A 825-kg jeep comes to rest from a speed of 87.5 km/h in a distance of 125 m. Co

Question

A 825-kg jeep comes to rest from a speed of 87.5 km/h in a distance of 125 m. Consider that the kepis initially traveling in the positive direction. (a) If the brakes are the only thing making the jeep come to a stop, calculate the force (in newtons, in a component along the direction of motion of the jeep) that the brakes apply on the jeep. (b) Suppose instead of braking that the jeep hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force, in newtons, exerted on the jeep in this case. (c) What is the ratio of the force on the jeep from the concrete to the braking force? F_c/F_b =

Explanation / Answer

u =87.5 km/h =87.5*1000m/3600s =24.03 m/s

v=0

And,v2 = u2 + 2as

=>0 = 24.032 +2a*125

=> a = -2.309 m/s2

(a) Fb =ma = 825*(-2.309) = -1905.55 N

(b) In this case, v2 = u2 + 2as

=>0 = 24.032 +2a*2

=>a = -144.36 m/s2

Fc =ma =825*(-144.36) = -119097.18 N

(c) Therfore Fc/Fb =119097.18/1905.55 = 62.5

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