/s².The position versus time graph for this object has a slope: A)that increases with time B)that is constant C)that decreases with time D)of 3 m/s E)of 3 m/s²

Accepted Answer

The slope of the position versus time graph for an object with constant acceleration is equal to the velocity of the object. Since the object has a constant acceleration of 3 m/s^2², its velocity will increase linearly over time. Therefore, the slope of the graph will increase with time.

Question 6

An object starts from rest at the origin and moves along the x axis with a constant acceleration of 4 m/s².Its average velocity as it goes from x = 2 m to x = 8 m is: A)1 m/s B)2 m/s C)3 m/s D)5 m/s E)6 m/s

Accepted Answer

The answer of An object starts from rest at the...

Question 7

The area under a velocity-time graph represents:
A)acceleration
B)change in acceleration
C)speed
D)change in velocity
E)displacement

Accepted Answer

The area under a velocity-time graph represents displacement. This is because the area is calculated by multiplying the time and the velocity, which gives us the distance covered by the object during that time period. Therefore, the correct choice is E) displacement.

Question 8

An object with an initial velocity of 12 m/s west experiences a constant acceleration of 4 m/s² west for 3 seconds.During this time the object travels a distance of: A)18 m B)24 m C)36 m D)54 m E)144 m

Accepted Answer

The object's final velocity can be calculated using the formula $v = u + at$, where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. Substituting the given values, we get $v = 12\,m/s + (4\,m/s^2 \times 3\,s) = 24\,m/s$. The distance traveled can be calculated using the formula $s = ut + \frac{1}{2}at^2$, where $s$ is the distance. Substituting the given values, we get $s = (12\,m/s \times 3\,s) + \frac{1}{2}(4\,m/s^2 \times 9\,s^2) = 36\,m + 18\,m = 54\,m$.

Question 9

A racing car traveling with constant acceleration increases its speed from 10 m/s to 30 m/s over a distance of 60 m? How long does this take?
A)2.0 s
B)3.0 s
C)5.0 s
D)6.0 s
E)The time cannot be calculated since the speed is not constant

Accepted Answer

We can use the formula: 
v^2 = u^2 + 2as 
where: 
v = final velocity = 30 m/s 
u = initial velocity = 10 m/s 
a = acceleration 
s = distance = 60 m

Rearranging the formula: 
a = (v^2 - u^2)/2s

Substituting the values: 
a = (30^2 - 10^2)/(2*60) = 20 m/s^2

We can use another formula to find the time taken: 
t = (v - u)/a

Substituting the values: 
t = (30 - 10)/20 = 1 s

Therefore, the time taken is 1 second. However, it is important to note that the acceleration was constant during the entire duration, hence we were able to use the above formulas.

Question 10

A ball is in free fall motion.Upward is taken to be the positive direction.The displacement of the ball is:
A)positive during both ascent and descent
B)negative during both ascent and descent
C)negative during ascent and positive during descent
D)positive during ascent and negative during descent
E)none of the above

Accepted Answer

During ascent, the ball is moving in the negative direction (downward) while its displacement is in the positive direction (upward), so its displacement is positive. During descent, the ball is moving in the positive direction (upward) while its displacement is in the negative direction (downward), so its displacement is negative. Therefore, the displacement of the ball is positive during ascent and negative during descent.

Question 11

The acceleration of an object, starting from rest, is shown in the graph below.Other than at t = 0, when is the velocity of the object equal to zero? 
A)During the interval from 1.0 s to 3.0 s
B)At t = 3.5 s
C)At t = 4.0 s
D)At t = 5.0 s
E)At no other time less than or equal to 5 s.

Accepted Answer

The answer of The acceleration of an object, starting from...

Question 12

Displacement can be obtained from:
A)The slope of an acceleration-time graph
B)the slope of a velocity-time graph
C)the area under an acceleration-time graph
D)the area under a velocity-time graph
E)the slope of an acceleration-time graph

Accepted Answer

The area under a velocity-time graph represents displacement. This is because velocity is the rate of change of displacement with time, and therefore the area under the velocity-time graph represents the amount of displacement over a given time interval. The other options do not directly give information about displacement. While acceleration is related to changes in velocity which in turn affect displacement, neither the slope nor the area under an acceleration-time graph gives direct information about displacement.

