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A particle is moving along the straight line OX and its distance x is in metres from O after t seconds from start is given by x = t3 – t2 – 5t. What will be the acceleration of the particle when it is at a distance 28 metres from O?

A particle is moving along the straight line OX and its distance x is in metres from O after t seconds from start is given by x = t3 – t2 – 5t. What will be the acceleration of the particle when it is at a distance 28 metres from O? Correct Answer 22 m/sec2

We have, x = t3 – t2 – 5t  ……….(1) When x = 28, then from (1) we get, t3 – t2 – 5t = 28 Or t3 – t2 – 5t – 28 = 0 Or (t – 4)(t2 + 3t +7) = 0 Thus, t = 4 Let v and f be the velocity and acceleration respectively of the particle at time t seconds. Then, v = dx/dt = d(t3 – t2 – 5t)/dt = 3t2 – 2t – 5 And f = dv/dt = d(3t2 – 2t – 5)/dt = 6t – 2 Therefore, the acceleration of the particle at the end of 4 seconds i.e., when the particle is at a distance of 28 metres from O, t = 4 = (6*4 – 2) m/sec2 = 22 m/sec2

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