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course Phy 121
9/25 10:18 pm
An object is given an unknown initial velocity up a long ramp on which its acceleration is known to have magnitude 12 cm/s^2. .206 seconds later it passes a point 7.4 cm up the ramp from its initial position.
What are its possible initial velocities, and what is a possible scenario for each?
What is the maximum distance the object travels up the ramp?
If the acceleration is -12cm/s^2:
`dv=a*`dt = -12cm/s^2 * .206s = -58cm/s (approx)
vAVe = `ds/`dt = 7.4/.206 = 36cm/s (approx)
(vf+v0)/2 = 36cm/s
vf+v0 = 72cm/s
vf-v0 = -58cm/s Add these two equations together
2vf = 14cm/s
vf= 7cm/s
vf+v0 = 72cm/s
7cm/s + v0 = 72cm/s
v0 = 65cm/s
In this case the ball travels up the ramp until it reaches 0 velocity
`dv = -65cm/s
a= `dv/`dt
`dt = `dv/a
`dt =- 65cm/s/-12cm/s^2 = 5.4 s
vAve =32.5cm/s
32.5cm/s *5.4s = 176cm (approx)
Another scenario is that the acceleraion is positive, but then I don't know how that is possible given that we are going up a ramp. I'll give it a shot though. But I don't know how we'd ever figure out how far it could go. I suppose it could go the full length of the ramp.
dv=a*`dt = 12cm/s^2 * .206s = 58cm/s (approx)
vAVe = `ds/`dt = 7.4/.206 = 36cm/s (approx)
(vf+v0)/2 = 36cm/s
vf+v0 = 72cm/s
vf-v0 = 58cm/s Add these two equations together
2vf = 130 cm/s
vf= 65cm/s
vf+v0 = 72cm/s
65cm/s +v0 = 72cm/s
v0 = 7cm/s
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Excellent reasoning.
However -12 * .206 isn't -58 (see your first step).
This would be very easy to fix, but since everything but that one number is right there is no need to do so or to resubmit.
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