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

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