Question

The parachute on a race car of weight 8150 N opens at the end of a quarter mile run when the car is traveling at 33.6 m/s. What total retarding force must be supplied by the parachute to stop the car in a distance of 1090 m?

Answer 1

Given that :

weight of the car, W = 8150 N

initial velocity of the car, v₀ = 33.6 m/s

we know that, W = mg

or     m = W / g                                                        { eq.1 }

where, g = acceleration due to gravity = 9.8 m/s²

inserting the values in above eq.

m = (8150 N) / (9.8 m/s²)

m = 831.6 kg

now, using equation of motion 3 :

v² = v₀² + 2 a s                                                   { eq.2 }

where, v = final velocitry = 0 m/s

s = distance travelled = 1090 m

inserting the values in eq.2,

(0 m/s)² = (33.6 m/s)² + 2 a (1090 m)

- (1128.9 m²/s²) = (2180 m) a

a = - (1128.9 m²/s²) / (2180 m)

a = -0.517 m/s²

The total retarding force must be supplied by the parachute to stop the car which will be given as ::

F_t = m a                                                                                { eq.3 }

inserting the values in eq.3,

F_t = (831.6 kg) (-0.517 m/s²)

F_t = - 429.93 N