The parachute on a race car of weight 8280 N opens at the end of a quarter mile run when the car is traveling at 31.2 m/s. What total retarding force must be supplied by the parachute to stop the car in a distance of 1060 m?
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Start with finding the required acceleration.
v^2 = u^2 + 2ad, where v = 0 (final velocity), u = 31.2 m/s (initial velocity) and d is the distance (some places use 's' in this formula instead)
0^2 = 31.2^2 +2*a*1060
Rearranging gives that the acceleration is -0.46 m/s/s (the negative sign just means it's in the opposite direction to the initial velocity.
Now it's just Newton's second law, but first you need to convert from weight to mass, by dividing by 9.8, so the mass of the car is 845 kg.
F = ma = 845*-0.46 = -388.7 N (again, the negative sign means the force acts in the opposite direction to the initial velocity).
v^2 = u^2 + 2ad, where v = 0 (final velocity), u = 31.2 m/s (initial velocity) and d is the distance (some places use 's' in this formula instead)
0^2 = 31.2^2 +2*a*1060
Rearranging gives that the acceleration is -0.46 m/s/s (the negative sign just means it's in the opposite direction to the initial velocity.
Now it's just Newton's second law, but first you need to convert from weight to mass, by dividing by 9.8, so the mass of the car is 845 kg.
F = ma = 845*-0.46 = -388.7 N (again, the negative sign means the force acts in the opposite direction to the initial velocity).
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F * d = 1/2 m v^2
so F = 1/2 m v^2 / d
m = 8280/g = 8280/9.8 ( approx)
v = 31.2 m/s
d = 1060 m
Just substitute to get your answer
so F = 1/2 m v^2 / d
m = 8280/g = 8280/9.8 ( approx)
v = 31.2 m/s
d = 1060 m
Just substitute to get your answer