Question

A particle, initially at rest, moves along the x-axis such that its acceleration at time t > 0 is given by a(t) = 6cos(t). At the time t = 0, its position is x = 6.
(a) Find the velocity and position functions for the particle.
(b) Find the values of t for which the particle is at rest.

Answer 1

a. By the fundamental theorem of calculus, the velocity function is

$v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du$

The particle starts at rest, so $v(0)=0$, and we have

$v(t)=\displaystyle\int_0^t6\cos u\,\mathrm du=6\sin u\bigg|_0^t=6\sin t$

Then the position function is

$x(t)=\displaystyle x(0)+\int_0^tv(u)\,\mathrm du$

with $x(0)=6$, so

$x(t)=6+\displaystyle\int_0^t6\sin u\,\mathrm du=6-6\cos u\bigg|_0^t=12-6\cos t$

b. The particle is at rest whenever $v(t)=0$; this happens for

$6\sin t=0\implies \sin t=0\implies t=n\pi$

where $n$ is any integer.