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

Approximately
`2.2\; {\rm m\cdot s^(-2)}`.

Step-by-step 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})}\\ =\; & (\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}`.