Question

(a) Calculate the force needed to bring a 1050 kg car to rest from a speed of 95.0 km/h in a distance of 130 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

here,

a)

mass of car , m = 1050 kg

initial speed , u = 95 km/h= 26.39 m/s

s1 = 130 m

let the accelration be a1

v^2 - u^2 = 2 *a1 * s1

0 - 26.39^2 = 2 * a1 *130

solving for a1

a1 = - 2.68 m/s^2

the force required , F1 = m1 * a1

F1 = - 2.68 * 1050 = - 2.81 * 10^3 N

b)

initial speed , u = 95 km/h= 26.39 m/s

s2 = 2 0m

let the accelration be a2

v^2 - u^2 = 2 *a2 * s2

0 - 26.39^2 = 2 * a2 *2

solving for a2

a2 = - 174.1 m/s^2

the force required , F2 = m2 * a2

F2 = - 174.1 * 1050 = - 1.83 * 10^5 N

c)

the ratio of force in part(b) to force in part(a) , R = F2 /F!

R = 65.1