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?

$9 s$ | |

$4 s$ | |

$8 s$ | |

$13 s$ | |

$16 s$ |

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)

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?

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)

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 $t = 4 \ s$?

$ 2 \skew{2.3}\hat{i} − 6 \skew{4.5}\hat{j} \ m/s$ | |

$2 \skew{2.3}\hat{i}+3 \skew{4.5}\hat{j} \ m/s$ | |

$4 \skew{2.3}\hat{i} −12 \skew{4.5}\hat{j} \ m/s$ | |

$6 \skew{2.3}\hat{i} +3 \skew{4.5}\hat{j} \ m/s$ | |

$10\skew{2.3} \hat{i} − 4 \skew{4.5}\hat{j} \ m/s$ |

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

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?

$160 \ m$ | |

$80 \ m$ | |

$120 \ m$ | |

$100 \ m$ | |

$200 \ m$ |

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 $s'= 4s = 4 × 40 = 160\ m$

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 $y = b(x-a)^2$)

$a(t) = \dfrac{4}{3} t$ | |

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

$a(t) = \dfrac{4}{3} (t-3)$ | |

$a(t) = \dfrac{2}{3} t^2$ | |

$a(t) = \dfrac{4}{3} t- \dfrac{2}{3}$ |

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) $

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

$8.62\ m$ | |

$25.4\ m$ | |

$34.0\ m$ | |

$3.42\ m$ | |

$10.0\ m$ |

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 $

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,

$\overrightarrow{v}_{BA} (t) = (8-t) \skew{2.3}\hat{i} -(t+1) \skew{4.5}\hat{j}$ | |

$\overrightarrow{v}_{BA} (t) = (4-t) \skew{2.3}\hat{i} +(t+1) \skew{4.5}\hat{j}$ | |

$\overrightarrow{v}_{BA} (t) = (8+t) \skew{2.3}\hat{i} -(t+1) \skew{4.5}\hat{j}$ | |

$\overrightarrow{v}_{BA} (t) = (8-t) \skew{2.3}\hat{i} +(t+3) \skew{4.5}\hat{j}$ | |

$\overrightarrow{v}_{BA} (t) = (t-8) \skew{2.3}\hat{i} -(t+1) \skew{4.5}\hat{j}$ |

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} $

$= 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 $B$?

The average velocity of $A$ is less than $B$ | |

The acceleration of $A$ is always $0$ | |

The instantaneous velocity of $B$ at $t = 3\ s$ is more than $5\ m/s$ | |

The velocity of $A$ and $B$ becomes zero once during their journey | |

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 $t=4 \ s$

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?

$\dfrac{2mg}{γ}$ | |

$\dfrac{mg}{γ}$ | |

$\dfrac{mg}{2}γ$ | |

$\dfrac{mg}{3γ}$ | |

$\dfrac{mg}{4γ}$ |

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

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 $t=4 \ s$?

$92 \ m$ | |

$85 \ m$ | |

$120 \ m$ | |

$72 \ m$ | |

$108 \ m$ |

Question 10 Explanation:

$\Delta r=\int_1^4 v(t) \ dt $

$=\int_1^4 (6t+3t^2 ) \ dt$

$=3t^2+t^3 |_1^4$

$=3(16-1)+(64-1)$

$=45+63$

$=108 m$

$=\int_1^4 (6t+3t^2 ) \ dt$

$=3t^2+t^3 |_1^4$

$=3(16-1)+(64-1)$

$=45+63$

$=108 m$

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 $)

$-61.3 \skew{2.3}\hat{i}+51.4 \skew{4.5}\hat{j} \ m/s$ | |

$61.3 \skew{2.3}\hat{i} +51.4 \skew{4.5}\hat{j} \ m/s$ | |

$-63.1 \skew{2.3}\hat{i} +31.4 \skew{4.5}\hat{j} \ m/s$ | |

$63.1 \skew{2.3}\hat{i} +31.4 \skew{4.5}\hat{j} \ m/s$ | |

$61.3 \skew{2.3}\hat{i} +31.4 \skew{4.5}\hat{j} \ m/s$ |

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 $

$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 $t = 4 \ s$?

$252.0 \skew{2.3}\hat{i} +120.2 \skew{4.5}\hat{j} \ m$ | |

$265.3 \skew{2.3}\hat{i} +133.9 \skew{4.5}\hat{j} \ m$ | |

$245.2 \skew{2.3}\hat{i} +125.6 \skew{4.5}\hat{j} \ m$ | |

$245.5 \skew{2.3}\hat{i} +114.4 \skew{4.5}\hat{j} \ m$ | |

$230.4 \skew{2.3}\hat{i} +125.6 \skew{4.5}\hat{j} \ m$ |

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,

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

$→\Delta r_x=61.3×4=245.2 \ m$

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

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

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?

$12.80 \ m/s$ | |

$13.41 \ m/s$ | |

$8.90 \ m/s$ | |

$6.70 \ m/s$ | |

$11.05 \ m/s$ |

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}$

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

$= 6.70 m/s$

$v^2= u^2+2as$ 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.

$\left(\dfrac{2h}{a}\right)^{1/3}$ | |

$\left(\dfrac{h}{a}\right)^{2/3}$ | |

$\left(\dfrac{a}{h}\right)^{1/3}$ | |

$\left(\dfrac{a}{h}\right)^{3/2}$ | |

$\left(\dfrac{h}{a}\right)^{1/3}$ |

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,

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

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.

$t = 2$ | |

$t = 1$ | |

$t = 0.5$ | |

$t = 3.2$ | |

$t = 0$ |

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 $t=1$

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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