Physics Help Forum basic velocity/acceleration problems

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 Aug 4th 2013, 06:12 PM #1 Junior Member   Join Date: Aug 2013 Posts: 3 basic velocity/acceleration problems Hello all, just had a few study questions for a midterm tomorrow. Not sure how to work through these, even though at this point in the semester I should probably have a better idea. Anyways, they are as follows, and thanks for any help... In the NBA Finals Russell Westbrook goes for a spot up two point jump shot and misses. Lebron James then goes for the rebound. He bends down a third of his height of 2.03m and jumps, reaching his maximum vertical leap height of 0.112m. (a) What is his velocity when he leaves the floor so that he can reach his maximum leap height? (b) How long does it take for him to jump? What is his acceleration when jumping? Also have another similar problem. A car is speeding down I-35 doing 41 m/s. He passes a police officer sitting in his car at rest. The police car starts accelerating at a=2m/s as soon as the speeding car goes by. (a) How long does it take for the police car to catch up with the speeding car? (b) How far do they travel?
 Aug 5th 2013, 04:25 AM #2 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 1. (a) For the first question, are you sure the leap height is correct? 0.112 m is rather weak... Anyway, you have the leap height and the height of the player, you can get the change in height. Then use: v^2 + u^2 = 2as v is the velocity of the player at the peak of the leap (it will be equal to zero at the top), u is the velocity of the player just before leaving the floor, a is the acceleration due to gravity, here equal to -9.81 m/s^2 (or to the appropriate figure your question asks you to use) and s is the change in height. (b) Use v^2 + u^2 = 2as again, this time, but this time, u will be equal to 0, v will be the velocity you got in (a), s will be a third of the player's height and you have to calculate a to get the acceleration. Then use this acceleration in either of: s = ut + 0.5 at^2 v = u + at to get the time. 2. (a) You will need to generate two equations, one for the car and one fort he police, in terms of distance. That of the car is obtained from s = ut (constant velocity), the other for the police: s = 0.5at^2 Equate both equations for the distance to be equal to each other after substituting the relevant values to get the time for the police to catch up with the car. (b) Substitute back this time into any of the two equations you formed in (a). Post your workings! __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?

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