Answer:
Approximately [tex]2.27\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
Let [tex]a[/tex] denote the acceleration, [tex]u[/tex] denote initial velocity, [tex]t[/tex] denote the duration of acceleration, and [tex]x[/tex] denote displacement.
Note that [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] since the automobile started from rest. Additionally, it is given that within [tex]t= 16\; {\rm s}[/tex] the displacement was [tex]0.29\; {\rm km}[/tex]. Apply unit conversion and ensure that the unit of distance is meters (same as the unit of distance in acceleration) [tex]x = 0.29\; {\rm km} = (0.29 \times 10^{3})\; {\rm m} = 290\; {\rm m}[/tex].
Make use of the SUVAT equation [tex]x = (1/2)\, a\, t^{2} + u\, t[/tex] to find [tex]a[/tex] in terms of [tex]u[/tex], [tex]t[/tex], and [tex]x[/tex]. Rearrange the equation to obtain:
[tex]\begin{aligned}\frac{1}{2}\, a\, t^{2} + u\, t = x\end{aligned}[/tex].
[tex]\begin{aligned}\frac{1}{2}\, a\, t^{2} = x - u\, t\end{aligned}[/tex].
[tex]\begin{aligned}a = \frac{2\, (x - u\, t)}{t^{2}}\end{aligned}[/tex].
Substitute in [tex]x = 290\; {\rm m}[/tex], [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex], and [tex]t = 16\; {\rm s}[/tex] and solve for [tex]a[/tex]:
[tex]\begin{aligned}a &= \frac{2\, (x - u\, t)}{t^{2}} \\ &= \frac{2\, (290\; {\rm m} - 0\; {\rm m\cdot s^{-1}} \times 16\; {\rm s})}{(16\; {\rm s})^{2}} \\ &\approx 2.27\; {\rm m\cdot s^{-2}\end{aligned}[/tex].
