The front 1.20 m of a 1300 kg car is designed as a "crumple zone" that collapses to absorb the shock of a coll

Question

The front 1.20 m of a 1300 kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision.

(a) If a car traveling 25.0 m/s stops uniformly in 1.20 m, how long does the collision last?

(b) What is the magnitude of the average force on the car?

(c) What is the acceleration of the car? Express the acceleration as a multiple of the acceleration of gravity.

Answer 1

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)/(25m/s)

t=0.096seconds.

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)(-25m/s)/(0.096s)

F=-338541N, or -3.39x10^5N 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

I think i may have made a mistake somewhere in the previous two questions, but I cant find it. If so I apologize. If you need any more help, contact me on yahoo messenger. My name is [e-mail]