Question

(a) Calculate the force needed to bring a 900 kg car to rest from a speed of 95.0 km/h in a distance of 105 m (a fairly typical distance for a nonpanic stop).

(b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a), i.e. find the ratio of the force in part(b) to the force in part(a).

Answer 1

Solution:

a) First find the acceration,

u = 95km/h = 26.4m/s

Using kinamatics,

v² - u² = 2aS

a = 0² - 26.4²/2*105 = - 3.316 m/s².

so, Using newtons law,

F_net = ma = 900*-3.316 = - 2984.5 N.

b) Similarly, find the acceration of the car,

Using kinamatics,

v² - u² = 2aS

a = 0² - 26.4²/2*2 = - 174.1 m/s².

so, Using newtons law,

F_net = ma = 900*-174.1 = - 1.57*10⁵ N.

Ration is,

F_b/F_a = 52.5.

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