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A particle moving in a straight line covers a distance of x cm in t second, where x = t^3 + 6t^2 – 15t + 18. What will be the acceleration of the particle at the end of 2 seconds?(a) 22cm/sec^2(b) 23cm/sec^2(c) 24cm/sec^2(d) 25cm/sec^2I have been asked this question in an interview for internship.I'm obligated to ask this question of Application of Derivative in section Application of Derivatives of Mathematics – Class 12

Answer»

Right choice is (c) 24cm/sec^2

Explanation: We have, X = t^3 + 6t^2 – 15t + 18

Let, ‘V’ be the velocity of the particle and ‘a’ be the acceleration of the particle at the end of t seconds. Then,

v = dx/dt = d/dt(t^3 + 6t^2 – 15t + 18)

So, v = 3t^2 + 12t – 15

Therefore, a = dv/dt = d/dt(3t^2 + 12t – 15)

So, a = 6t + 12

Thus, acceleration of the particle at the end of 2 seconds is,

[dv/dt]t = 2 = 6(2) + 12 = 24cm/sec^2.


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