Question

A particle moves along the x-axis so that at any time t, t ≥ 0, its acceleration is a(t) = -4sin(2t). If the velocity of the particle at t = 0 is v(0) = 7 and its position at t = 0 is x(0) = 0, then its position at time t is x(t) = ?
A. sin(2t) + 5t
B. sin(2t) + 7t
C. sin(2t) + 9t
D. 16sin(2t) + 7t

Answer 1

The velocity of the particle is given by

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

Since $a(t)=-4\sin2t$ and $v(0)=7$, we get

$\displaystyle\int_0^ta(u)\,\mathrm du=\int_0^t-4\sin2u\,\mathrm du=2\cos2u\bigg|_0^t=2\cos2t-2$

$\implies v(t)=2\cos2t+5$

Similarly, the position function is obtained via

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

We know $v(t)$ and we're told that $x(0)=0$, so

$\displaystyle\int_0^tv(u)\,\mathrm du=\int_0^t(2\cos2u+5)\,\mathrm du=\sin2u+5u\bigg|_0^t=\sin2t+5t$

$\implies x(t)=\sin2t+5t$

making the answer A.