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The distance transversed by a particle moving along a straight line is given by x = 180t +50 t 2 metre . find (1) the initial velocity of a particle. (2) the velocity at the end of 4s and (3) the acceleration of the particle |
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Answer» - Initial velocity = 180 m/s Velocity at 4 s = 580 m/s Acceleration = 100 m/s^2 ◆ Explanation - Displacement of PARTICLE is GIVEN by - x = 180t + 50t^2 Differentiate w.r.t. t, dx/dt = 180 + 50×2t V = 180 + 100t Acceleration of the particle is given by - dv/dt = 0 + 100 a = 100 m/s^2 Initial velocity of particle is - v1 = 180 + 100×0 v1 = 180 m/s Velocity of particle at 4 s is - v2 = 180 + 100×4 v2 = 580 m/s |
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