Question

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.02t3 (a) find the velocity at time t (in ft/s). v(t) = (b) what is the velocity after 1 second(s)? v(1) = ft/s (c) when is the particle at rest? t = s (smaller value) t = s (larger value)

Answer 1

The distance (ft) traveled by the particle at time t (s) is
s(t) = 0.01 t⁴ - 0.02 t³

Part (a)
The velocity at time t is
v(t) = 0.04t³ - 0.06t²  ft/s

Part (b)
After 1 s, the velocity is
v(1) = 0.04 - 0.06 = - 0.02 ft/s

Part (c)
When the particle is at rest, the velocity is zero. The time when this happens is given by
0.04t³ - 0.06t² = 0
t²(0.04t - 0.06) = 0
The graph shown below presents a clear picture of the motion.

Answer:
t = 0 (smaller value) or t = 1.5 s (larger value)