course Phy 231
An object is given an unknown initial velocity up a long ramp on which its acceleration is known to have magnitude 10 cm/s^2. .123 seconds later it passes a point 6.6 cm up the ramp from its initial position. •What are its possible initial velocities, and what is a possible scenario for each?
Answer: ds = v0*dt + .5*a*dt^2, v0*dt = ds - .5adt^2, v0=(ds-.5adt^2)/(dt) = (6.6cm-.5(10cm/s/s)(.015s^2))/.123s = 53.0cm/s. dt = dv/acc = 53.0cm/s / 10cm/s/s = 5.3s.
53.0 cm/s is the initial velocity for the .123 s interval, not the change in velocity.
-53.0 cm/s is the change in velocity from the start to the position of rest. However with a positive acceleration, this leads to a negative `dt. The ball would have been at rest 5.3 s earlier, assuming the acceleration had remained uniform during this interval.
There is another possible solution. Only the magnitude of the acceleration was given, so you may equally well assume acceleration - 10 cm/s^2. This will yield a slightly different initial velocity, and the time to rest will in this case be positive .
The ball starts at the bottom of the ramp at a velocity of 53.0cm/s. It then takes 5.3 seconds to get to its maximum distance up the ramp.
• What is the maximum distance the object travels up the ramp?
Answer: ave vel = ds/dt, ave vel = (53.0cm/s + 0cm/s)/2 = 26.5cm/s, ds=ave vel*dt = 26.5cm/s * 5.3s = 140.45 cm.
Good, but there are a couple of difficulties with your solution, which should be easy to remedy.
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