A particle is moving along the straight line OX and its distance x is in metres from O after t seconds from start is given by x = t^3 – t^2 – 5t. What will be the acceleration of the particle when it is at a distance 28 metres from O?

Category: QuestionsA particle is moving along the straight line OX and its distance x is in metres from O after t seconds from start is given by x = t^3 – t^2 – 5t. What will be the acceleration of the particle when it is at a distance 28 metres from O?
Editor">Editor Staff asked 11 months ago

A particle is moving along the straight line OX and its distance x is in metres from O after t seconds from start is given by x = t^3 – t^2 – 5t. What will be the acceleration of the particle when it is at a distance 28 metres from O?
 
(a) 20 m/sec^2
 
(b) 22 m/sec^2
 
(c) 24 m/sec^2
 
(d) 26 m/sec^2
 
The question was asked by my college professor while I was bunking the class.
 
My question comes 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 
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience

1 Answers
Editor">Editor Staff answered 11 months ago

Right option is (b) 22 m/sec^2
 
Best explanation: We have, x = t^3 – t^2 – 5t  ……….(1)
 
When x = 28, then from (1) we get,
 
t^3 – t^2 – 5t = 28
 
Or t^3 – t^2 – 5t – 28 = 0
 
Or (t – 4)(t^2 + 3t +7) = 0
 
Thus, t = 4
 
Let v and f be the velocity and acceleration respectively of the particle at time t seconds. Then,
 
v = dx/dt = d(t^3 – t^2 – 5t)/dt
 
= 3t^2 – 2t – 5
 
And f = dv/dt = d(3t^2 – 2t – 5)/dt
 
= 6t – 2
 
Therefore, the acceleration of the particle at the end of 4 seconds i.e., when the particle is at a distance of 28 metres from O,
 
[f]t = 4 = (6*4 – 2) m/sec^2
 
= 22 m/sec^2