📝 Average vs instantaneous velocity calculus (18 MCQs)

📖 From Calculus • 3. The Derivation • 18 questions available

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What is Average vs instantaneous velocity calculus?

Definition:
Average velocity measures total displacement over a time interval, while instantaneous velocity is the limit of average velocity as the time interval shrinks to zero, providing the exact speed and direction at a precise moment in time during motion.

Example:
If $s(t) = t^2$ , average velocity from $t=1$ to $t=3$ is $\frac{9-1}{2} = 4$ , but instantaneous at $t=2$ is $v(2) = 4$ .

Reason:
Distinguishing these helps analyze motion accurately, as average values mask variations that instantaneous rates reveal, crucial for physics and engineering dynamics problems.

Easy

Medium

Hard

📝 All Average vs instantaneous velocity calculus MCQs

Q1. What is the definition of average velocity over the interval [t1, t2]?

A.Δs/Δt ✅

B.lim_{Δt→0} Δs/Δt

C.s/t

D.v^2

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Average velocity is defined as the total displacement Δs divided by the elapsed time Δt, giving the ratio Δs/Δt. It characterizes how far the object moves on average between the two times, independent of any instantaneous changes. Hence it does not reflect the speed at any particular instant.

Q2. Which expression correctly represents the instantaneous velocity of a particle at time t0 using the limit definition?

A.v = lim_{Δt→0} (s(t0+Δt) - s(t0)) / Δt

B.v = Δs / Δt

C.v = s(t0) / t0

D.v = ds/dt evaluated at t0 ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The instantaneous velocity is the derivative of the position function evaluated at the specific time, commonly written as ds/dt|_{t=t0}. This derivative is precisely the limit of the difference quotient as Δt approaches zero, capturing the instantaneous rate of change of position.

Q3. Using the limit definition, what is the instantaneous velocity of a particle whose position is s(t)=t^2 at t=3?

A.3

B.6 ✅

C.9

D.12

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Applying the definition v = lim_{h→0} ( (t+h)^2 - t^2 ) / h at t=3 gives v = lim_{h→0} (6h + h^2)/h = 6 + h → 6. Thus the instantaneous velocity at t=3 equals 6 units per time, matching the derivative 2t evaluated at t=3.

Q4. How does the slope of the tangent line to the graph of s(t) at t0 relate to the instantaneous velocity at that time?

A.It equals the average velocity over any interval containing t0.

B.It is the reciprocal of the instantaneous velocity.

C.It is exactly the instantaneous velocity. ✅

D.It provides the acceleration, not the velocity.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The tangent line slope at a point on the position‑time graph represents the derivative ds/dt at that point, which is precisely the instantaneous velocity. Therefore the slope of the tangent line and the instantaneous velocity are numerically identical, both describing the instantaneous rate of change of position.

Q5. If an object moves with constant speed 8 m/s, which statement about its average and instantaneous velocities over any interval is true?

A.Both are equal to 8 m/s. ✅

B.Average is 8 m/s, instantaneous varies.

C.Instantaneous is 8 m/s, average varies.

D.Neither can be determined without direction.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: When speed is constant, the magnitude of velocity does not change, so the average velocity (displacement divided by time) equals the constant speed, and the instantaneous velocity (the derivative) also equals that same constant value. Direction is assumed consistent, making both quantities identical.

Q6. For a particle with position s(t)=4t^2, what happens to its instantaneous velocity when the time t is doubled?

A.It remains the same.

B.It doubles. ✅

C.It quadruples.

D.It halves.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The instantaneous velocity is the derivative v=ds/dt=8t. If t is replaced by 2t, the new velocity becomes 8(2t)=16t, which is exactly twice the original value 8t. Hence doubling the time doubles the instantaneous velocity.

Q7. Given s(t)=t^3, why is the instantaneous velocity at t=0 equal to zero even though the particle is moving?

A.Because the derivative of t^3 at 0 is undefined.

B.Because the limit of ((h)^3)/h as h→0 equals 0. ✅

C.Because displacement is zero at that instant.

D.Because speed cannot be negative.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The instantaneous velocity uses the limit v = lim_{h→0} ( (0+h)^3 - 0 ) / h = lim_{h→0} h^2 = 0. Although the particle’s position changes for non‑zero times, the rate of change at the exact instant t=0 is zero, reflecting a momentary pause in motion.

Q8. Compare the average velocity on the interval [1,3] for s(t)=t^2 with the instantaneous velocity at t=2. Which is larger?

A.Average is larger.

B.Instantaneous is larger.

C.They are equal. ✅

D.Cannot compare without more info.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The average velocity over [1,3] is (9‑1)/(3‑1)=4. The instantaneous velocity at t=2 is the derivative 2t evaluated at 2, giving 4 as well. Since both values are the same, neither is larger; they are equal.

Q9. Using the limit definition, which step is essential to eliminate the 0/0 indeterminate form for f(x)=√x at x₀=4?

A.Multiply numerator and denominator by √(x₀+h) - √x₀.

B.Rationalize the numerator by multiplying by the conjugate. ✅

C.Apply L'Hôpital's rule directly.

