Unit 1: Kinematics
AP Physics C: Mechanics – Comprehensive Review
Calculus-Based
1. Motion with Calculus
Q1: A particle’s position is given by \( x(t) = 3t^2 – 12t + 4 \). At what time \( t \) is the particle’s velocity zero?
Find velocity by differentiating position: \( v(t) = \displaystyle\frac{dx}{dt} = 6t – 12 \).
Set \( v(t) = 0 \):
\( 6t – 12 = 0 \Rightarrow 6t = 12 \Rightarrow t = 2 \text{ s} \).
Correct Answer: B
Set \( v(t) = 0 \):
\( 6t – 12 = 0 \Rightarrow 6t = 12 \Rightarrow t = 2 \text{ s} \).
Correct Answer: B
Q2: An object accelerates from rest according to \( a(t) = 6t \). What is its position at \( t = 2 \text{ s} \) if \( x(0) = 0 \)?
Step 1: Find velocity \( v(t) = \int 6t \, dt = 3t^2 \).
Step 2: Find position \( x(t) = \int 3t^2 \, dt = t^3 \).
At \( t = 2 \): \( x(2) = 2^3 = 8 \text{ m} \).
Correct Answer: B
Step 2: Find position \( x(t) = \int 3t^2 \, dt = t^3 \).
At \( t = 2 \): \( x(2) = 2^3 = 8 \text{ m} \).
Correct Answer: B
Q3: A particle moves with velocity \( v(t) = e^{2t} \). What is the acceleration at \( t = 0 \)?
Acceleration is the derivative of velocity: \( a(t) = \displaystyle\frac{dv}{dt} = 2e^{2t} \).
At \( t = 0 \): \( a(0) = 2e^{0} = 2(1) = 2 \text{ m/s}^2 \).
Correct Answer: C
At \( t = 0 \): \( a(0) = 2e^{0} = 2(1) = 2 \text{ m/s}^2 \).
Correct Answer: C
Graphical
2. Graphical Analysis
Q1: The slope of a velocity-time graph represents:
By definition, the instantaneous rate of change of velocity is acceleration: \( a = \displaystyle\frac{dv}{dt} \). This is represented by the slope of the \( v-t \) graph.
Correct Answer: C
Correct Answer: C
Q2: If the area under a velocity-time graph is negative, the object has:
The integral of velocity with respect to time represents displacement: \( \Delta x = \int v \, dt \). A negative area indicates displacement in the negative direction.
Correct Answer: B
Correct Answer: B
Q3: A position-time graph is linear. This implies:
A linear \( x-t \) graph means the slope (velocity) is constant. If velocity is constant, its derivative (acceleration) is zero.
Correct Answer: C
Correct Answer: C
Multi-Dimensional
3. Motion in Two or Three Dimensions
Q1: A projectile is launched at speed \( v_0 \) at angle \( \theta \). If the initial speed is doubled to \( 2v_0 \), how does the maximum height change?
Maximum height formula: \( H = \displaystyle\frac{(v_0 \sin\theta)^2}{2g} \).
Since \( H \propto v_0^2 \), doubling the speed results in \( 2^2 = 4 \) times the height.
Correct Answer: D
Since \( H \propto v_0^2 \), doubling the speed results in \( 2^2 = 4 \) times the height.
Correct Answer: D
Q2: The position vector is \( \vec{r}(t) = (2t^2)\hat{i} + (5t)\hat{j} \). What is the magnitude of the velocity at \( t = 1 \text{ s} \)?
\( \vec{v}(t) = \displaystyle\frac{d\vec{r}}{dt} = (4t)\hat{i} + 5\hat{j} \).
At \( t = 1 \): \( \vec{v} = 4\hat{i} + 5\hat{j} \).
Magnitude \( |\vec{v}| = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \text{ m/s} \).
Correct Answer: C
At \( t = 1 \): \( \vec{v} = 4\hat{i} + 5\hat{j} \).
Magnitude \( |\vec{v}| = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \text{ m/s} \).
Correct Answer: C
Q3: A plane flies North at 200 m/s relative to air. Wind blows East at 50 m/s. The speed relative to ground is:
Using vector addition: \( \vec{v}_{ground} = \vec{v}_{plane/air} + \vec{v}_{air/ground} \).
Because North and East are perpendicular: \( v = \sqrt{200^2 + 50^2} \).
Correct Answer: C
Because North and East are perpendicular: \( v = \sqrt{200^2 + 50^2} \).
Correct Answer: C
Q4: At the peak of a projectile trajectory, which is true?
At the peak, vertical velocity \( v_y = 0 \), leaving only horizontal velocity \( v_x \). Acceleration \( g \) is strictly vertical. Thus, they are perpendicular.
Correct Answer: C
Correct Answer: C
Q5: A swimmer heads directly across a river (1.5 m/s flow) at 2 m/s relative to water. Resultant speed is:
The velocities are perpendicular: \( v = \sqrt{2^2 + 1.5^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5 \text{ m/s} \).
Correct Answer: B
Correct Answer: B
Q6: Acceleration is \( \vec{a} = 2\hat{i} + 6t\hat{j} \). If it starts from rest, the y-position at \( t=1 \) is:
Y-component only: \( a_y = 6t \).
\( v_y = \int 6t \, dt = 3t^2 \).
\( y = \int 3t^2 \, dt = t^3 \).
At \( t = 1 \), \( y = 1^3 = 1 \text{ m} \).
Correct Answer: A
\( v_y = \int 6t \, dt = 3t^2 \).
\( y = \int 3t^2 \, dt = t^3 \).
At \( t = 1 \), \( y = 1^3 = 1 \text{ m} \).
Correct Answer: A
Kinematics Formulas Recap
Velocity\( \vec{v} = \displaystyle\frac{d\vec{r}}{dt} \)
Acceleration\( \vec{a} = \displaystyle\frac{d\vec{v}}{dt} \)
X-Projectile\( x = v_{0x}t \)
Y-Projectile\( y = v_{0y}t – \frac{1}{2}gt^2 \)
Relative\( \vec{v}_{AB} = \vec{v}_A – \vec{v}_B \)
Magnitude\( |\vec{v}| = \sqrt{v_x^2 + v_y^2} \)
