After 22 s, the car has velocity
[tex]v=\left(1.7\dfrac{\rm m}{\mathrm s^2}\right)(22\,\mathrm s)=36.4\dfrac{\rm m}{\rm s}[/tex]
In this time, it will have traveled a distance of
[tex]\dfrac12\left(1.7\dfrac{\rm m}{\mathrm s^2}\right)(22\,\mathrm s)^2=411.4\,\mathrm m[/tex]
Over the next 29 s, the car moves at a constant velocity of 36.4 m/s, so that it covers a distance of
[tex]\left(36.4\dfrac{\rm m}{\rm s}\right)(29\,\mathrm s)=1055.6\,\mathrm m[/tex]
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 [tex]t[/tex] is
[tex]v(t)=36.4\dfrac{\rm m}{\rm s}+\left(-5.8\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
The car comes to rest when [tex]v(t)=0[/tex]:
[tex]36.4-5.8t=0\implies t=6.3[/tex]
That is, it comes to rest about 6.3 s after the first 51 s. In this interval, it will have traveled
[tex]\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[/tex]
so that after 57.3 s, the total distance traveled by the car is 1581.2 m, or about 1.6 km.
