📝 Instantaneous velocity using derivative (16 MCQs)

📖 From Calculus • 3. The Derivation • 16 questions available

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What is Instantaneous velocity using derivative?

Definition:
Instantaneous velocity is obtained by taking the derivative of the position function with respect to time, yielding the exact velocity at any given moment, which represents the limit of average velocities over increasingly smaller time intervals approaching zero.

Example:
If position $s(t) = 5t^2$ , then velocity $v(t) = s'(t) = 10t$ , so at $t=3$ , $v(3) = 30$ meters per second.

Reason:
This application demonstrates the practical utility of derivatives in physics, allowing precise prediction of object motion and analysis of acceleration patterns in kinematic studies.

Easy

Medium

Hard

📝 All Instantaneous velocity using derivative MCQs

Q1. For the particle with position function $s(t)=1+5t-2t^{2}$ , at what time does the particle change its direction of motion?

A.t=1

B.t=1.5

C.t=1.25 ✅

D.t=2

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The direction changes when instantaneous velocity is zero. Differentiating gives $v(t)=5-4t$ . Setting $v(t)=0$ yields $t=5/4=1.25$ . At this instant the velocity sign switches, indicating a reversal of motion.

Q2. Particle A follows $s_{A}(t)=t^{2}$ and Particle B follows $s_{B}(t)=2t^{2}$ . Which particle has the greater instantaneous speed at $t=3$ ?

A.Particle B ✅

B.Particle A

C.Both have the same speed

D.Cannot be determined

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Instantaneous speed is the magnitude of the derivative. $v_{A}(t)=2t$ gives $v_{A}(3)=6$ . $v_{B}(t)=4t$ gives $v_{B}(3)=12$ . Since 12 > 6, Particle B moves faster at $t=3$ .

Q3. Which of the following best describes instantaneous velocity?

A.Average change in position over a long interval

B.Total distance traveled divided by total time

C.Slope of the tangent line to the position‑time curve at a specific instant ✅

D.A constant that depends only on the unit system

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Instantaneous velocity is the derivative of position with respect to time, which geometrically equals the slope of the tangent line to the position‑time graph at the instant considered.

Q4. Given $s(t)=t^{3}-6t^{2}+9t$ , on which interval(s) is the particle accelerating?

t<2

0<t<2

C.Never

t>2

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Acceleration is the derivative of velocity. First $v(t)=3t^{2}-12t+9$ , then $a(t)=6t-12$ . The acceleration is positive when $6t-12>0\Rightarrow t>2$ . Positive acceleration means the speed is increasing on $(2,\infty)$ .

Q5. For the motion described by $s(t)=4\sin t$ , what is the average velocity on the interval $[0,\pi]$ ?

A.0 ✅

B.4

C.-4

D.2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Average velocity equals $\frac{s(\pi)-s(0)}{\pi}$ . Since $\sin\pi=0$ and $\sin0=0$ , the numerator is $0$ . Thus the average velocity is $0/\pi=0$ .

Q6. If the instantaneous velocity function $v(t)$ is a non‑zero constant, what must the position function $s(t)$ be?

A.Quadratic function

B.Exponential function

C.Sinusoidal function

D.Linear function ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: A constant velocity means $v(t)=k$ where $k\neq0$ . Integrating $ds/dt=k$ gives $s(t)=kt+C$ , a linear function of time. Any higher‑order term would produce a varying velocity.

Q7. For the position function $s(t)=e^{2t}$ , what is the instantaneous velocity at $t=0$ ?

A.1

B.0

C.2 ✅

e^{2}

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Differentiate: $v(t)=\frac{d}{dt}e^{2t}=2e^{2t}$ . Evaluating at $t=0$ yields $v(0)=2e^{0}=2$ .

Q8. Particle A has velocity $v_{A}(t)=3t$ and Particle B has velocity $v_{B}(t)=t^{2}$ . Which particle has traveled a greater distance from $t=0$ to $t=2$ ?

A.Particle A ✅

B.Particle B

C.Both travel the same distance

D.Insufficient information

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Distance traveled equals the integral of speed. $\int_{0}^{2}3t\,dt=3\cdot(2^{2}/2)=6$ . $\int_{0}^{2}t^{2}\,dt=8/3\approx2.67$ . Since 6 > 2.67, Particle A covers more ground.

Q9. Why does the limit definition $v(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}{h}$ give the instantaneous velocity?

