two particles P an Q move on a line of greatest slope of a smooth inclined plane. the particles start at the same instant and from the same point, each with speed 1.3 m/s. initially P moves down the plane and Q moves up the plane. the distance between the particles t seconds after they start to move is d metres. 1. show that d= 2.6t when t=2.5 the difference in the vertical height of the particles is 1.6m. find: 2. the acceleration of the particles down the plane, 3. the distance travelled by P when Q is at its highest point.
this can be done by applying maths. you just need to use this physical concept: v=s/t do you need further help?
What is the angle of the plane to the horizontal?
the angle is not given
note that it's speed. the angle isn't really need. basically it's |dw:1354616449530:dw|
speed of approach * e = speed of separation?
but what is meant by greatest slope?
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