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

Answer:

86.10941 m

Explanation:

Force = F = $8.18\times 10^3\ N$

m = Mass = $1.15\times 10^3\ kg$

u = Initial velocity = 35 m/s

v = Final velocity = 0

Work done

$W=F\times s$

$W=\frac{1}{2}m(v^2-u^2)\\\Rightarrow Fs=\frac{1}{2}m(v^2-u^2)\\\Rightarrow s=\frac{\frac{1}{2}m(v^2-u^2)}{F}\\\Rightarrow s=\frac{\frac{1}{2}1.15\times 10^3(0^2-35^2)}{8.18\times 10^3}\\\Rightarrow s=-86.10941$

The car should brake 86.10941 m before the traffic