A 1 300-kg automobile moving at 21 m/s has the brakes applied with a deceleration of 3.0 m/s².How far does the car travel before it stops?

Question

Quizplus · Accepted Answer

To find the stopping distance, we can use the formula $d = \frac{v^2}{2a}$, where $v = 21 \, \text{m/s}$ is the initial velocity, and $a = 3.0 \, \text{m/s}^2$ is the deceleration. Plugging in the values, we get $d = \frac{(21)^2}{2 \times 3} = \frac{441}{6} = 73.5 \, \text{m}$, which is approximately $7.4 \times 10^1 \, \text{m}$.

A 1 300-Kg Automobile Moving at 21 M/s Has the Brakes