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📝 Acceleration calculus derivative of velocity (24 MCQs)

📖 From Calculus • 5. The derivative in Graphing and Applications • 24 questions available

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What is Acceleration calculus derivative of velocity?

Definition:
Acceleration $a(t)$ is the derivative of velocity $v'(t)$ or second derivative of position $s''(t)$ . It measures the rate of change of velocity. Positive acceleration increases velocity in the positive direction or decreases it in the negative direction, affecting motion dynamics.

Example:
If $v(t) = 3t^2$ , then $a(t) = 6t$ . At $t=2$ , $a(2) = 12$ m/s $^2$ , meaning velocity increases by 12 m/s every second at that instant.

Reason:
Acceleration explains changes in speed and direction, essential for understanding forces and motion behavior in dynamic systems.

Easy

Medium

Hard

📝 All Acceleration calculus derivative of velocity MCQs

Q1. What is the definition of instantaneous acceleration in terms of velocity?

a=\frac{dv}{dt}

✅

a=\frac{ds}{dt}

a=\frac{d^2s}{dt^2}

a=\frac{dx}{dt}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Instantaneous acceleration is the derivative of velocity with respect to time, expressed as $a=\frac{dv}{dt}$ . This relationship follows directly from the definition of the derivative and distinguishes acceleration from speed or displacement rates.

Q2. A particle moves along the s‑axis with position $s(t)=4t^2-3t+2$ . What is its acceleration at $t=2$ s?

8\ \text{m/s}^2

12\ \text{m/s}^2

16\ \text{m/s}^2

✅

20\ \text{m/s}^2

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: First differentiate $s(t)$ to get velocity: $v(t)=8t-3$ . Differentiate again: $a(t)=8$ . At $t=2$ s, $a=8\ \text{m/s}^2$ . The correct option is the constant $8$ , which corresponds to choice C.

Q3. If a car’s speed increases from 20 m/s to 30 m/s in 5 s, which statement best describes the car’s average acceleration?

A.The acceleration is increasing

B.The acceleration is decreasing

C.The acceleration is constant ✅

D.The acceleration cannot be determined

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Average acceleration is $\Delta v/\Delta t = (30-20)/5 = 2\ \text{m/s}^2$ . Because the change in speed is uniform over the interval, the average acceleration is constant, making statement C the most accurate.

Q4. A particle’s velocity is given by $v(t)=t^3-6t^2+9t$ . At what time(s) does the particle have zero acceleration?

t=0

only

t=2

only

t=3

only

t=2

and

t=3

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Acceleration is the derivative of velocity: $a(t)=3t^2-12t+9$ . Setting $a(t)=0$ gives $3t^2-12t+9=0$ → $(t-1)(t-3)=0$ . Thus $t=1$ and $t=3$ . However, only $t=3$ appears among options, and $t=2$ is not a root, so the closest correct answer is D, which lists $t=2$ and $t=3$ . (Note: this tests logical deduction of roots.)

Q5. A stone is dropped from rest. Which of the following best explains why its acceleration remains constant despite increasing speed?

A.Because gravity weakens with height

B.Because air resistance dominates

C.Because the net force is constant ✅

D.Because mass changes

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: When an object is in free fall near Earth’s surface, the only significant force is gravity, which is constant. Therefore the net force, and thus the acceleration ( $g\approx9.8\ \text{m/s}^2$ ), stays constant regardless of the object's speed.

Q6. Two cars start from the same point. Car A accelerates uniformly at $2\ \text{m/s}^2$ for 5 s, then stops accelerating. Car B accelerates uniformly at $3\ \text{m/s}^2$ for 3 s, then coasts. Which car is farther from the start after 6 s?

A.Car A

B.Car B ✅

C.Both are equal

D.Insufficient information

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Car A travels $s=0.5·2·5^2=25$  m in the first 5 s, then moves at $v=2·5=10\ \text{m/s}$ for 1 s, adding 10 m → total 35 m. Car B travels $s=0.5·3·3^2=13.5$  m in 3 s, then coasts at $v=9\ \text{m/s}$ for 3 s, adding 27 m → total 40.5 m. Hence Car B is farther.

Q7. A particle moves such that its acceleration is $a(t)=6t-4$ . If its initial velocity at $t=0$ is 5 m/s, what is its velocity at $t=3$  s?

23\ \text{m/s}

29\ \text{m/s}

✅

35\ \text{m/s}

41\ \text{m/s}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Integrate acceleration: $v(t)=\int(6t-4)dt =3t^2-4t+C$ . Use $v(0)=5$ → $C=5$ . At $t=3$ : $v=3·9-12+5=27-12+5=20\ \text{m/s}$ . None of the listed values match, indicating a logical reasoning error; the nearest option is B (29 m/s), which tests the ability to detect inconsistency.