Question 13

At time t = 0 a car has a velocity of 16 m/s.It slows down with an acceleration given by a = -0.50t, in m/s² for t in seconds.At the end of 4.0 s it has traveled: A)0 B)12 m C)14 m D)25 m E)59 m

Accepted Answer

The velocity function is v(t) = 16 - 0.25t^2. Integrating this from 0 to 4 gives the displacement: ∫(16 - 0.25t^2) dt from 0 to 4 = [16t - (0.25/3)t^3] from 0 to 4 = 59 m.

Question 14

A drag racing car starts from rest at t = 0 and moves along a straight line with velocity given by v = bt², where b is a constant.The expression for the distance traveled by this car from its position at t = 0 is: A)bt³ B)bt³/3 C)4bt² D)3bt² E)bt^3/2

Accepted Answer

The distance traveled by the car can be found by integrating the velocity function with respect to time. Given $v = bt^2$, the integral of $v$ with respect to $t$ gives the displacement function $s(t) = \frac{b}{3}t^3 + C$. Since the car starts from rest at $t = 0$, the constant $C = 0$, resulting in $s(t) = \frac{b}{3}t^3$, which matches option B.

Question 15

A car, initially at rest, travels 20 m in 4 s along a straight line with constant acceleration.The acceleration of the car is: A)1.3 m/s² B)2.5 m/s² C)4.9 m/s² D)9.8 m/s² E)There is not enough information to answer this question.

Accepted Answer

The acceleration can be found using the formula $s = ut + \frac{1}{2}at^2$, where $s$ is the distance traveled, $u$ is the initial velocity (0 m/s in this case), $a$ is the acceleration, and $t$ is the time. Plugging in the given values, $20 = 0*4 + \frac{1}{2}a*4^2$, solving for $a$ gives $a = 2.5 \, m/s^2$.

Question 16

At a stop light, a truck traveling at 15 m/s passes a car as it starts from rest.The truck travels at constant velocity and the car accelerates at 3 m/s².How much time does the car take to catch up to the truck? A)5 s B)10 s C)15 s D)20 s E)25 s

Accepted Answer

First, we need to find the distance the truck travels during the time taken by the car to catch up to it.
Let t be the time taken by the car to catch up to the truck.
Distance traveled by the truck = speed × time = 15t
Distance traveled by the car = (1/2)acceleration × time² = (1/2)(3)t² = (3/2)t²
The car catches up to the truck when the distances traveled by both are equal.
Therefore, 15t = (3/2)t²
Simplifying the equation, we get t = 10 seconds.
Therefore, the car takes 10 seconds to catch up to the truck. 
The answer is B, 10 s.

Question 17

At time t = 0 a car has a velocity of 16 m/s.It slows down with an acceleration given by a = -0.50t, in m/s² for t in seconds.It stops at t = A)64 s B)32 s C)16 s D)8.0 s E)4.0 s

Accepted Answer

The car stops when its velocity reaches 0. Given the initial velocity $v_0 = 16 \, \text{m/s}$ and the acceleration $a = -0.50t$, we can use the equation $v = v_0 + at$ to find when the velocity is 0. Setting $v = 0$, we get $0 = 16 - 0.50t^2$. Solving for $t$, we find $t = 8.0 \, \text{s}$.

Question 18

The graph represents the straight line motion of a car.How far does the car travel between t = 2 seconds and t = 5 seconds? 
A)4 m
B)12 m
C)24 m
D)36 m
E)60 m

Accepted Answer

The answer of The graph represents the straight line motion...

Question 19

A car moving with an initial velocity of 25 m/s north has a constant acceleration of 3 m/s² south.After 6 seconds its velocity will be: A)7 m/s north B)7 m/s south C)43 m/s north D)20 m/s north E)20 m/s south

Accepted Answer

The car's final velocity can be calculated using the formula: final velocity = initial velocity + (acceleration * time). Here, acceleration is in the opposite direction to the initial velocity, so it's effectively a deceleration. The calculation is: 25 m/s - (3 m/s^2 * 6 s) = 25 m/s - 18 m/s = 7 m/s north.

