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

Expert's answer

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

S(t)=v0t−at22S(t) = v_0 t - \frac{a t^2}{2}, v(t)=v0−atv(t) = v_0 - a t.

At the moment of time, when the car stops completely v(t′)=v0−at′=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 t′t' is given (L=800m)L = 800m), therefore: L=S(t′)=v02a−v022a=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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