The 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 don't 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.
A ball moves at 33m/s through a wall of donuts 31 cm thick. It comesout of the donut wall at 18 m/s. How much time does it spend in the wall of donuts? How thick would the wall of dounts need to be to stop the ball entirely?
Is the answer 1.3 secondsand the wall would need to be 27 cm thick?
2 answers
Assuming 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
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
1 answer