D.Substitute h=0 before simplifying.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For the expression lim_{h→0} (√(4+h)‑√4)/h, multiplying numerator and denominator by the conjugate √(4+h)+√4 transforms the numerator into (4+h‑4)=h, cancelling the h in the denominator and removing the 0/0 form. This rationalization is the key step.

Q10. A particle’s average velocity over [0,h] is given by (h+2)/h. What is its instantaneous velocity at t=0?

A.0

B.2

C.Does not exist (infinite) ✅

D.1

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The instantaneous velocity is the limit of the average velocity as h→0: lim_{h→0}(h+2)/h = lim_{h→0}(1+2/h). As h approaches zero, 2/h grows without bound, so the limit diverges to infinity. Hence the instantaneous velocity does not exist as a finite number.

Q11. Given two position functions s₁(t)=t² and s₂(t)=t²+5t, which has the larger instantaneous velocity at t=2?

A.s₁ has larger instantaneous velocity.

B.Both have the same instantaneous velocity.

C.s₂ has larger instantaneous velocity. ✅

D.Cannot determine without acceleration.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Instantaneous velocity is the derivative: v₁=2t, v₂=2t+5. At t=2, v₁=4 while v₂=4+5=9. Therefore s₂(t) yields the larger instantaneous velocity at that instant.

Q12. Explain why the tangent line to y=1/x at the point (2, 0.5) has a negative slope.

A.Because the derivative of 1/x is -1/x², which is negative for positive x. ✅

B.Because the function is decreasing, so slope must be positive.

C.Because the point lies in the second quadrant.

D.Because the limit of secant lines yields a positive value.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The derivative of y=1/x is dy/dx = -1/x². At x=2 (a positive value), the derivative evaluates to -1/4, a negative number. This negative derivative is precisely the slope of the tangent line at that point, indicating the line descends as x increases.

Q13. If instantaneous velocity at t=5 s is 12 m/s and acceleration is constant 2 m/s², what was the average velocity over the interval [5,7] s?

A.12 m/s

B.13 m/s

C.14 m/s ✅

D.16 m/s

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: With constant acceleration a=2, velocity varies linearly: v(t)=12+2(t‑5). At t=7, v=12+2·2=16 m/s. The average velocity over [5,7] is (v₅+v₇)/2 = (12+16)/2 = 14 m/s.

Q14. For s(t)=t³‑6t²+9t, find the time t in [0,3] where instantaneous velocity equals the average velocity on [0,3].

A.t=1 only

B.t=3 only

C.t=1 or t=3 ✅

D.No such t

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The average velocity on [0,3] is (s(3)‑s(0))/3 = 0/3 = 0. Instantaneous velocity v=ds/dt = 3t²‑12t+9. Setting v=0 gives 3t²‑12t+9=0 → t²‑4t+3=0 → (t‑1)(t‑3)=0, so t=1 or t=3 satisfy the condition.

Q15. Derive the expression for instantaneous velocity of s(t)=e^{2t} using the limit definition, then evaluate it at t=0.

A.2e^{2t}, and at t=0 it equals 2. ✅

B.e^{2t}, and at t=0 equals 1.

C.2e^{t}, and at t=0 equals 2.

D.2t e^{2t}, and at t=0 equals 0.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Using the definition v = lim_{h→0} (e^{2(t+h)}‑e^{2t})/h = e^{2t}·lim_{h→0} (e^{2h}‑1)/h = e^{2t}·2, so v=2e^{2t}. At t=0, this gives v=2e^{0}=2.

Q16. Assuming linear acceleration, if instantaneous velocity at t=1 s is 4 m/s and at t=3 s is 12 m/s, what is the instantaneous velocity at t=2 s?

A.6 m/s

B.8 m/s ✅

C.10 m/s

D.12 m/s

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Linear acceleration implies constant rate of change, so velocity varies linearly. The slope is (12‑4)/(3‑1)=8/2=4 m/s². Starting from 4 m/s at t=1, after one second the velocity increases by 4 m/s, giving v(2)=8 m/s.

Q17. For the position function s(t)=2/t, compute the limit defining instantaneous velocity at t=2 and interpret its sign.

A.-0.5, indicating motion toward decreasing x. ✅

B.0.5, indicating motion toward increasing x.

C.-2, indicating acceleration.

D.0, indicating rest.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The derivative v = ds/dt = lim_{h→0} (2/(2+h)‑2/2)/h = -2/(2²) = -0.5. The negative sign shows the particle’s x‑coordinate is decreasing as time increases, i.e., it moves in the negative x‑direction at t=2.

Q18. Explain how the concept of instantaneous velocity extends to vector‑valued motion, and why the scalar limit of speed may not exist at a cusp.

A.The scalar limit of speed always exists at a cusp.

B.Instantaneous velocity is a vector defined by componentwise limits, while speed may be undefined if direction changes abruptly. ✅

C.Both velocity and speed are undefined at any non‑differentiable point.

D.Vector‑valued motion does not require limits.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For vector‑valued position r(t), instantaneous velocity is the vector derivative dr/dt, obtained by taking limits of each component separately. At a cusp, the direction of motion changes instantaneously, so the derivative vector may exist, but its magnitude (speed) can fail to have a limit because the direction shift causes the speed’s scalar limit to be ambiguous or infinite.)

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