Question

. A sports car is advertised as being able to go from 0 to 60 in 6.00 seconds. If 60 mi/h is equal to 27 m/s, what is the sports car's average acceleration?​

Answer 1

Answer:

Approximately $2.2\; {\rm m\cdot s^{-2}}$.

Explanation:

Acceleration is the rate at which velocity changes.

In this example, velocity has changed from $v_{0} = 0\; {\rm m\cdot s^{-1}}$ to $v_{1} = 27\; {\rm m\cdot s^{-1}}$. The total change in velocity is:

$\begin{aligned}\Delta v &= v_{1} - v_{0} \\ &= 27\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}} \\ &= 27\; {\rm m\cdot s^{-1}} \end{aligned}$.

This change happened over a period of $\Delta t = 6.00\; {\rm s}$. Therefore:

$\begin{aligned} & (\text{avg. acceleration}) \\ =\; & (\text{avg. rate of change in velocity}) \\ =\; & \frac{(\text{change in velocity})}{(\text{time required})}\\ =\; & \frac{\Delta v}{\Delta t} \\ \approx\; & \frac{27\; {\rm m\cdot s^{-1}}}{6.00\; {\rm s}} \\ \approx \; & 2.2\; {\rm m\cdot s^{-2}} \end{aligned}$.