## Skidding

Consider a 500 kg motorcycle is moving at 20 m/s. At time t = 0, the rider pulls the emergency brake, causing the bike to skid. If the static friction between the rubber wheels and the concrete is 0.7, and the kinetic friction is 0.3, what is the motorcycle's speed at t = 2 seconds? Hint
Because the bike is sliding, kinetic friction is relevant:
$$F_{friction}=F_{normal}\times \mu_k$$$where $$\mu_k$$ is the kinetic friction constant and $$F_{normal}$$ is the normal force. Hint 2 Solve for the normal force: $$F=mass\times acceleration$$$
Because the bike is sliding, kinetic friction is relevant:
$$F_{friction}=F_{normal}\times \mu_k$$$where $$\mu_k$$ is the kinetic friction constant and $$F_{normal}$$ is the normal force. Since $$F=mass\times acceleration$$ , the normal force is: $$F_{normal}=ma=(500kg)(9.8m/s^2)=4,900\:N$$$
Then the friction force is:
$$F_{friction}=(4,900N)(0.3)=1,470\:N$$$To find the deceleration due to friction: $$F_{f}=1,470N=ma\rightarrow a=\frac{F_f}{m}=\frac{1,470N}{500kg}=2.94\:m/s^2$$$
After $$t=2$$ , the speed will be reduced to:
$$v_{final}=v_{initial}+a(t)$$$$$v_{final}=20m/s+(-2.94m/s^2)(2s)=14.12\:m/s$$$
14.12 m/s