Practice Questions: Calculus and Motion

$$\begin{align*} \text{Velocity is the derivative of position:} \\ v(t) &= s'(t) \\ v(t) &= 6t - 2 \\ \\ \text{At } t = 2: \\ v(2) &= 6(2) - 2 \\ &= 12 - 2 \\ &= 10 \text{ m/s} \end{align*}$$

Answer: The velocity at $t = 2$ seconds is $10$ m/s.

$$\begin{align*} \text{a) Velocity function:} \\ v(t) &= s'(t) \\ &= 3t^2 - 12t + 9 \\ \\ \text{b) Object is at rest when } v(t) = 0: \\ 3t^2 - 12t + 9 &= 0 \\ 3(t^2 - 4t + 3) &= 0 \\ 3(t - 1)(t - 3) &= 0 \\ t = 1 \text{ or } &t = 3 \end{align*}$$

Answers:

• a) $v(t) = 3t^2 - 12t + 9$ m/s
• b) The object is at rest at $t = 1$ second and $t = 3$ seconds

$$\begin{align*} \text{a) Find velocity:} \\ v(t) &= h'(t) = -10t + 20 \\ v(1) &= -10(1) + 20 = 10 \text{ m/s} \\ \\ \text{b) Maximum height when } v(t) = 0: \\ -10t + 20 &= 0 \\ t &= 2 \text{ seconds} \\ \\ \text{c) Maximum height:} \\ h(2) &= -5(2)^2 + 20(2) + 2 \\ &= -20 + 40 + 2 \\ &= 22 \text{ meters} \end{align*}$$

Answers:

• a) Velocity at $t = 1$ is $10$ m/s (moving upward)
• b) Maximum height reached at $t = 2$ seconds
• c) Maximum height is $22$ meters

$$\begin{align*} \text{a) Acceleration is the derivative of velocity:} \\ s(t) &= 2t^3 - 15t^2 + 24t \\ v(t) &= s'(t) = 6t^2 - 30t + 24 \\ a(t) &= v'(t) = s''(t) = 12t - 30 \\ \\ \text{b) At } t = 3: \\ a(3) &= 12(3) - 30 \\ &= 36 - 30 \\ &= 6 \text{ m/s}^2 \end{align*}$$

Answers:

• a) $a(t) = 12t - 30$ m/s²
• b) Acceleration at $t = 3$ is $6$ m/s²

$$\begin{align*} \text{Car moves forward when } v(t) > 0: \\ v(t) &= s'(t) \\ &= 3t^2 - 18t + 24 \\ &= 3(t^2 - 6t + 8) \\ &= 3(t - 2)(t - 4) \\ \\ \text{Critical points: } t = 2 \text{ and } t = 4 \\ \\ \text{Test intervals:} \\ t < 2: \quad v(0) &= 24 > 0 \quad \text{(forward)} \\ 2 < t < 4: \quad v(3) &= 3(1)(-1) = -3 < 0 \quad \text{(backward)} \\ t > 4: \quad v(5) &= 3(3)(1) = 9 > 0 \quad \text{(forward)} \end{align*}$$

Answer: The car moves forward when $t < 2$ or $t > 4$ (i.e., $t \in [0; 2) \cup (4; \infty)$).

$$\begin{align*} \text{a) Velocity is zero when } v(t) = 0: \\ t^2 - 4t + 3 &= 0 \\ (t - 1)(t - 3) &= 0 \\ t = 1 \text{ or } &t = 3 \\ \\ \text{b) Acceleration:} \\ a(t) &= v'(t) \\ &= 2t - 4 \\ \\ a(1) &= 2(1) - 4 \\ &= -2 \text{ m/s}^2 \end{align*}$$

Answers:

• a) Velocity is zero at $t = 1$ second and $t = 3$ seconds
• b) Acceleration at $t = 1$ is $-2$ m/s² (decelerating)

$$\begin{align*} \text{a) Direction changes when velocity changes sign:} \\ v(t) &= s'(t) = -3t^2 + 24t - 36 \\ &= -3(t^2 - 8t + 12) \\ &= -3(t - 2)(t - 6) \\ \\ v(t) = 0 \text{ at } &t = 2 \text{ and } t = 6 \\ \\ \text{b) At } t = 1: \\ v(1) &= -3(1)^2 + 24(1) - 36 \\ &= -3 + 24 - 36 = -15 \text{ (negative)} \\ \\ a(t) &= v'(t) = -6t + 24 \\ a(1) &= -6(1) + 24 = 18 \text{ (positive)} \\ \\ \text{Since } v < 0 \text{ and } a > 0, \text{ they have opposite signs,} \\ \text{so the particle is slowing down.} \end{align*}$$

Answers:

