Answer:
Acceleration: approximately [tex]33.56\; {\rm m\cdot s^{-2}}[/tex].
Final velocity: approximately [tex]164.9\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
Let [tex]x[/tex] denote displacement. Let [tex]a[/tex] denote the acceleration of the vehicle. Let [tex]t[/tex] denote the duration of the acceleration. Let [tex]u[/tex] denote the initial velocity of the vehicle.
In this question, it is given that [tex]x = 405.0\; {\rm m}[/tex], [tex]t = 4.913\; {\rm s}[/tex], and [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] since the vehicle started from rest. Acceleration [tex]a[/tex] is to be found.
Since acceleration [tex]a[/tex] is constant, apply the SUVAT equation [tex]x = (1/2)\, a\, t^{2} + u\, t[/tex] and solve for acceleration [tex]a[/tex].
Note that since [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex], the SUVAT equation becomes [tex]x = (1/2)\, a\, t^{2}[/tex].
Rearrange and solve this equation for [tex]a[/tex]:
[tex]\begin{aligned}a &= \frac{2\, x}{t^{2}} \\ &= \frac{2 \, (405.0\; {\rm m})}{(4.913\; {\rm s})^{2}} \\ &\approx 33.558\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
Let [tex]v[/tex] denote the velocity of the vehicle after the acceleration. Apply the SUVAT equation [tex]v = u + a\, t[/tex] to find this velocity:
[tex]\begin{aligned}v &= u + a\, t \\ &= (0\; {\rm m\cdot s^{-1}}) + (33.558\; {\rm m\cdot s^{-2}})\, (4.913\; {\rm s}) \\ &\approx 164.9\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