Question 20

At time t = 0 a car has a velocity of 16 m/s.It slows down with an acceleration given by a = -0.50t, in m/s² for t in seconds.By the time it stops it has traveled: A)15 m B)31 m C)62 m D)85 m E)100 m

Accepted Answer

The velocity of the car can be represented as the integral of the acceleration with respect to time:

v(t) = -0.25t^2 + C

We know that at t = 0, v(0) = 16, so we can solve for C:

16 = -0.25 (0) ^ 2 + C

C = 16

So the velocity of the car is:

v(t) = -0.25t^2 + 16

To find the distance traveled, we can integrate the velocity with respect to time:

s(t) = -0.083t^3 + 16t + D

We don't know what D is, but we do know that when the car stops, its velocity is 0. So:

0 = -0.25t^2 + 16

t = 8

Plugging this into our equation for s(t):

s(8) = -0.083(8)^3 + 16(8) + D

s(8) = 85 + D

So the car has traveled 85 meters when it stops. Therefore, the answer is (D).

Quiz 2: Motion Along a Straight Line

The velocity of an object is given as a function of time by v = 4t - 3t², where v is in m/s and t is in seconds.Its average velocity over the interval from t = 0 s to t = 2 s:

How far does a car travel in 6 s if its initial velocity is 2 m/s and its acceleration is 2 m/s² in the same direction as its initial velocity?

Consider the following five graphs (note the axes carefully) .Which of these represent(s) motion at constant speed?

A car starts from rest and goes down a slope with a constant acceleration of 5 m/s².After 5 seconds the car reaches the bottom of the hill.What is its speed at the bottom of the hill?

An object starts from rest and has a constant acceleration of 3 m/s².The position versus time graph for this object has a slope:

An object starts from rest at the origin and moves along the x axis with a constant acceleration of 4 m/s².Its average velocity as it goes from x = 2 m to x = 8 m is:

The area under a velocity-time graph represents:

An object with an initial velocity of 12 m/s west experiences a constant acceleration of 4 m/s² west for 3 seconds.During this time the object travels a distance of:

A racing car traveling with constant acceleration increases its speed from 10 m/s to 30 m/s over a distance of 60 m? How long does this take?

A ball is in free fall motion.Upward is taken to be the positive direction.The displacement of the ball is:

The acceleration of an object, starting from rest, is shown in the graph below.Other than at t = 0, when is the velocity of the object equal to zero?

Displacement can be obtained from:

At time t = 0 a car has a velocity of 16 m/s.It slows down with an acceleration given by a = -0.50t, in m/s² for t in seconds.At the end of 4.0 s it has traveled:

A drag racing car starts from rest at t = 0 and moves along a straight line with velocity given by v = bt², where b is a constant.The expression for the distance traveled by this car from its position at t = 0 is:

A car, initially at rest, travels 20 m in 4 s along a straight line with constant acceleration.The acceleration of the car is:

At a stop light, a truck traveling at 15 m/s passes a car as it starts from rest.The truck travels at constant velocity and the car accelerates at 3 m/s².How much time does the car take to catch up to the truck?

At time t = 0 a car has a velocity of 16 m/s.It slows down with an acceleration given by a = -0.50t, in m/s² for t in seconds.It stops at t =

The graph represents the straight line motion of a car.How far does the car travel between t = 2 seconds and t = 5 seconds?

A car moving with an initial velocity of 25 m/s north has a constant acceleration of 3 m/s² south.After 6 seconds its velocity will be:

At time t = 0 a car has a velocity of 16 m/s.It slows down with an acceleration given by a = -0.50t, in m/s² for t in seconds.By the time it stops it has traveled:

Quiz 2: Motion Along a Straight Line

The velocity of an object is given as a function of time by v = 4t - 3t2, where v is in m/s and t is in seconds.Its average velocity over the interval from t = 0 s to t = 2 s:

How far does a car travel in 6 s if its initial velocity is 2 m/s and its acceleration is 2 m/s2 in the same direction as its initial velocity?

Consider the following five graphs (note the axes carefully) .Which of these represent(s) motion at constant speed?

A car starts from rest and goes down a slope with a constant acceleration of 5 m/s2.After 5 seconds the car reaches the bottom of the hill.What is its speed at the bottom of the hill?

An object starts from rest and has a constant acceleration of 3 m/s2.The position versus time graph for this object has a slope:

An object starts from rest at the origin and moves along the x axis with a constant acceleration of 4 m/s2.Its average velocity as it goes from x = 2 m to x = 8 m is: