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

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This interesting question is from Calculus Application in division Application of Calculus of Mathematics – Class 12

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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