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A motorcycle,travelling east starts from rest ,moves in a straight line with a constant acceleration and covers a distance of 64m in 4s.calculate (a)its acceleration(b)it's final velocity(c)at what time the motorcycle had covered half the total distance.(d)what distance the motorcycle had covered in half the total time |
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Answer» Answer: (a) 8 m/s^2(B) 32 m/s(c) 2.82 s ( approx.)(d) 16 mExplanation: Given: constant acceleration ( => we can USE the equations of motion)u = 0 m/s s = 64 m t = 4s Since he is travelling in one direction only, every vector quantity will be +ve Average velocity = TOTAL displacement / total time Here, since the motorcycle has linear motion in one direction, DISTANCE = displacement = 64m Avg. velocity = 64 /4 = 16 m/s Also, for a given time, avg. velocity = ( u + v) /2 => 16 = ( 0 + v) /2 => v = 32 m/s => final velocity = 32 m/s a = ( v - u) /t = 32/4 = 8 m/s^2 => acceleration = 8 m/s^2 Now , s = 1/2 x 64 = 32m a = 8 m/s^2 u = 0 m/s t = ? s = ut + 1/2at^2 => 32 = 0 + 1/2 x 8 x t^2 => 32 = 4t^2 => t^2 = 8 => t = 2.82 seconds ( approx) The motorcyclist covers half the total distance in 2.82 seconds. Now, t = 1/2 x 4 = 2 seconds a = 8 m/s^2 u = 0 m/s s = ? s = ut + 1/2at^2 => s = 1/2 x 8 x 2 x 2 => s = 16m The motorcyclist covered a distance of 16m in exactly 1/2 the total time. |
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