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 distance traversed before it comes to rest?

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 distance traversed before it comes to rest?
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 distance traversed before it comes to rest?
 
(a) -173/27
 
(b) 173/27
 
(c) -175/27
 
(d) 175/27
 
I got this question during an interview.
 
This interesting question is from Calculus Application in division 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

Correct choice is (c) -175/27
 
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
 
Again the particle comes to rest when v = 0
 
Or 3t^2 – 2t – 5 = 0
 
Or (3t – 5)(t + 1) = 0
 
Or t = 5/3, -1
 
As, t > 0, so, t = 5/3
 
Therefore, the distance traversed by the particle before it comes to rest
 
= [(5/3)^3 – (5/3)^2 – 5(5/3)] m   [putting t = 5/3 in (1)]
 
= -175/27