AP Physics C Mechanics: Unit 1 Practice Test — Kinematics

AP Physics C: Mechanics practice test 1.

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Question 1

John starts running from $x = 0\ m$. His speed as a function of time $t$ is given by $v(t) = \dfrac{1}{20} \ t(16-t)\ m/s$. For which of the below time instants is John’s speed a maximum?

A
$9 s$
B
$4 s$
C
$8 s$
D
$13 s$
E
$16 s$
Question 1 Explanation: 
Substitute the time instants in the formula for each option,

Option A $→ v(t=9) = \dfrac{1}{20} \ 9(16-9) = 3.15\ m/s$

Option B $→ v(t=4)= \dfrac{1}{20} \ 4(16-4) = 2.40\ m/s$

Option C $→ v(t=8)= \dfrac{1}{20} \ 8(16-8) = 3.20\ m/s$

Option D $→ v(t=13)= \dfrac{1}{20} \ 13(16-13) = 1.95\ m/s$

Option E $→ v(t=16)= \dfrac{1}{20} \ 16(16-16) = 0.00\ m/s$

Maximum speed is when $t=8\ s$ (Option C)
Question 2

Which position-time graph (in 1D) shows an average speed of $5\ m/s$ over the time interval when the object is in motion?

A
B
C
D
E
Question 2 Explanation: 
Average Speed is equal to the magnitude of Average Velocity

And, Average Velocity $=\dfrac{r(t_2 ) - r(t_1 )}{t_2 - t_1 }$

Check each option,

Option A $→$ Average Velocity $= \dfrac{4 - 0}{4 - 0} = 1.00\ m/s$

Option B $→$ Average Velocity $= \dfrac{4 - 0}{3 - 0} = 1.33\ m/s$

Option C $→$ Average Velocity $= \dfrac{1 - 4}{1 - 0} = -3.00\ m/s$

Option D $→$ Average Velocity $= \dfrac{4 - 0}{2 - 0} = 2.00\ m/s$

Option E $→$ Average Velocity $= \dfrac{0 - 5}{2 - 1} = -5.00\ m/s$

Thus, option E shows an average speed of $5\ m/s$

(Speed is magnitude of velocity)
Question 3

Lilly’s position vector $\overrightarrow{r}$ is a function of time $t$ given by the relation,

$\overrightarrow{r} (t)=2t \skew{2.5}\hat{i} + (5-t^2 ) \skew{4}\hat{j} m$

What is Lilly’s average velocity as she moves from $t = 2 \ s$ to $t = 4 \ s$?

A
$ 2 \skew{2.3}\hat{i} − 6 \skew{4.5}\hat{j} \ m/s$
B
$2 \skew{2.3}\hat{i}+3 \skew{4.5}\hat{j} \ m/s$
C
$4 \skew{2.3}\hat{i} −12 \skew{4.5}\hat{j} \ m/s$
D
$6 \skew{2.3}\hat{i} +3 \skew{4.5}\hat{j} \ m/s$
E
$10\skew{2.3} \hat{i} − 4 \skew{4.5}\hat{j} \ m/s$
Question 3 Explanation: 
$\overrightarrow{r} (t)=2t \skew{2}\hat{i}+(5-t^2 ) \skew{5}\hat{j} $

So, $\overrightarrow{r} (t=2)=4 \skew{2}\hat{i}+ \skew{5}\hat{j} \ and \ \overrightarrow{r} (t=4)=8 \skew{2}\hat{i} -11 \skew{5}\hat{j}$

Displacement$ (\Delta \overrightarrow{r} )=\overrightarrow{r} (t=4)-\overrightarrow{r} (t=2)$
                                  $=(8 \skew{2}\hat{i} -11 \skew{5}\hat{j} )-(4 \skew{2}\hat{i} + \skew{5}\hat{j} )$
                                  $=4 \skew{2}\hat{i} -12 \skew{4}\hat{j}$

Time interval $(\Delta t)=4-2=2$

Average velocity $=\dfrac{1}{2} (4 \skew{2}\hat{i} -12 \skew{5}\hat{j})=2 \skew{2}\hat{i} -6 \skew{5}\hat{j}$
Question 4

An object of mass $2\ kg$, starting from rest, accelerates uniformly for $5\ s$. If it covered a distance of $40\ m$ during this time interval, what distance would it cover in the same time if its acceleration was $4$ times larger than the initial acceleration?