A.Measures average over a finite interval

B.Approximates slope of secant line as interval shrinks ✅

C.Ignores behavior near t

D.Works only for linear functions

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The expression computes the slope of the secant line between $(t,s(t))$ and $(t+h,s(t+h))$ . As $h$ approaches zero, the secant approaches the tangent, whose slope is precisely the instantaneous rate of change, i.e., the velocity.

Q10. The position of a particle is given by $s(t)=t^{4}-8t^{2}+16$ . At which times does the instantaneous velocity equal zero, and what does the sign of the velocity indicate about the particle’s motion?

t=-2,0,2

; velocity is positive for

t<-2

and

t>2

and negative between ✅

t=-2,0,2

; velocity changes from positive to negative at each zero

t=0

only; velocity always positive otherwise

D.No real times; velocity never zero

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Differentiate: $v(t)=4t^{3}-16t=4t(t^{2}-4)=4t(t-2)(t+2)$ . Zeros at $-2,0,2$ . Test intervals: for $t<-2$ and $t>2$ , $v>0$ (motion forward); between $-2$ and $0$ and between $0$ and $2$ , $v<0$ (motion backward).

Q11. If a particle’s velocity is $v(t)=6t-5$ and its position at $t=0$ is $s(0)=3$ , what is the position function $s(t)$ ?

3t^{2}-5t+3

✅

3t^{2}-5t

3t^{2}+5t+3

6t^{2}-5t+3

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Integrate velocity: $s(t)=\int(6t-5)dt=3t^{2}-5t+C$ . Use $s(0)=3$ to find $C=3$ . Hence $s(t)=3t^{2}-5t+3$ .

Q12. Consider a motion where instantaneous velocity is $v(t)=k t^{n}$ with $k>0$ and integer $n$ . How does increasing the exponent $n$ affect the shape of the position function $s(t)$ for large $t$ ?

A.Position grows slower because higher powers diminish the effect of

k

B.Position becomes periodic

C.Position grows faster, approaching a polynomial of degree

n+1

✅

D.Position remains linear regardless of

n

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Integrating $v(t)$ gives $s(t)=\frac{k}{n+1}t^{\,n+1}+C$ . As $n$ increases, the dominant term is of higher degree, so for large $t$ the position grows more rapidly, following a polynomial of degree $n+1$ .

Q13. The limit definition of instantaneous velocity does not exist at $t=1$ for $s(t)=|t-1|$ . What does this indicate about the particle’s motion at that instant?

A.The particle is at rest

B.The particle has a well‑defined speed but undefined direction

C.The particle experiences a sudden change in direction (a cusp) ✅

D.The particle moves with constant acceleration

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The absolute‑value function has a sharp corner at $t=1$ ; the left‑hand and right‑hand derivatives are $-1$ and $+1$ . Because the derivative does not exist, the instantaneous velocity is undefined, indicating an abrupt reversal of direction.

Q14. Compare the instantaneous speeds of $s_{1}(t)=t^{3}$ and $s_{2}(t)=t^{3}+t$ as $t$ becomes very large. Which statement is true?

A.Both speeds become identical because the lower‑order term becomes negligible ✅

s_{2}

always has a larger speed by a constant amount

s_{1}

eventually surpasses

s_{2}

D.Speeds diverge to infinity at different rates

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Derivatives are $v_{1}=3t^{2}$ and $v_{2}=3t^{2}+1$ . As $t\to\infty$ , the added $+1$ is negligible compared with $3t^{2}$ ; thus the speeds approach each other, becoming effectively identical for large $t$ .

Q15. What is the definition of instantaneous velocity?

A.The limit of average velocity as the time interval approaches zero ✅

B.The total distance traveled divided by total time

C.The maximum speed reached during motion

D.The derivative of acceleration

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Instantaneous velocity is defined as the derivative of position with respect to time, equivalently the limit of the average velocity $\frac{s(t+h)-s(t)}{h}$ as the interval $h$ shrinks to zero.

Q16. Which formula correctly expresses instantaneous velocity using a limit?

v(t)=\frac{s(t+h)-s(t)}{h}

v(t)=\int_{0}^{t}s&#039;(u)\,du

v(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}{h}

✅

v(t)=s(t)\cdot h

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The precise definition uses a limit: $v(t)=\displaystyle\lim_{h\to0}\frac{s(t+h)-s(t)}{h}$ . This captures the instantaneous rate of change of position, distinguishing it from the average rate over a finite interval.

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