Tag: Kinematics

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? (A) \( 0 \text{ s} \) (B) \( 2 \text{ s} \) (C) \( 4 \text{ s} \) (D) \( 6 \text{ s} \) 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 Q2: An object accelerates from rest according to \( a(t) = 6t \). What is its position at \( t = 2 \text{ s} \) if \( x(0) = 0 \)? (A) \( 4 \text{ m} \) (B) \( 8 \text{ m} \) (C) \( 12 \text{ m} \) (D) \( 24 \text{ m} \) 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 Q3: A particle moves with velocity \( v(t) = e^{2t} \). What is the acceleration at \( t = 0 \)? (A) \( 0 \text{ m/s}^2 \) (B) \( 1 \text{ m/s}^2 \) (C) \( 2 \text{ m/s}^2 \) (D) \( 4 \text{ m/s}^2 \) 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 Graphical 2. Graphical Analysis Q1: The slope of a velocity-time graph represents: (A) Position (B) Displacement (C) Acceleration (D) Jerk 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 Q2: If the area under a velocity-time graph is negative, the object has: (A) Negative acceleration (B) Negative displacement (C) Negative position (D) A decreasing speed 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 Q3: A position-time graph is linear. This implies: (A) Acceleration is constant and non-zero (B) Velocity is changing (C) Acceleration is zero (D) The object is at rest A linear \( x-t \) graph means the slope (velocity) is constant. If velocity is constant, its derivative (acceleration) is zero. 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? (A) It doubles (B) It remains the same (C) It increases by a factor of \( \sqrt{2} \) (D) It increases by a factor of 4 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 Q2: The position vector is \( \vec{r}(t) = (2t^2)\hat{i} + (5t)\hat{j} \). What is the magnitude of the velocity at \( t = 1 \text{ s} \)? (A) \( 3 \text{ m/s} \) (B) \( \sqrt{29} \text{ m/s} \) (C) \( \sqrt{41} \text{ m/s} \) (D) \( 9 \text{ m/s} \) \( \vec{v}(t) = \displaystyle\frac{d\vec{r}}{dt} = (4t)\hat{i} + 5\hat{j} \). 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 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: (A) \( 150 \text{ m/s} \) (B) \( 250 \text{ m/s} \) (C) \( \sqrt{200^2 + 50^2} \text{ m/s} \) (D) \( 200 \text{ m/s} \) 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 Q4: At the peak of a projectile trajectory, which is true? (A) Velocity is zero (B) Acceleration is zero (C) Velocity is perpendicular to acceleration (D) Velocity is parallel to acceleration 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 Q5: A swimmer heads directly across a river (1.5 m/s flow) at 2 m/s relative to water. Resultant speed is: (A) \( 0.5 \text{ m/s} \) (B) \( 2.5 \text{ m/s} \) (C) \( 3.5 \text{ m/s} \) (D) \( 2.0 \text{ m/s} \) 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 Q6: Acceleration is \( \vec{a} = 2\hat{i} + 6t\hat{j} \). If it starts from rest, the y-position at \( t=1 \) is: (A) \( 1 \text{ m} \) (B) \( 2 \text{ m} \) (C) \( 3 \text{ m} \) (D) \( 6 \text{ m} \) 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 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\( |\vec{v}| = \sqrt{v_x^2 + v_y^2} \)