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InterviewSolution
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A particle moving with constant acceleration of 2m/s^(2) due west has an initial velocity of 9 m/s due east. Find the distace covered in the fifth second of its motion. |
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Answer» Solution :Initial velocity u = +9 m/s Acceleration a = - 2 m/`s^(2)` In this problem, acceleration.s direction is opposite to the velocity.s direction. Let .t. be the TIME taken by the particle to reach a point where it makes a turn along the straight LINE. wehave , V = u + at O = 9 - 2 t We get , t = 4.5s Now let us find the distace convered in `(1)/(2)` second i.e. from 4.5 to 5 second Let u at t = 4.5 sec. Then distance convered in `(1)/(2) ` s. ` s= (1)/(2) at^(2)` s `= (1)/(2) xx 2 xx [ (1)/(2)]^(2)` `= (1)/(4)` m TOTAL distance COVERED in fifth second of its motion is given by `S_(0) = 2s = 2 ((1)/(4)) = (1)/(2) `m . 
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