A particle is moving in a straight line is at a distance x from a fixed point in the straight line at time t seconds, where x = 2t^3 – 12t + 11. What is the acceleration of the particle at the end of 2 seconds?

Category: QuestionsA particle is moving in a straight line is at a distance x from a fixed point in the straight line at time t seconds, where x = 2t^3 – 12t + 11. What is the acceleration of the particle at the end of 2 seconds?
Editor">Editor Staff asked 11 months ago

A particle is moving in a straight line is at a distance x from a fixed point in the straight line at time t seconds, where x = 2t^3 – 12t + 11. What is the acceleration of the particle at the end of 2 seconds?
 
(a) 22 cm/sec^2
 
(b) 24 cm/sec^2
 
(c) 26 cm/sec^2
 
(d) 28 cm/sec^2
 
This question was addressed to me in an international level competition.
 
Enquiry is from Calculus Application in portion 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

Right option is (a) 22 cm/sec^2
 
Explanation: We have, x = 2t^3 – 12t + 11  ……….(1)
 
Let v and f be the velocity and acceleration respectively of the particle at time t seconds.
 
Then, v = dx/dt = d(2t^3 – 12t + 11)/dt
 
= 6t^2 – 12  ……….(2)
 
And f = dv/dt = d(6t^2 – 12)/dt
 
= 12t   ……….(3)
 
Putting the value of t = 2 in (3),
 
Therefore, the displacement of the particle at the end of 2 seconds,
 
12t = 12(2)
 
= 24 cm/sec^2