Question

The space s described in time t by a particle moving in a straight line is given by s = t⁵ – 40t³ + 30t² + 80t – 250 Find the minimum value of acceleration.

Answer 1

Given : 

The Distance(S) covered by a particle in time t is given by 

⇒ S = t⁵ - 40t³ + 30t² + 80t - 250 

We know that acceleration of a particle is given by \(\frac{d^2S}{dt^2}\).

We need acceleration to be minimum, 

We know that for maxima and minima,

⇒ t² = 4 

⇒ t = 2 

(∵ Time cannot be negative) 

Differentiating ‘a’ again,

We get minimum for t = 2 sec, 

The corresponding acceleration at t = 2 sec is, 

⇒ a = 20(2)³ - 240(2) + 60 

⇒ a = 160 - 480 + 60 

⇒ a = - 260 

∴ The minimum acceleration is - 260.