A particle is moving along a straight line, according to the equation given, where s is the distance meter oriented between the positionof the particle source and at time t seconds. Find the time...

A particle is moving along a straight line, according to the equation given, where s is the distance meter oriented between the position

of the particle source and at time t seconds. Find the time at which the instantaneous acceleration is zero and at this moment oriented determine the distance from the origin and the instantaneous velocity of the particle.

`s(t)=9t^2-2sqrt{2t+1}` , `t>=0` 

Expert Answers
jeew-m eNotes educator| Certified Educator

Velocity is the rate of change of the position and the acceleration is the rate of change of velocity. If velocity function is v(t) and acceleration function is a(t) then;

`v(t) = (ds(t))/dt`

`a(t) = (dv(t))/dt`

 

v(t)

`= (ds(t))/dt`

`= (d(9t^2-2sqrt(2t+1)))/dt`

`= 18t-2*1/(2sqrt(2t+1))*2`

`= 18t-2/(sqrt(2t+1))`

 

`v(t) = 18t-2/(sqrt(2t+1))`

 

a(t)

`= (dv(t))/dt`

`= (d(18t-2/(sqrt(2t+1))))/dt`

`= 18-2*(-1/(2sqrt(2t+1))*2)/(2t+1)`

`= 18+2/((2t+1)sqrt(2t+1))`

 

`a(t) = 18+2/((2t+1)sqrt(2t+1))`

 

When accelaration is 0;

a(t) = 0

`18+2/((2t+1)sqrt(2t+1)) = 0`

  `((2t+1)sqrt(2t+1)) = -9`

 `((2t+1)sqrt(2t+1))^2 = (-9)^2`

             `(2t+1)^3 = 81`

                 `2t+1 = 81^(1/3)`

                 `2t+1 = 4.326`

                        `t = 1.66`

 

Insantanius velocity `= (v(t))_(t=1.66) = 18(1.66)-2/(sqrt(2(1.66)+1))`

                             = 28.918 m/s

 

`s(t) = 9t^2-2sqrt(2t+1)`

`s(0) = 9(0)-2sqrt(0+1) = -1`

 

Distance from origin `= (s(t))_(t=1.66)-s(0) = (9(1.66)^2-2sqrt(2(1.66)+1))+1`

`= 21.643m`

 

So when accelaration is zero the velocity of particle is 28.918m/s and distance from origin is 21.643m