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October 1, 2014

October 1, 2014

Posted by **George** on Saturday, January 10, 2009 at 6:19pm.

Is the answer 1.3 secondsand the wall would need to be 27 cm thick?

- Physics -
**bobpursley**, Saturday, January 10, 2009 at 6:27pmThe average velocity in the donuts is

(33+18)/2 or 25.5m/s

avgvelocity=distance/time

you know avgvelocity, and distance, solve for time.

I dont get your time.

Now, for the thickness. find the acceleration of the ball in the donuts

Vf^2=Vi^2 + 2ad (solve for a)

now use the same equation with that a to solve for d if Vf is zero.

- Physics -
**drwls**, Saturday, January 10, 2009 at 6:28pmAssuming uniform deceleration, which may not be valid, the average speed going through is 25.5 m/s and the time is 0.31 m/25.5 m/s = 0.012 s

You are quite far off. Did you forget to express the wall thickness in meters?

Your last answer is also incorrect. If the ball goes through 31 cm that quickly, while losing less than half its speed, why would a thinnner wall stop it?

Assume speed reduction is proportional to thickness.

15/33 = 31/x

x is the required thickness

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