1. A car is moving at a relatively slow speed, let\'s say a good running speed f
ID: 1603801 • Letter: 1
Question
1. A car is moving at a relatively slow speed, let's say a good running speed for a short distance sprint, when it hits a solid wall (top figure).
The car comes to a complete stop, with a somewhat shorter front section now. The driver, who was not wearing a seat belt, tries to hold himself off the steering wheel.
Let us find the necessary average force acting on the 98-kg driver that is necessary to bring him to a complete stop before hitting the steering wheel or windshield. This is also the force the driver needs to exert on the steering wheel (Newton's 3rd Law).
The distance that is available to achieve this, as we will see impossible feat, is the original distance to the steering wheel plus the amount by which the front section shortens. For now, let's assume a probably too generous distance of 1 m.
I want you to work this out two different ways:
Use your knowledge about motion at constant acceleration to find the acceleration, and then use Newton's 2nd Law to find the magnitude of the force.
Use the Work-Kinetic Energy Theorem to find the magnitude of the force.
Of course, you should expect to get the same answer with both methods. Below answer with the force in pounds (lbs).
2.
Same problem as before, just with a speed of 23 mph and a total distance for the driver to slow down of 2 ft. And, we don't know the driver's mass.
Again find the magnitude of the average force necessary to bring the driver to a complete stop. Express this force in multiples of the driver's weight, mg. For example, if the answer should be 1.5 times the weight, answer 1.5mg.
Explanation / Answer
1.
d = stopping distance = 1
Vi = initial speed
Vf = final speed = 0
a = acceleration
using the equation
Vf2 = Vi2 + 2 a d
a = (Vf2 - Vi2 )/(2d)
a = (Vf2 - Vi2 )/2 = - Vi2 /2
m = mass = 98 kg
using newton's second law
F = ma
F = 98 (Vf2 - Vi2 )/2 = 49 (Vf2 - Vi2 ) = - 49 Vi2
using work-change in kinetic energy theorem
F d = (0.5) m (Vf2 - Vi2)
F d = (0.5) m (- Vi2)
F (1) = (0.5) (98) (- Vi2)
F = - 49 Vi2
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