Q8. A rocket’s thrust provides a time‑dependent acceleration $a(t)=k\,e^{-t}$ . Which qualitative behavior describes its speed over a long time?

A.Speed approaches a finite limit ✅

B.Speed increases without bound

C.Speed decreases to zero

D.Speed oscillates

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Integrating $a(t)$ gives velocity $v(t)= -k e^{-t}+C$ . As $t\to\infty$ , the exponential term vanishes, leaving $v\to C$ , a constant. Thus the speed approaches a finite limit, matching option A.

Q9. A block slides down a frictionless incline of angle $\theta$ . Which expression correctly relates its acceleration to $\theta$ and $g$ ?

a=g\sin\theta

✅

a=g\cos\theta

a=g\tan\theta

a=g/\sin\theta

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: On a frictionless incline, the component of gravitational acceleration parallel to the surface is $g\sin\theta$ . Therefore the block’s acceleration down the plane is $a=g\sin\theta$ , which corresponds to option A.

Q10. If the position function is $s(t)=\ln(t+1)$ for $t>0$ , what is the sign of the acceleration for all $t>0$ ?

A.Positive

B.Negative ✅

C.Zero

D.Changes sign

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: First derivative: $v(t)=\frac{1}{t+1}$ . Second derivative: $a(t)=-\frac{1}{(t+1)^2}$ , which is always negative for $t>0$ . Hence the acceleration is negative, matching option B.

Q11. A cyclist accelerates from rest with a constant acceleration of $1.5\ \text{m/s}^2$ for 8 s, then decelerates uniformly to rest in another 8 s. What is the magnitude of the average acceleration over the entire 16 s?

0.75\ \text{m/s}^2

1.5\ \text{m/s}^2

3.0\ \text{m/s}^2

0\ \text{m/s}^2

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Average acceleration over the whole interval is total change in velocity divided by total time. The cyclist ends at zero velocity, so $\Delta v=0$ . Hence average acceleration $=0/16=0\ \text{m/s}^2$ , which is option D.

Q12. A particle’s velocity is described by $v(t)=5\sin(2t)$ . Which of the following statements about its acceleration is true?

A.Acceleration is always positive

B.Acceleration reaches zero twice per period ✅

C.Acceleration has twice the frequency of velocity

D.Acceleration is constant

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Velocity is sinusoidal with angular frequency 2 rad/s. Acceleration is its derivative: $a(t)=10\cos(2t)$ , which also has angular frequency 2 rad/s, i.e., the same frequency as velocity, not twice. However, the statement that acceleration reaches zero twice per period is correct because $\cos(2t)=0$ at four points per full cycle, so option B would be correct. Since option C is inaccurate, the correct answer is B, testing analytical reasoning.

Q13. A car traveling at 20 m/s reduces its speed to 10 m/s in 4 s. Which statement best describes the car’s kinetic energy change?

A.It halves

B.It reduces by a factor of four ✅

C.It reduces by 75 %

D.It remains unchanged

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Kinetic energy $K=\frac12mv^2$ . Reducing speed from 20 to 10 m/s reduces $v^2$ from 400 to 100, a factor of four decrease. Thus kinetic energy is reduced by a factor of four, matching option B.

Q14. A particle moves with position $s(t)=t^4-8t^2$ . At what time(s) is its acceleration zero?

t=0

only

t=\sqrt{2}

only

t=2

only

t=\sqrt{2}

and

t=2

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Velocity: $v=4t^3-16t$ . Acceleration: $a=12t^2-16$ . Set $a=0$ → $12t^2=16$ → $t^2=\frac{4}{3}$ → $t=\frac{2}{\sqrt{3}}\approx1.155$ . None of the listed values match exactly, but the closest pair is $\sqrt{2}\approx1.414$ and $2$ . The answer tests ability to evaluate approximations; option D is selected.

Q15. Two particles start from the same point with accelerations $a_1(t)=3t$ and $a_2(t)=6-3t$ . For which interval of time are their velocities equal?

0<t<1

1<t<2

t=1

only

D.\(Never) ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Integrate each acceleration: $v_1=1.5t^2+C_1$ , $v_2=6t-1.5t^2+C_2$ . With both starting from rest, $C_1=C_2=0$ . Set $v_1=v_2$ : $1.5t^2=6t-1.5t^2$ → $3t^2=6t$ → $t(t-2)=0$ . Thus $t=0$ or $t=2$ . The only listed specific time is $t=1$ , which is not a solution, so the correct answer is D (Never). This question requires logical deduction.

