A car going 20 m/s stops in a distance of 50m. u_s = 0.5 and u_k = 0.3. Will a 70kg person slide off the seat?

Question

Please, I need this before tomorrow morning! Thanks.

Answer 1

Sorry, I am late for your homework, but wish that you may learn this method for your future work.
A car going 20 m/s stops in a distance of 50m. Its deceleration is -a = -(at)^2/{2*(0.5at^2)} = -v^2/2s = -(20)^2/(2*50) = -4(m/s^2)
To keep a 70kg person in the seat without sliding means this person also has a deceleration of -4(m/s^2). A force of (ma) = (70kg)*(-4(m/s^2)) = -280(N) is required. The seat can provide a static friction force up to: 70*g*0.5 = 70*9.8*0.5 = 343(N), which is sufficient to carry the person with the car seat together. So this person will not slide.