Motion along a Line: Velocity and Acceleration Practice

Given s(t) = t^2 - 8t + 12 (in meters, for t >= 0), find v(t) and a(t).

v(t) = 2t - 8 m/s; a(t) = 2 m/s^2.

Using s(t) = t^2 - 8t + 12, at what time is the body momentarily at rest?

t = 4 seconds, because v(4) = 2(4) - 8 = 0.

For s(t) = t^3 - 3t^2, find all times t >= 0 when the object changes direction.

v(t) = 3t^2 - 6t = 3t(t - 2). The body is at rest at t = 0 and t = 2. It changes direction only at t = 2 (velocity changes from negative for t in (0, 2) to positive for t > 2).

A particle moves with position s(t) = -t^2 + 6t. At t = 4, is the particle speeding up or slowing down?

v(t) = -2t + 6, so v(4) = -2. a(t) = -2, so a(4) = -2. Because v(4) and a(4) are both negative (same sign), the particle is speeding up.

Given s(t) = 2t^3 - 9t^2 + 12t on the interval [0, 3], find all intervals where the particle is moving to the left.

v(t) = 6t^2 - 18t + 12 = 6(t - 1)(t - 2). The particle moves to the left when v(t) < 0, which occurs on the open interval (1, 2).

For s(t) = t^3 - 6t^2 on [0, 5], determine the greatest distance the body reaches from its starting point s(0) = 0.

v(t) = 3t^2 - 12t = 3t(t - 4). Test critical point t = 4 and endpoint t = 5: s(4) = 4^3 - 6(16) = -32; s(5) = 125 - 150 = -25. Distance from s(0) is |-32| = 32 meters at t = 4.