(a) For the first 6.2 s, the car has velocity at time t given by
[tex]v(t)=13.5\dfrac{\rm m}{\rm s}+\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
so that after 6.2 s, it attains a velocity of
[tex]v(6.2\,\mathrm s)=25.28\dfrac{\rm m}{\rm s}[/tex]
For any time t after 6.2 s, its velocity is given by
[tex]v(t)=25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
which tells us the velocity only falls from this point onward. This means the maximum speed is 25.28 m/s, or about 25.3 m/s.
(b) Solve for t (after 6.2 s) that makes v(t) = 0 :
[tex]25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t=0[/tex]
[tex]\implies t\approx21.067\,\mathrm s[/tex]
It takes the car about 21.2 s to come to a rest, so the car travels a total of about 6.2 s + 21.2 s = 27.4 s.
(c) For the first 6.2 s, the car undergoes a displacement at time t of
[tex]x(t)=\left(13.5\dfrac{\rm m}{\rm s}\right)(6.2\,\mathrm s)+\dfrac12\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)(6.2\,\mathrm s)^2[/tex]
[tex]\implies x\approx120.218\,\mathrm m[/tex]
For time t beyond 6.2 s, its displacement is
[tex]x(t)=120.218\,\mathrm m+\left(25.28\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
The car comes to a rest after 21.2 s (accelerating at a rate of -1.2 m/s^2), so that its total displacement is
[tex]x(21.2\,\mathrm s)\approx386.49\,\mathrm m[/tex]
so the car travels a total distance of about 387 m.