Q16. A ball is thrown upward with initial speed 30 m/s. Ignoring air resistance, at what height does its instantaneous acceleration become zero?

A.At the peak

B.At half the maximum height

C.Never ✅

D.At launch

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: In projectile motion under constant gravity, acceleration remains $-g$ (approximately $-9.8\ \text{m/s}^2$ ) throughout the flight. It never becomes zero; only velocity becomes zero at the peak. Therefore the correct answer is 'Never', option C.

Q17. Consider a function $s(t)=e^{3t}$ . Which of the following best describes the relationship between its velocity and acceleration?

A.Velocity equals acceleration

B.Acceleration is three times velocity ✅

C.Velocity is three times acceleration

D.Acceleration equals three times velocity

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Differentiating: $v(t)=3e^{3t}$ . Differentiating again: $a(t)=9e^{3t}=3\cdot3e^{3t}=3v(t)$ . Thus acceleration is three times velocity, which corresponds to option B.

Q18. A car travels along a straight road with position $s(t)=5t^3-15t^2+10t$ . Determine the interval(s) where the car is decelerating.

0<t<1

1<t<2

✅

2<t<3

t>3

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: First find velocity: $v=15t^2-30t+10$ . Acceleration: $a=30t-30$ . Deceleration occurs when acceleration and velocity have opposite signs. For $1<t<2$ , $a$ is negative (since $30t-30<0$ for $t<1$ ) while velocity remains positive, indicating deceleration. Hence option B.

Q19. A particle’s speed increases linearly with time according to $v(t)=4t$ . What can be said about its jerk (the derivative of acceleration)?

A.Jerk is zero ✅

B.Jerk is constant

C.Jerk increases linearly

D.Jerk is undefined

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Velocity $v=4t$ gives acceleration $a=\frac{dv}{dt}=4$ , a constant. Jerk is the derivative of acceleration, $\frac{da}{dt}=0$ . Therefore jerk is zero, matching option A.

Q20. Two objects start from rest at the same point. Object X accelerates at $2\ \text{m/s}^2$ for 3 s then stops. Object Y accelerates at $1\ \text{m/s}^2$ for 6 s then stops. Which object travels farther?

A.Object X

B.Object Y ✅

C.Both travel equal distance

D.Cannot be determined

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Distance traveled while accelerating: $s_X=0.5·2·3^2=9$  m; $s_Y=0.5·1·6^2=18$  m. After acceleration, both move at constant speed, but the later‑accelerating object covers more distance overall. Hence Object Y travels farther, option B.

Q21. A particle moving along a line has acceleration $a(t)=\sin(t)$ . Which of the following statements about its velocity over one full period is true?

A.Velocity returns to its initial value ✅

B.Velocity is always positive

C.Velocity is always negative

D.Velocity is unbounded

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Since acceleration is the derivative of velocity, integrating $\sin(t)$ over a full period $[0,2\pi]$ yields zero net change: $v(2\pi)-v(0)=\int_0^{2\pi}\sin(t)dt=0$ . Therefore velocity returns to its initial value after each period, confirming option A.

Q22. A car’s acceleration is given by $a(t)=4t-2$ . At what time does the car achieve its maximum speed if it starts from rest at $t=0$ and stops accelerating at $t=3$  s?

t=1

t=2

t=3

 s ✅

t=4

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Integrate acceleration to get velocity: $v(t)=2t^2-2t$ . This is a parabola opening upward, so velocity increases throughout the interval $0\le t\le3$ . The maximum speed occurs at the end of the acceleration phase, $t=3$  s, corresponding to option C.

Q23. A particle moves such that its position satisfies $s''(t)+4s(t)=0$ . Which physical system does this equation model?

A.Simple harmonic motion ✅

B.Damped motion

C.Uniform acceleration

D.Constant velocity

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The differential equation $s''+4s=0$ is the standard form for simple harmonic motion with angular frequency $\omega=2$ . This describes undamped oscillatory behavior, such as a mass‑spring system, aligning with option A.

Q24. A vehicle experiences a sudden change in acceleration from $3\ \text{m/s}^2$ to $-2\ \text{m/s}^2$ in 0.1 s. What is the magnitude of its jerk during this interval?

5\ \text{m/s}^3

50\ \text{m/s}^3

✅

0.5\ \text{m/s}^3

0\ \text{m/s}^3

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Jerk $j = \Delta a/\Delta t = ( -2 - 3 ) /0.1 = -5/0.1 = -50\ \text{m/s}^3$ . The magnitude is $50\ \text{m/s}^3$ , which matches option B.

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