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A particle is moving in a straight line and its distance s cm from a fixed point in the line after t seconds is given by s = 12t – 15t2 + 4t3. What will be the distance between the two positions of the particle at two times, when the velocity is instantaneously 0?

A particle is moving in a straight line and its distance s cm from a fixed point in the line after t seconds is given by s = 12t – 15t2 + 4t3. What will be the distance between the two positions of the particle at two times, when the velocity is instantaneously 0? Correct Answer 27/4 cm

We have, s = 12t – 15t2 + 4t3 ……….(1) Differentiating both side of (1) with respect to t we get, (ds/dt) = 12 – 30t + 12t2 Clearly, the velocity is instantaneously zero, when (ds/dt) = 12 – 30t + 12t2 = 0 Or 12 – 30t + 12t2 = 0 Or (2t – 1)(t – 2) = 0 Thus, t = 2 or t = ½ Putting the value t = 2 and t = ½ in (1), We get, when t = 2 then s = (s1) = 12(2) – 15(2)2 + 4(2)3 = -4. When t = ½ we get, s = (s2) = 12(1/2) – 15(1/2)2 + 4(1/2)3 = 11/4. Thus, the distance between the two positions of the particle at two times, when the velocity is instantaneously 0 = s2 – s1 = 11/4 – (-4) = 27/4 cm.

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