Question

At time t = 0 a car has a velocity of 16 m/s. It slows down with an acceleration given by –0.50t, in m/s^2 for t in seconds. By the time it stops it has traveled: A) 15 m B) 31 m C) 62 m D) 85 m E) 100 m

Answer 1

The car's velocity at time $t$ is

$v(t)=16\dfrac{\rm m}{\rm s}+\displaystyle\int_0^t\left(\left(-0.50\frac{\rm m}{\mathrm s^2}\right)u\right)\,\mathrm du=16\dfrac{\rm m}{\rm s}+\left(-0.25\dfrac{\rm m}{\mathrm s^2}\right)t^2$

It comes to rest at

$v(t)=0\implies16\dfrac{\rm m}{\rm s}=\left(0.25\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=8.0\,\mathrm s$

Its velocity over this period is positive, so that the total distance the car travels is

$\displaystyle\int_0^{8.0}v(t)\,\mathrm dt=\left(16\dfrac{\rm m}{\rm s}\right)(8.0\,\mathrm s)+\frac13\left(-0.25\dfrac{\rm m}{\mathrm s^2}\right)(8.0\,\mathrm s)^3=\boxed{85\,\mathrm m}$

so the answer is D.