Answer to Question #88366 in Mechanical Engineering for Umar faruq

Question

A particle moves along a straight line such that at any instant t in seconds,the displacement s in metres is given by the equation s=t^3+15t^2+63t-40.
Find the distance covered by the starting point to the point when the particle is momentary at rest.
Find the velocity of the particle when acceleration is zero

Answer 1

Answer:-

(a)  Find the distance covered by the starting point to the point when the particle is momentary at rest

(b)  Find the velocity of the particle when acceleration is zero

So, velocity (v) = 0

 And acceleration (a) = 0

Now,

               S = t^3+15t^2+63t-40         ………………….(1)       

     velocity (v) =(ds/dt) = 3t^2 + 30t +63  ………………….(2)

acceleration (a) = [dv/dt]

                          =6t + 30

When , acceleration (a) = 0

               6t +30 = 0

              t = -30/6 = -5 second

From equation (1)..

distance (s) = t^3+15t^2+63t-40         

   S = (-5^3) + 15 (-5^2) + 63 (-5)-40

 S = -125 +375 – 315 – 40

(a) Distance (S) =[ -105 m ] … Ans..

From equation (2)..

When (a=0) i.e, at t = -5 sec

Velocity (v) = 3t^2 + 30t +63

               = 3 (-5^2) + 30 (-5) + 63

               = 75 -150 + 63  

               = -12  m/s

(b) Velocity (V) = [-12 m/s] …. Ans..