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📝 Speeding up slowing down motion calculus (16 MCQs)

📖 From Calculus • 5. The derivative in Graphing and Applications • 16 questions available

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What is Speeding up slowing down motion calculus?

Definition:
An object speeds up when velocity and acceleration have the same sign ( $v(t)a(t) > 0$ ) and slows down when they have opposite signs ( $v(t)a(t) < 0$ ). This rule determines whether the magnitude of velocity (speed) is increasing or decreasing at any given time.

Example:
If $v(t) = -2$ and $a(t) = -3$ , both negative, so the object speeds up. If $v(t) = -2$ and $a(t) = 3$ , it slows down.

Reason:
Sign agreement indicates force acting in direction of motion (speeding up), while disagreement indicates resistance (slowing down), clarifying motion trends.

Easy

Medium

Hard

📝 All Speeding up slowing down motion calculus MCQs

Q1. In rectilinear motion, when is a particle said to be speeding up?

A.When its speed is decreasing

B.When its speed is constant

C.When its speed is increasing ✅

D.When its velocity is zero

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Speeding up means the magnitude of velocity (speed) is growing. Therefore the particle is speeding up precisely when its speed is increasing, regardless of direction. The other choices describe opposite or neutral situations, making option C the correct definition.

Q2. If the acceleration a(t) is zero over an interval, what can be concluded about the particle's motion during that interval?

A.It is at rest

B.It moves with constant velocity ✅

C.It is speeding up

D.It is slowing down

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When acceleration is zero, the derivative of velocity is zero, so velocity does not change. The particle continues moving at whatever speed it already has, which could be zero or non‑zero, but the speed remains constant. Hence the motion is uniform with constant velocity.

Q3. For a particle with velocity $v(t) = -2t+5$ and acceleration $a(t) = -2$ , during which time interval does the particle speed up?

0<t<2.5

t>2.5

✅

t<0

D.All times

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Speeding up occurs when velocity and acceleration share the same sign. Here $a(t)=-2$ is always negative. The velocity becomes negative when $t>2.5$ . Thus, for $t>2.5$ both $v$ and $a$ are negative, giving the same sign and a rising speed magnitude.

Q4. A particle moves leftward with velocity $v(t) = -3t^2+6t$ and acceleration $a(t) = -6t+6$ . Determine the time intervals where the particle is slowing down.

0<t<1

1<t<2

✅

t>2

D.No interval

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Slowing down requires opposite signs of velocity and acceleration. For $0<t<2$ the velocity is positive, while acceleration changes sign at $t=1$ . Between $1$ and $2$ the acceleration is negative, giving opposite signs and thus the particle slows down only on $(1,2)$ .

Q5. If a particle's speed is observed to increase from $t=3$  s to $t=5$  s, which of the following must be true about the signs of its velocity and acceleration on that interval?

A.Both are positive

B.Both are negative

C.They have the same sign (both positive or both negative) ✅

D.They have opposite signs

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: An increasing speed means the magnitude of velocity is growing. This occurs whenever velocity and acceleration point in the same direction, i.e., share the same sign, regardless of whether that direction is positive or negative. Hence the only necessary condition is that the signs coincide.

Q6. During an interval, a particle has velocity $v(t) = 4$  m/s and acceleration $a(t) = -2$  m/s². What is happening to the particle's speed?

A.Speed increasing

B.Speed decreasing ✅

C.Speed constant

D.Cannot determine

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The velocity is positive while the acceleration is negative, giving opposite signs. Opposite signs indicate that the speed magnitude is decreasing, so the particle is slowing down despite moving forward.

Q7. A particle's velocity is given by $v(t)=t^3-3t$ . At which times does the particle transition from speeding up to slowing down?

t = -\sqrt{3},\; -1,\; 0,\; 1,\; \sqrt{3}

t = -1,\; 0,\; 1

t = -\sqrt{3},\; \sqrt{3}

t = 0

only ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Transitions occur whenever either the velocity or the acceleration changes sign, because the relative sign determines speed behavior. Here $v=0$ at $t=0,\pm\sqrt{3}$ and $a=0$ at $t=\pm1$ . All five points are moments where the sign relationship may switch, marking a transition.

Q8. Compare the motion of two particles: Particle A has $v=5$  m/s and $a=2$  m/s², Particle B has $v=-5$  m/s and $a=-2$  m/s². Which statement correctly describes their speed behavior?

A.Both are speeding up ✅

B.Both are slowing down

C.A is speeding up, B is slowing down

D.A is slowing down, B is speeding up

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Speeding up requires velocity and acceleration to share the same sign. Particle A has both positive, and Particle B has both negative, so each pair has matching signs. Consequently, both particles increase the magnitude of their velocity, i.e., they are speeding up.

