Kinematics from displacement function A particle moves on a straight line with displacement S(t) = 4 t^3 + 3 t^2 − 10 (S in m, t in s). Starting from rest at t = 0, how long does it take for its velocity to reach 18 m/s?
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A0.5 s
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B1.0 s
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C1.5 s
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D2.0 s
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ENot attainable
Answer
Correct Answer: 1.0 s
Explanation
Introduction / Context:This problem exercises basic calculus in kinematics: velocity is the time derivative of displacement. Given S(t) explicitly, we compute v(t), apply the initial condition “from rest”, and solve for the time when v reaches a specified value.
Given Data / Assumptions:
- S(t) = 4 t^3 + 3 t^2 − 10 (meters).
- Velocity v(t) = dS/dt (m/s).
- Starts from rest at t = 0 (v = 0).
- Find t when v = 18 m/s, with t ≥ 0.
Concept / Approach:
Differentiation gives the velocity function. We then set v = 18 and solve the resulting quadratic. Only the physically meaningful positive root is accepted because time is measured forward from the instant the particle is at rest (t = 0).
Step-by-Step Solution:
Compute velocity: v(t) = dS/dt = 12 t^2 + 6 t.Check initial rest: v(0) = 0 (satisfied).Set target velocity: 12 t^2 + 6 t = 18.Divide by 6: 2 t^2 + t = 3.Rearrange: 2 t^2 + t − 3 = 0.Solve quadratic: t = [−1 ± √(1 + 24)] / 4 = (−1 ± 5) / 4.Physical root: t = 1 s (discard t = −1.5 s).Verification / Alternative check:
Substitute t = 1 s into v(t): 12(1)^2 + 6(1) = 18 m/s, confirming the result.
Why Other Options Are Wrong:
0.5 s, 1.5 s, 2.0 s give velocities not equal to 18 m/s; “Not attainable” is incorrect because a valid positive solution exists.
Common Pitfalls:
Forgetting to differentiate S(t); using the negative root; arithmetic errors when solving the quadratic.
Final Answer:
1.0 s