Question

Please help! It's due in one hour!

An automobile manufacturer claims that its product will travel 0.29 km in 16 s, starting from rest. What is the magnitude of the constant acceleration required to do this? Answer in units of m/s2.

Answer 1

Answer:

Approximately $2.27\; {\rm m\cdot s^{-2}}$.

Explanation:

Let $a$ denote the acceleration, $u$ denote initial velocity, $t$ denote the duration of acceleration, and $x$ denote displacement.

Note that $u = 0\; {\rm m\cdot s^{-1}}$ since the automobile started from rest. Additionally, it is given that within $t= 16\; {\rm s}$ the displacement was $0.29\; {\rm km}$. Apply unit conversion and ensure that the unit of distance is meters (same as the unit of distance in acceleration) $x = 0.29\; {\rm km} = (0.29 \times 10^{3})\; {\rm m} = 290\; {\rm m}$.

Make use of the SUVAT equation $x = (1/2)\, a\, t^{2} + u\, t$ to find $a$ in terms of $u$, $t$, and $x$. Rearrange the equation to obtain:

$\begin{aligned}\frac{1}{2}\, a\, t^{2} + u\, t = x\end{aligned}$.

$\begin{aligned}\frac{1}{2}\, a\, t^{2} = x - u\, t\end{aligned}$.

$\begin{aligned}a = \frac{2\, (x - u\, t)}{t^{2}}\end{aligned}$.

Substitute in $x = 290\; {\rm m}$, $u = 0\; {\rm m\cdot s^{-1}}$, and $t = 16\; {\rm s}$ and solve for $a$:

$\begin{aligned}a &= \frac{2\, (x - u\, t)}{t^{2}} \\ &= \frac{2\, (290\; {\rm m} - 0\; {\rm m\cdot s^{-1}} \times 16\; {\rm s})}{(16\; {\rm s})^{2}} \\ &\approx 2.27\; {\rm m\cdot s^{-2}\end{aligned}$.