AP Physics C Mechanics · Question of the Day

AP Physics C Mechanics Question of the Day

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Friday, August 21, 2026

The velocity of a particle moving in one dimension is described by the equation $v(t) = (12t - 3t^2) \text{ m/s}$ . What is the particle's acceleration at the instant it momentarily stops?

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Question of the Day

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The velocity of a particle moving in one dimension is described by the equation $v(t) = (12t - 3t^2) \text{ m/s}$ . What is the particle's acceleration at the instant it momentarily stops?

$-12 \text{ m/s}^2$ (correct answer)
$0 \text{ m/s}^2$
$-24 \text{ m/s}^2$
$12 \text{ m/s}^2$

Explanation: First, find the time when the particle momentarily stops by setting $v(t) = 0$ . This gives $12t - 3t^2 = 3t(4 - t) = 0$ , so $t=0 \text{ s}$ and $t=4 \text{ s}$ . Acceleration is the derivative of velocity: $a(t) = \frac{dv}{dt} = 12 - 6t$ . At $t=4 \text{ s}$ , the acceleration is $a(4) = 12 - 6(4) = 12 - 24 = -12 \text{ m/s}^2$ .