Question

A particle moves in a straight line and its velocity v at time t seconds is given byv = 3t^2 – 4t + 5 cm/second. What will be the acceleration of the particle during first 3 seconds after the start?

(a) 10cm/sec^2

(b) 12cm/sec^2

(c) 14cm/sec^2

(d) 16cm/sec^2

This question was posed to me in unit test.

Question is from Calculus Application topic in division Application of Calculus of Mathematics – Class 12

Answer 1

The correct choice is (c) 14cm/sec^2

Explanation: Let f be the acceleration of the particle in time t seconds. Then,

f = dv/dt = d(3t^2 – 4t + 5)/dt

= 6t – 4    ……….(1)

Therefore, the acceleration of the particle at the end of 3 seconds,

= [f]t = 3 = (6*3 – 4) cm/sec^2

= 14cm/sec^2