Question #130936
A motor car of mass 800kg travelling at 20ms(on top of it there is -1)is brought to rest by brakes in 100m.
Calculate the average braking force required
1
Expert's answer
2020-08-28T09:33:25-0400

Let us write equations of motion of the car, assuming it moves with constant retardation (and hence constant braking force):

S(t)=v0tat22S(t) = v_0 t - \frac{a t^2}{2}, v(t)=v0atv(t) = v_0 - a t.

At the moment of time, when the car stops completely v(t)=v0at=0v(t') = v_0 - a t' = 0, from where the time it takes to completely stop is t=v0at' = \frac{v_0}{a}.

The horizontal distance covered in time tt' is given (L=800m)L = 800m), therefore: L=S(t)=v02av022a=v022aL = S(t') = \frac{v_0^2}{a} - \frac{v_0^2}{2 a} = \frac{v_0^2}{2 a}, from where the acceleration is a=v022La = \frac{v_0^2}{2 L}, and from second Newton's law a=Fma = \frac{F}{m}, so v022L=Fm\frac{v_0^2}{2 L} = \frac{F}{m}, from where the force is F=mv022L=1600NF = \frac{m v_0^2}{2 L} = 1600 N.


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