A
$160 \ m$
B
$80 \ m$
C
$120 \ m$
D
$100 \ m$
E
$200 \ m$
Question 4 Explanation: 
$s=ut+\dfrac{1}{2} \ at^2 $

For an object starting from rest, $u = 0$

$→s∝a$

Thus, when $a'=4a$ then $s'= 4s = 4 × 40 = 160\ m$
Question 5

The graph of the velocity $v$, as a function of time $t$, is shown below.

Which option would show the correct relation for the acceleration $a(t)$?
(Hint: The graph is a parabola of the form $y = b(x-a)^2$)

A
$a(t) = \dfrac{4}{3} t$
B
$a(t) = \dfrac{2}{3} (t-3)^2$
C
$a(t) = \dfrac{4}{3} (t-3)$
D
$a(t) = \dfrac{2}{3} t^2$
E
$a(t) = \dfrac{4}{3} t- \dfrac{2}{3}$
Question 5 Explanation: 
Using the hint: $v(t)=\dfrac{2}{3} (t-3)^2$

Since, $a(t) = \dfrac{dv}{dt}$

$a(t) =\dfrac{2}{3} \dfrac{d}{dt} (t-3)^2$

        $ = \dfrac{2}{3} \ 2(t-3) $

        $ = \dfrac{4}{3} (t-3) $
Question 6

A projectile is launched at an angle of $20°$ with a speed of $10\ m/s$. Another projectile is launched at an angle of $40°$ with the same speed. The difference between the range of the two projectiles is,

(Use $g=10\ m/s^2$)

A
$8.62\ m$
B
$25.4\ m$
C
$34.0\ m$
D
$3.42\ m$
E
$10.0\ m$
Question 6 Explanation: 
Range $=\dfrac{v_o^2}{g} \ \sin2θ$

For $θ = 20°$, Range $= 10 \sin⁡ 40°$

For $θ = 40°$, Range $= 10 \sin ⁡80°$

Difference $= 10 \ (sin⁡ 80°-\sin40° )$

                 $= 10 \ (0.342) $

                 $= 3.42\ m $
Question 7

The velocity of particle $A$ is given by $\overrightarrow{v}_A (t)=(t-2) \skew{2.5}\hat{i} +5 \skew{4}\hat{j},$ and the velocity of particle $B$ is given by $\overrightarrow{v}_B (t)=6 \skew{2.5}\hat{i} + (4-t) \skew{4}\hat{j}.$ The velocity of $B$, as seen by $A$, is given by,

A
$\overrightarrow{v}_{BA} (t) = (8-t) \skew{2.3}\hat{i} -(t+1) \skew{4.5}\hat{j}$
B
$\overrightarrow{v}_{BA} (t) = (4-t) \skew{2.3}\hat{i} +(t+1) \skew{4.5}\hat{j}$
C
$\overrightarrow{v}_{BA} (t) = (8+t) \skew{2.3}\hat{i} -(t+1) \skew{4.5}\hat{j}$
D
$\overrightarrow{v}_{BA} (t) = (8-t) \skew{2.3}\hat{i} +(t+3) \skew{4.5}\hat{j}$
E
$\overrightarrow{v}_{BA} (t) = (t-8) \skew{2.3}\hat{i} -(t+1) \skew{4.5}\hat{j}$
Question 7 Explanation: 
$\overrightarrow{v}_{BA} (t)=\overrightarrow{v}_{B} $(t)-$\overrightarrow{v}_A (t) $

               $= 6 \skew{2}\hat{i}+(4-t) \skew{5}\hat{j} -(t-2) \skew{2}\hat{i} -5 \skew{5}\hat{j} $

               $= (8-t) \skew{2}\hat{i} -(1+t) \skew{5}\hat{j} $
Question 8

Which of the below statements is TRUE for the position-time graphs of two particles, $A$ and $B$?

