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?
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.
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
To find the time spent in the wall of donuts, you can first calculate the acceleration experienced by the ball while passing through the wall. Using the equation of motion v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance, you can rearrange the equation to solve for the acceleration:
a = (v^2 - u^2) / (2s)
Plugging in the given values: u = 33 m/s, v = 18 m/s, and s = 31 cm (which needs to be converted to meters as 0.31 m):
a = (18^2 - 33^2) / (2 * 0.31)
Simplifying the expression, you get:
a ≈ -54.49 m/s^2
Notice that the acceleration is negative because the ball is decelerating as it passes through the wall.
Now, to calculate the time spent in the wall, you can use the equation of motion v = u + at, where t is the time:
18 = 33 + (-54.49)t
Rearranging the equation, you get:
54.49t = 33 - 18
54.49t = 15
t ≈ 0.28 seconds
So, the ball spends approximately 0.28 seconds in the wall of donuts.
To determine the thickness of the wall of donuts needed to stop the ball entirely, you need to find the distance traveled while experiencing deceleration. You can use the equation of motion v^2 = u^2 + 2as once again, but this time v = 0 since the ball comes to a stop:
0 = 33^2 + 2(-54.49)s
Simplifying the equation:
1782.57s = 1089
s ≈ 0.61 meters
To convert this distance to centimeters, multiply by 100:
0.61 * 100 = 61 cm
Therefore, the wall of donuts would need to be approximately 61 cm thick to stop the ball entirely. So, the answer you provided for the thickness of 27 cm is incorrect.