Question

An automobile accelerates from rest at 1.7 m/s 2 for 22 s. The speed is then held constant for 29 s, after which there is an acceleration of −5.8 m/s 2 until the automobile stops. What total distance was traveled? Answer in units of km.

Answer 1

After 22 s, the car has velocity

$v=\left(1.7\dfrac{\rm m}{\mathrm s^2}\right)(22\,\mathrm s)=36.4\dfrac{\rm m}{\rm s}$

In this time, it will have traveled a distance of

$\dfrac12\left(1.7\dfrac{\rm m}{\mathrm s^2}\right)(22\,\mathrm s)^2=411.4\,\mathrm m$

Over the next 29 s, the car moves at a constant velocity of 36.4 m/s, so that it covers a distance of

$\left(36.4\dfrac{\rm m}{\rm s}\right)(29\,\mathrm s)=1055.6\,\mathrm m$

so that after the first 51 s, the car will have moved 1467 m.

After the 29 s interval of constant speed, the car's negative acceleration kicks in, so that its velocity at time $t$ is

$v(t)=36.4\dfrac{\rm m}{\rm s}+\left(-5.8\dfrac{\rm m}{\mathrm s^2}\right)t$

The car comes to rest when $v(t)=0$:

$36.4-5.8t=0\implies t=6.3$

That is, it comes to rest about 6.3 s after the first 51 s. In this interval, it will have traveled

$\left(36.4\dfrac{\rm m}{\rm s}\right)(6.3\,\mathrm s)+\dfrac12\left(-5.8\dfrac{\rm m}{\mathrm s^2}\right)(6.3\,\mathrm s)^2=114.2\,\mathrm m$

so that after 57.3 s, the total distance traveled by the car is 1581.2 m, or about 1.6 km.