A particle moving in a straight line covers a distance of x cm in t second, where x = t3 + 6t2 – 15t + 18. What will be the velocity of the particle at the end of 2 seconds?a)20 cm/secb)22 cm/secc)21 cm/secd)23 cm/secCorrect answer is option 'C'. Can you explain this answer?

Question

A particle moving in a straight line covers a distance of x cm in t second, where x = t3 + 6t2 &ndash; 15t + 18. What will be the velocity of the particle at the end of 2 seconds?a)20 cm/secb)22 cm/secc)21 cm/secd)23 cm/secCorrect answer is option 'C'. Can you explain this answer?

Answer

The distance covered by the particle in t seconds can be represented by the equation x = t^3 - 6t^2.

Answer

We have, x = t³ + 6t² – 15t + 18
Let, v be the velocity of the particle at the end of t seconds. Then, v = dx/dt = d/dt(t³ + 6t² – 15t + 18)
So, v = 3t² + 12t – 15
Thus, velocity of the particle at the end of 2 seconds is, [dx/dt]_{t = 2} = 3(2)² + 12(2) – 15 = 21cm/sec.

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