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?

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:

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