course Phy 241
An object is given an unknown initial velocity up a long ramp on which its acceleration is known to have magnitude 13 cm/s^2. .179 seconds later it passes a point 5 cm up the ramp from its initial position. •What are its possible initial velocities, and what is a possible scenario for each?
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• What is the maximum distance the object travels up the ramp?
The average velocity can first be found by taking 5 cm and dividing by .179 seconds to yield 27.9 seconds this would be 27.9 cm/s
. If a=`dv/`dt then `dv must equal 13*.179 or 2.327.
Vf and V0 are unknowns so we can use the fact that Vf-V0=2.327 and let Vf=V0+2.327. Then we can plug that into the equation (Vf+v0)/2 to equal 27.9.
Therefore the equation to solve for V0 looks like {(2.327 +V0)+v0}/2=27.9. Solve for v0 to obtain 26.7 cm/s.
If the acceleration is constant along the ramp, the maximum distance gone would be the distance of the ramp.
Excellent reasoning. However only the magnitude of the acceleration was given. If the ball travels up the ramp, the acceleration will have a sign opposite that of the displacement. So either the 5 cm is negative, or the acceleration is negative.
If we choose the displacement to be positive, then, the acceleration is negative and the change in velocity is negative. This leads to {(-2.327 +V0)+v0}/2=27.9, with solution indicating v0 = 29.1 cm/s, approximately.
The maximum distance up the ramp will occur over a time interval characterized by v0 = 29.1 cm/s, vf = 0 and a = -13 cm/s^2. You can solve this situation for `ds and, if you wish, `dt.
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