A
The average velocity of $A$ is less than $B$
B
The acceleration of $A$ is always $0$
C
The instantaneous velocity of $B$ at $t = 3\ s$ is more than $5\ m/s$
D
The velocity of $A$ and $B$ becomes zero once during their journey
E
The acceleration of $B$ never becomes negative
Question 8 Explanation: 
Check each option,

Option A $→$ False, because displacement of $A$ is more than $B$ and the time interval of $A$ is less than $B$

Option B $→$ False, because velocity of $A$ keeps changing

Option C $→$ False, because velocity of $B$ at $t=3 \ s$ is about $\dfrac{2-0}{4-2}=1 \ m/s$

Option D $→$ True, because velocity of $A$ is $0 \ m/s$ in the interval $(3,4)$ while $B$ is $0 \ m/s$ in the interval $(4,6)$

Option E $→$ False, because velocity of $B$ decreases after $t=4 \ s$
Question 9

The velocity $v(t)$ of a particle when passing through a medium is given by the formula,

$v(t) = \dfrac{mg}{γ} (1-e^{\dfrac{-γt}{m}} )$

Where $g$ is the acceleration due to gravity, $γ$ is a constant, $m$ is the mass of the particle, and $t$ is the time. What is the maximum velocity that the particle will reach when passing through the medium?

A
$\dfrac{2mg}{γ}$
B
$\dfrac{mg}{γ}$
C
$\dfrac{mg}{2}γ$
D
$\dfrac{mg}{3γ}$
E
$\dfrac{mg}{4γ}$
Question 9 Explanation: 
For maximum velocity, $e^{\dfrac{-γt}{m}}$ should be as small as possible

Which is true when $t$ is large enough

For large $t, v(t) = \dfrac{mg}{γ} (1-0) = \dfrac{mg}{γ}$
Question 10

A particle moving in 1D has velocity $v(t) = 6t + 3t^2 \ m/s$. What is the magnitude of displacement $\Delta r$ of the particle as it moves from $t = 1 \ s$ to $t=4 \ s$?

A
$92 \ m$
B
$85 \ m$
C
$120 \ m$
D
$72 \ m$
E
$108 \ m$
Question 10 Explanation: 
$\Delta r=\int_1^4 v(t) \ dt $

      $=\int_1^4 (6t+3t^2 ) \ dt$

      $=3t^2+t^3 |_1^4$

      $=3(16-1)+(64-1)$

      $=45+63$

      $=108 m$
Question 11

Questions 11 and 12 are based on the below information:

A projectile is launched, as shown in the diagram below.


What will be the velocity of the projectile after $t = 2 \ s$?

(Use $g = 10 \ m/s^2 $)

A
$-61.3 \skew{2.3}\hat{i}+51.4 \skew{4.5}\hat{j} \ m/s$
B
$61.3 \skew{2.3}\hat{i} +51.4 \skew{4.5}\hat{j} \ m/s$
C
$-63.1 \skew{2.3}\hat{i} +31.4 \skew{4.5}\hat{j} \ m/s$
D
$63.1 \skew{2.3}\hat{i} +31.4 \skew{4.5}\hat{j} \ m/s$
E
$61.3 \skew{2.3}\hat{i} +31.4 \skew{4.5}\hat{j} \ m/s$
Question 11 Explanation: 
The horizontal $(x)$ component of the velocity is unchanged and is given by,
$v_x = 80 \cos ⁡40° = 61.3 \ m/s $

The vertical $(y)$ component of the velocity changes as,
$v_y = v_{yo}-gt$

Where $v_{yo} = 80 \sin ⁡40° = 51.4 \ m/s$
$v_y = 51.4-10×2 = 31.4 \ m/s$

Thus, the velocity vector is equal to,
$61.3 \skew{2}\hat{i}+31.4 \skew{5}\hat{j} \ m/s $
Question 12

A projectile is launched, as shown in the diagram below.


What is the displacement of the projectile after $t = 4 \ s$?

