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A particle is moving along a straight line, according to the equation given, where s is...

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juniorsilvamath | Student, Undergraduate | Honors

Posted September 5, 2012 at 9:25 PM via web

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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` 

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 7, 2012 at 12:42 AM (Answer #1)

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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

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