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.