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Question

Mathematics Question on Differential Calculus

The distance ‘s’ in meters travelled by a particle in ‘t’ seconds is given by s = 2t3318t+53.\frac{2t^3}{3}-18t+\frac{5}{3}. The acceleration when the particle comes to rest is

A

12 m2/sec.

B

3 m2/sec.

C

18 m2/sec.

D

10 m2/sec.

Answer

12 m2/sec.

Explanation

Solution

The correct answer is (A) : 12 m2/sec.