The front 1.20 m of a 1,300-kg car is designed as a \"crumple zone\" that collap

Question

The front 1.20 m of a 1,300-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a car traveling 27.0 m/s stops uniformly in 1.20 m, how long does the collision last? Incorrect: Your answer is incorrect. Use the relation between velocity, acceleration, and distance to find the distance the car travels in order to slow at uniform acceleration to a stop. s (b) What is the magnitude of the average force on the car? N (c) What is the magnitude of the acceleration of the car? Express the acceleration as a multiple of the acceleration of gravity. g

Explanation / Answer

For part A:

We know the stopping distance, 1.20m, we know the final velocity, 0m/s (its stopped), and the starting velocity, 25m/s. Thats all the information needed to solve for time.

d=t(v2+v1)/2 Rearange to solve for t, and remove the v2^2 as its zero
t=2d/v1 Plug in and solve
t=2(1.20m)/(27m/s)
t=0.0888 seconds.

Part B:

This is a question about impulse.
Ft=m(v2-v1)
v2=zero, so we can get rid of that, and bring t over to the side:
F=m(-v1)/t
PLug in and solve
F=(1300kg)(-27m/s)/(0.0888s)
F=-394914.5N, or -3.94x10^5N which correct significant digits
The reason it is negative is because it is a force opposite the direction of motion, slowing it down.

Part C:

Acceleration of the car can now be found in two seperate ways: using Newtons law, F=ma, or using kinematics, a=(v2-v1)/t. I like to avoid using numbers I solved myself, because if you make a mistake in the previous question you end up making a mistake in this one too, so I would use a=(v2-v1)t

v2^2=v1^2+2ad drop the v2, its zero, and solve for a
a=-v1^2/2d
a=10.4m/s

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