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A particle has initial velocity `12 m//s`. It moves due to a constant force along the line of velocity which initially produces retardation of `8 m//s^(2)`. Then :A. the distance travelled in first 3 second is 0 mB. the distance travelled in first 3 second is `36 m`C. the distance travelled in first 3 second is `18 m`D. displacement in first 3 second is 0 m |
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Answer» Correct Answer - C::D `u = 12 m//s` `a = -8 //s^(2)` Displacement `vec(S) = vec(u) t + (1)/(2) vec(a) t^(2)` `= 12 xx 3 + (1)/(2) xx (-8) xx 3^(2) = 0` Distance `v = u + at` `0 = 12 - 8t` `t = (3)/(2)s at t = (3)/(2)s` velocity becomes zero & direction of motion reverses Distance `t = 0 " to " t = (3)/(2)s` `S = ut + (1)/(2) at^(2) = 9 m` Distance `t (3)/(2)s "to" 3s` `S = 9m` Total distance `= 9 + 9 = 18 m` |
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