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 distance travelled by the particle during first 3 seconds after the start?

(a) 21 cm

(b) 22 cm

(c) 23 cm

(d) 24 cm

The question was asked in an online quiz.

This interesting question is from Calculus Application in section Application of Calculus of Mathematics – Class 12

Answer 1

Right choice is (d) 24 cm

The best explanation: Let, x be the distance travelled by the particle in time t seconds.

Then, v = dx/dt = 3t^2 – 4t + 5

Or ∫dx = ∫ (3t^2 – 4t + 5)dt

So, on integrating the above equation, we get,

x = t^3 – 2t^2 + 5t + c where, c is a constant.   ……….(1)

Therefore, the distance travelled by the particle at the end of 3 seconds,

= [x]t = 3 – [x]t = 0

= (3^3 – 2*3^2 + 5*3 + c) – c [using (1)]

= 24 cm.