Question

Use the worked example above to help you solve this problem. The driver of a 1.15 103 kg car traveling on the interstate at 35.0 m/s (nearly 80.0 mph) slams on his brakes to avoid hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the brakes are applied, a constant kinetic friction force of magnitude 8.18 103 N acts on the car. Ignore air resistance. At what minimum distance should the brakes be applied to avoid a collision with the other vehicle?

Answer 1

86.10941 m

Step-by-step explanation:

Force = F =
`8.18* 10^3\ N`

m = Mass =
`1.15* 10^3\ kg`

u = Initial velocity = 35 m/s

v = Final velocity = 0

Work done

`W=F* s`

`W=(1)/(2)m(v^2-u^2)\\\Rightarrow Fs=(1)/(2)m(v^2-u^2)\\\Rightarrow s=((1)/(2)m(v^2-u^2))/(F)\\\Rightarrow s=((1)/(2)1.15* 10^3(0^2-35^2))/(8.18* 10^3)\\\Rightarrow s=-86.10941`

The car should brake 86.10941 m before the traffic