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?

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 acceleration of 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 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
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

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