Question

A car traveling at 91.0 km/h approaches the turn off for a restaurant 30.0 m ahead. If the driver slams on the brakes with the an acceleration of -6.40m/s^2what will her stopping distance

Answer 1

Answer: 49.92 m

Explanation:

In this situation the following equation will be useful:

$V^{2}=V_{o}^{2} +2 a d$

Where:

$V=0 m/s$ is the final velocity of the car, when it finally stops

$V_{o}=91 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=25.27 m/s$ is the initial velocity of the car

$a=-6.4 m/s^{2}$ is the constant acceleration of the car after the driver slams on the brakes

$d$ is the stopping distance

Isolating $d$:

$d=\frac{-V_{o}^{2}}{2a}$

$d=\frac{-(25.27 m/s)^{2}}{2(-6.4 m/s^{2})}$

$d=41.919 m \approx 41.92 m$