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A particle starts its motion from the origin and at time t its velocity parallel to coordinate axes remain (2t+3) and (4t) respectively.find the equation of its path.
August 26, 2016 at 3:51pm
If you have velocity, you have to integrate it to get displacement at different time, which also known as equation of its path. velocity vector = <2t+3, 4t> By integrating velocity, displacement vector = <2(t^2)/2 + 3t + c, 4(t^2)/2 + c> simplifying, you will get: displacement vector = <t^2 + 3t + c, 2t^2 + c> At t = 0, particle is at origin. So, x = 0, y = 0. By substituting t = 0, (0)^2 + 3(0) + c = 0 c = 0 2(0^2) + c = 0 c = 0 displacement or equation of its path = <t^2 + 3t, 2t^2>
August 27, 2016 at 1:25pm
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