• a) Particle changes direction at $t = 2$ seconds and $t = 6$ seconds
• b) At $t = 1$, the particle is slowing down (velocity and acceleration have opposite signs)

$$\begin{align*} \text{a) Velocity function:} \\ v(t) &= h'(t) = -10t + 100 \\ v(5) &= -10(5) + 100 = 50 \text{ m/s} \\ \\ \text{b) Rocket hits ground when } h(t) = 0: \\ -5t^2 + 100t &= 0 \\ -5t(t - 20) &= 0 \\ t = 0 \text{ or } &t = 20 \\ \text{(At } t = 0 \text{, it's at launch, so } t = 20\text{)} \\ \\ \text{c) Velocity at } t = 20: \\ v(20) &= -10(20) + 100 \\ &= -100 \text{ m/s} \end{align*}$$

Answers:

• a) Velocity after 5 seconds is $50$ m/s
• b) Rocket hits the ground at $t = 20$ seconds
• c) Velocity when hitting ground is $-100$ m/s (negative indicates downward motion)

$$\begin{align*} \text{Find velocity and acceleration:} \\ v(t) &= s'(t) = t^2 - 4t + 3 \\ a(t) &= v'(t) = 2t - 4 \\ \\ \text{a) When is acceleration zero?} \\ a(t) &= 0 \\ 2t - 4 &= 0 \\ t &= 2 \\ \\ \text{Velocity at } t = 2: \\ v(2) &= (2)^2 - 4(2) + 3 \\ &= 4 - 8 + 3 = -1 \text{ m/s} \\ \\ \text{b) Position at } t = 2: \\ s(2) &= \frac{(2)^3}{3} - 2(2)^2 + 3(2) + 5 \\ &= \frac{8}{3} - 8 + 6 + 5 \\ &= \frac{8}{3} + 3 \\ &= \frac{17}{3} \text{ meters} \approx 5.67 \text{ meters} \end{align*}$$

Answers:

• a) Velocity when acceleration is zero: $-1$ m/s
• b) Position when acceleration is zero: $\frac{17}{3}$ meters (or $5.67$ meters)

$$\begin{align*} \text{a) Find velocities:} \\ v_A(t) &= s_A'(t) = 2t - 4 \\ v_B(t) &= s_B'(t) = -2t + 6 \\ \\ \text{When do they have equal velocity?} \\ v_A(t) &= v_B(t) \\ 2t - 4 &= -2t + 6 \\ 4t &= 10 \\ t &= 2.5 \text{ seconds} \\ \\ \text{b) Positions at } t = 2.5: \\ s_A(2.5) &= (2.5)^2 - 4(2.5) + 3 \\ &= 6.25 - 10 + 3 \\ &= -0.75 \text{ meters} \\ \\ s_B(2.5) &= -(2.5)^2 + 6(2.5) - 5 \\ &= -6.25 + 15 - 5 \\ &= 3.75 \text{ meters} \end{align*}$$

Answers:

• a) Particles have the same velocity at $t = 2.5$ seconds
• b) At that time, particle A is at $-0.75$ m and particle B is at $3.75$ m

$$\begin{align*} \text{a) Maximum height when } v(t) = 0: \\ v(t) &= h'(t) = -10t + 15 \\ -10t + 15 &= 0 \\ t &= 1.5 \text{ seconds} \\ \\ h(1.5) &= -5(1.5)^2 + 15(1.5) + 50 \\ &= -11.25 + 22.5 + 50 \\ &= 61.25 \text{ meters} \\ \\ \text{b) Stone hits ground when } h(t) = 0: \\ -5t^2 + 15t + 50 &= 0 \\ t^2 - 3t - 10 &= 0 \\ (t - 5)(t + 2) &= 0 \\ t = 5 \text{ (reject } &t = -2\text{)} \\ \\ \text{c) Velocity at } t = 5: \\ v(5) &= -10(5) + 15 \\ &= -35 \text{ m/s} \end{align*}$$

Answers:

• a) Maximum height is $61.25$ meters
• b) Stone hits ground at $t = 5$ seconds
• c) Velocity when hitting ground is $-35$ m/s (downward)

$$\begin{align*} \text{a) Velocity:} \\ v(t) &= s'(t) = 2 \cdot \pi\cos(\pi t) = 2\pi\cos(\pi t) \\ \\ v(0.5) &= 2\pi\cos(\pi \cdot 0.5) \\ &= 2\pi\cos\left(\frac{\pi}{2}\right) \\ &= 2\pi \cdot 0 = 0 \text{ m/s} \\ \\ \text{b) Acceleration (using } \frac{d}{dt}[\cos(\pi t)] = -\pi\sin(\pi t)\text{):} \\ a(t) &= v'(t) = 2\pi \cdot (-\pi\sin(\pi t)) \\ &= -2\pi^2\sin(\pi t) \\ \\ a(0.5) &= -2\pi^2\sin\left(\frac{\pi}{2}\right) \\ &= -2\pi^2 \cdot 1 \\ &= -2\pi^2 \approx -19.74 \text{ m/s}^2 \end{align*}$$

Answers:

• a) Velocity at $t = 0.5$ is $0$ m/s (particle momentarily at rest)
• b) Acceleration at $t = 0.5$ is $-2\pi^2 \approx -19.74$ m/s²