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course Phy 121
October 26, 2012 @ 12:41 PM
At clock time t = 4 sec, a ball rolling straight down a hill is moving at 5 m/s and is 78 m from the top of the hill, while at clock time t = 9 sec it is moving at 7.5 m/s and is 114 m from the top of the hill. What is its average velocity during this time? What is its average acceleration during this time? Is it possible that the acceleration is uniform?t1 = 4 seconds; t2 = 9 seconds
v1 = 5 m/s; v2 = 7.5 m/s
x1 = 78 m; x2 = 114 m
Acceleration is .5 m/s^2
Average velocity is 6.25 m/s
Acceleration = (v2 - v1) / (t2 - t1) = .5 m/s^2
Average velocity = ((5 m/s + 7.5 m/s))/2 = 6.25 m/s
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I get confused with average velocity and the average of the initial and final velocities because in the given problems it says average velocity so I tjouhjy it would be the average of both the initial and final velocity instead of ds/dt = velocity.
t1 = 4 seconds; t2 = 9 seconds
v1 = 5 m/s; v2 = 7.5 m/s
x1 = 78 m; x2 = 114 m
What is the average velocity of this time?
Since we are trying to find the average velocity of the given time which I believe we are talking about the time interval of 4s - 9s then we would find the average velocity by..
x1 = 78 m
x2 = 114 m
t1 = 4s
t2 = 9s
ds/dt = vAve
(114 m - 78 m) / (9s - 4s) = vAve
(36 m) / (5s) = vAve
average velocity is 7.2 m/s for the given time..
What is the average acceleration of this time?
The average acceleration for this time is..
v1 = 5 m/s
v2 = 7.5 m/s
t1 = 4s
t2 = 9s
aAve = dv/dt = (7.5 m/s - 5 m/s) / (9s - 4s)
aAve = (2.5 m/s) / (5s)
aAve = .5 m/s^2
The average acceleration for the given time is .5 m/s^2
Is it possible that the acceleration is uniform?
No it isn't possible that the acceleration is uniform because let us say that at the beginning of the hill it is 0 m/s with t = 0
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You can't say this. No information is given about where the ball is or what its velocity is at t = 0. You are only given information for t = 4 sec and t = 9 sec.
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Your reasoning is good, and would answer the question if your assumption was correct. It's only your assumption that has to be discounted, along with the conclusions you have drawn from it.
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If acceleration is constant then the average velocity will be equal to the average of the initial and final velocities.
You calculated the average average of the initial and final velocities in your original submissino and got 6.25 m/s.
You calculated the average velocity here and got 7.2 m/s.
The conclusion is that, since the two are not equal, the acceleration could not be constant.
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And the difference from the initial and the second time is 4s and 5 m/s
To get the acceleration of that given time interval is
(5 m/s) / 4s
Then the answer is 1.25 m/s^2
Which is the same acceleration as the second time interval which is .5 m/s^2
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Average velocity is defined to be the average rate of change of position with respect to clock time.
You have calculated the average of the initial and final velocities, not the average velocity.
If acceleration is uniform, the two will be identical.
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Since we don't know how it started in the beginning, we can't say if the acceleration could be uniform.""
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&#This looks good. See my notes. Let me know if you have any questions. &#