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

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

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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.