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A human being cannot tolerate acceleration of much more than a few hundred m/s2

ID: 2181001 • Letter: A

Question

A human being cannot tolerate acceleration of much more than a few hundred m/s2 without sustaining serious permanent injury. If the acceleration is sustained for more than a few seconds, the maximum tolerable acceleration is much less. Suppose a car is initially moving at 29.7 m/s and then is quickly stopped in a collision.

What is the minimum distance the driver must be allowed to move forward during the deceleration? (Air bags are an improvement over seat belts because the distance the driver moves forward is greater with a compressible air bag than with a seat belt. This does not mean, however, that you would want to move forward in the car without air bags.)

Explanation / Answer

) Draw a picture of a car with two distinct points in time. The first is the car at constant velocity. The second point is the car stopped, in the collision with v= 0. Now list your known values. We know v1= 29.7 m/s and v2= 0 m/s. We know the acceleration= -9.80 x 10^1 m/s/s.The negative sign implicates an deceleration. We just need to know the time period between points 1 and 2. Acceleration= change in velocity/change in time. Rearrange to show that change in time= change in velocity/acceleration= v2-v1/a. (0-29.7)/-9.8 x 10^1 m/s/s= approximately 0.30 seconds. This is the minimum time needed to stop so that this deceleration can be reached. It is the minimum simply because it is the first time that can reach a deceleration of -9.8 x 10^1. b) We know the time it took to stop is 0.30 seconds. We know the deceleration is -9.8 x 10^1 m/s/s. Let's use the kinematic equation s (displacement) = 0.5 x [u (first velocity) + v (second velocity)] x t (time)= 1/2 (u+ v) (t)= 0.5 x (29.7 +0) (0.30)=4.45m approximately. This is the minimum distance the CAR has to move forward. To determine the driver's condition inside the car involves making assumptions and estimating, something that I don't think I could help you with.

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