Q9. Evaluate the statement: 'If acceleration is positive, the particle must be speeding up.' Is it always true?

A.Yes, always

B.No, only when velocity is positive

C.No, only when velocity is negative

D.No, depends on direction of motion ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: A positive acceleration increases the numerical value of velocity, but whether the speed (the magnitude) increases depends on the sign of the velocity. If the particle moves leftward (negative velocity), a positive acceleration actually reduces the speed. Thus the statement is not universally true.

Q10. Differentiate between the effects of a constant positive acceleration on a particle moving leftward versus rightward. Which of the following correctly describes the speed change in each case?

A.Leftward speed decreases, rightward speed increases ✅

B.Both speeds increase

C.Leftward speed increases, rightward speed decreases

D.Both speeds decrease

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: When the particle moves leftward its velocity is negative; a positive acceleration makes the velocity less negative, lowering its magnitude, so the speed decreases. Conversely, for rightward motion the velocity is positive and a positive acceleration makes it more positive, raising the speed.

Q11. Given the position function $s(t)=t^3-6t^2$ , which of the following correctly identifies the intervals where the particle is speeding up?

0<t<2

and

t>4

✅

2<t<4

t<0

only

D.All

t

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The velocity is $v(t)=3t^2-12t=3t(t-4)$ and the acceleration is $a(t)=6t-12=6(t-2)$ . Both are negative on $0<t<2$ and both are positive for $t>4$ ; in each case the signs match, indicating the speed is increasing (speeding up) on those intervals.

Q12. For the particle with $s(t)=t^4-8t^2$ , determine the intervals where the particle is slowing down. Choose the correct interval(s).

(-\infty,-2)\cup\left(-\sqrt{\tfrac{4}{3}},0\right)\cup\left(\sqrt{\tfrac{4}{3}},2\right)

(-\infty,-2)

only

(0,2)

only

D.All

t

✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The velocity is $v=4t(t^2-4)$ and the acceleration is $a=4(3t^2-4)$ . Slowing down occurs when $v$ and $a$ have opposite signs. Analyzing sign changes shows opposite signs on $(-\infty,-2)$ , $(-\,\sqrt{4/3},0)$ , and $(\sqrt{4/3},2)$ , giving the combined interval listed.

Q13. Compare the speed vs. time graphs for a particle that speeds up then slows down with those for a particle that constantly accelerates. Which feature distinguishes them?

A.Presence of a maximum speed ✅

B.Linear speed increase

C.Constant speed

D.No change in speed

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A particle that first speeds up and then slows down must reach a peak speed before decreasing, producing a maximum point on the speed‑time graph. A constantly accelerating particle shows a monotonic increase with no such peak. The existence of a maximum speed is the distinguishing characteristic.

Q14. If a particle's velocity is always positive and its acceleration changes sign from positive to negative, what happens to its speed?

A.Speed continuously increases

B.Speed reaches a maximum then decreases ✅

C.Speed remains constant

D.Speed decreases then increases

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: With a positive velocity, a positive acceleration makes the speed grow. When the acceleration becomes negative, it starts reducing the velocity magnitude, causing the speed to peak at the moment the acceleration switches sign and then to decline afterward.

Q15. Apply the rule for speeding up to the function $v(t)= -t^2+4t-3$ . For which intervals does the particle speed up?

1<t<2

only

t>3

only

1<t<2

and

t>3

✅

t<1

only

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The acceleration is $a(t)=-2t+4$ . Speeding up requires $v$ and $a$ to share the same sign. $v$ is positive on $(1,3)$ ; $a$ is positive on $(-\infty,2)$ and negative afterward. The overlapping regions where signs match are $(1,2)$ (both positive) and $(t>3)$ (both negative), giving the combined answer.

Q16. Synthesize the concepts of velocity sign and acceleration sign to predict the motion of a particle whose velocity is given by $v(t)=\sin t$ and acceleration by $a(t)=\cos t$ . During which intervals in $[0,2\pi]$ is the particle speeding up?

(0,\frac{\pi}{2})

and

(\pi,\frac{3\pi}{2})

✅

(\frac{\pi}{2},\pi)

only

(0,\pi)

only

D.None of the above

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Speeding up occurs when velocity and acceleration have the same sign. $\sin t$ and $\cos t$ are both positive on $(0,\pi/2)$ and both negative on $(\pi,3\pi/2)$ . In these two intervals the signs match, so the particle speeds up there.

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