A bullet traveling at 580 m/s strikes a block of wood withthickness .25 meters.

Question

A bullet traveling at 580 m/s strikes a block of wood withthickness .25 meters. The bullet travels all the way throughthe wood. When it leaves the wood it is traveleing 220 m/s.

A.What is the bullets average velocity when it is in theblock of wood?

B.What is the acceleration, assumed constant, while thebullet is in the wood?

That is the problem. I am not really clear on how tofigure it out, but below is my attempt.

A. The average velocity would be the Final v- Inital v whichwould be 220-580=-360 or 360 m/s

B. This would be the average velocity which is -360 / thicknessof .25 = -1440 m/s2

I hope that I am doing that problem right. If not any helpwould be appreciated to explain where I am going wrong. Thanks.

Explanation / Answer

acceleartion=[v*v-u*u]/2s=[220*220-580*580]/2*0.25=48400-336400/0.5=-576000m/sec2 time =t=v-u/a=220-580/-576000=0.000625sec average velocity=total displacement/totaltime=0.25/0.000625=400m/sec

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