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?

Category: QuestionsA 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?
Editor">Editor Staff asked 11 months ago

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
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1 Answers
Editor">Editor Staff answered 11 months ago

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.