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

Step-by-step 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 (km)/(h) (1000 m)/(1 km) (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=(-V_(o)^(2))/(2a)`

`d=(-(25.27 m/s)^(2))/(2(-6.4 m/s^(2)))`

`d=41.919 m \approx 41.92 m`