Question

(a) Calculate the force needed to bring a 900 kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a nonpanic stop). N(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). (force in part (b) / force in part (a))Additional Materials

Answer 1

(a)

m= mass = 900 kg

vi =initial speed = 90 km/h = 25m/s

vf= final speed = 0 (rest)

d = distance = 120 m

Apply:

vf^2 - vi^2 = 2ad

Rearrange for acceleration

a = (vf^2 - vi^2) / 2d

a = (0 - 25^2 ) / 2 (120)

a = -2.6m/s^2

Calculate force:

F= ma = 900kg (-2.6 m/s^2 ) = -2,340 N

A negative sign means that the force is applied against the motion direction.

b)

d = 2 m

a = ( 0 - 25^2 ) / 2 (2) = -156.25 N

F = ma = 900 (-156.25 ) = -140,625 N

It is larger than the force exerted previously

(force in part (b) / force in part (a)) = -140,625 / -2340 = 60.09