A
$252.0 \skew{2.3}\hat{i} +120.2 \skew{4.5}\hat{j} \ m$
B
$265.3 \skew{2.3}\hat{i} +133.9 \skew{4.5}\hat{j} \ m$
C
$245.2 \skew{2.3}\hat{i} +125.6 \skew{4.5}\hat{j} \ m$
D
$245.5 \skew{2.3}\hat{i} +114.4 \skew{4.5}\hat{j} \ m$
E
$230.4 \skew{2.3}\hat{i} +125.6 \skew{4.5}\hat{j} \ m$
Question 12 Explanation: 
$\Delta r_x=v_x×\Delta t $

$→\Delta r_x=61.3×4=245.2 \ m$

$\Delta r_y=v_yo \Delta t- \dfrac{1}{2} g\Delta t^2$

$→\Delta r_y=51.4×4-\dfrac{1}{2}×10×4^2=125.6 \ m$

Thus, the displacement vector is equal to,

$245.2 \skew{2}\hat{i} +125.6 \skew{5}\hat{j} \ m$
Question 13

A block starts decelerating at $3 \ m/s^2$ from time $ t = 0 \ s$, and covers $30 \ m$ before it stops. Which option is the closest to the average speed of the block?

A
$12.80 \ m/s$
B
$13.41 \ m/s$
C
$8.90 \ m/s$
D
$6.70 \ m/s$
E
$11.05 \ m/s$
Question 13 Explanation: 
Using $a=-3 m/s^2, s=30 m, v=0 m/s $ in the equation

$v^2= u^2+2as$ gives,

$0 = u^2-2×3×30$

$u^2 = 180→u=3 \sqrt{20} \ m/s$

$t = \dfrac{v - u}{a}$

$= \dfrac{0 - 3 \sqrt{20}}{- 3}$

$= \sqrt{20} \ s$

Average speed $ = \dfrac{Distance}{Time}$

                        $= \dfrac{30}{ \sqrt{20}}$

                        $= 6.70 m/s$

Question 14

An object is released from rest inside a medium, as shown below.

It experiences a downward acceleration $a(t) = 6at$ while moving through the medium, where $a$ is a constant. Find the time interval $\Delta t$ required for the object to fall through height $h$ in the medium.

A
$\left(\dfrac{2h}{a}\right)^{1/3}$
B
$\left(\dfrac{h}{a}\right)^{2/3}$
C
$\left(\dfrac{a}{h}\right)^{1/3}$
D
$\left(\dfrac{a}{h}\right)^{3/2}$
E
$\left(\dfrac{h}{a}\right)^{1/3}$
Question 14 Explanation: 
$v(t)-v(0)= ∫_0^t 6at' \ dt' = 3at^2$

Since, $v(0)=0→v(t)=3at^2$

$r(\Delta t)-r(0)=\int_0^{\Delta t} 3at^2 \ dt=a(\Delta t)^3 $

Since, $r(0)=0 \ and \ r(\Delta t)=h →h=a\Delta t^3$

Inverting the relation gives,

$\Delta t=(\dfrac{h}{a})^{\dfrac{1}{3}}$
Question 15

Two boats, $I$ and $II$, are moving in a river as follows:

$I: \overrightarrow{r}(t)=8t \skew{2.5}\hat{i} +(7-2t) \skew{4}\hat{j} $

$II: \overrightarrow{r}(t)=4t^2 \skew{2.5}\hat{i} +2(t-t^2 ) \skew{4}\hat{j} $

Find the time instant $t$ when the two boats are traveling with the same velocity.

A
$t = 2$
B
$t = 1$
C
$t = 0.5$
D
$t = 3.2$
E
$t = 0$
Question 15 Explanation: 
For $I$: velocity $\overrightarrow{v}=8 \skew{2}\hat{i} -2 \skew{5}\hat{j} $

For $II$: velocity $\overrightarrow{v}=8t \skew{2}\hat{i} +2(1-2t) \skew{5}\hat{j} $

For equal velocities, $8t=8→t=1$

[Also, $2(1-2t)=-2→t=1$]

Which confirms that time at which the velocities of the two boats are equal is $t=1$
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