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Answer to Question #78002 in Mechanics | Relativity for Khyu
Question #78002
A particle moving in a straight line with constant acceleration passes in succession A,B,C. The time taken from A to B is t and from B to C is 2t, AB = a, BC = b. Prove that the acceleration is (b-2a)÷(3t^2) and find the velocity at B. Assuming the initial velocity equals to u.
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