Question

At time t equals​0, a particle is located at the point ​(2​,1​,6​). It travels in a straight line to the point ​(9​,8​,5​), has speed 8 at ​(2​,1​,6​) and constant acceleration 7 Bold i plus 7 Bold j minus Bold k. Find an equation for the position vector r​(t) of the particle at time t.

Answer 1

`r(t)=<(7)/(2)t^2+(56)/(3√(11))t+2,(7)/(2)t^2+(56)/(3√(11))t+1,-(1)/(2)t^2-(8)/(3√(11))t+6>`

Explanation:

We are given that

Initial position of particle =
`r_0=r(0)=<2,1,6>`

Initial speed of the particle = 8 m/s

Acceleration=7 i+7j-k=<7,7,-1>

We have to find the equation of position vector r(t) of the particle at time t.

We know that

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

`dv=adt`

`dv=(7i+7j-k)dt`

Integrating on both sides then, we get

`v(t)=<7t,7t-t>+v_0`

Where
`v_0=,v_1,v_2,v_3>`

`v_0=v(0)=8`

`\mid <v_1,v_2,v_3>\mid=8`

Since, the particle travel in straight line to (9,8,5) and (9,8,5)-(2,1,6)=<7,7,-1>

There exist a constant s such that

`v_1=7s, v_2=7s, v_3=-s`

`√(v^2_1+v^2_2+v^2_3)=8`

`√(49s^2+49s^2+s^2)=8`

`√(99s^2)=8`

`99s^2=64`

`s^2=(64)/(99)`

`s=\sqrt{(64)/(99)}=(8)/(3√(11))`

`v_0=<(56)/(3√(11)),(56)/(3√(11)),-(8)/(3√(11))>`

`v(t)=<7t,7t,-t>+<(56)/(3√(11)),(56)/(3√(11)),-(8)/(3√(11))>`

`v(t)=<7t+(56)/(3√(11)),7t+(56)/(3√(11)),-t-(8)/(3√(11))>`

`dr=vdt`

Integrating on both sides then we get

`r(t)=\int <7t+(56)/(3√(11)),7t+(56)/(3√(11)),-t-(8)/(3√(11))> dt`

`r(t)= <(7)/(2)t^2+(56)/(3√(11))t,(7)/(2)t^2+(56)/(3√(11))t,-(t^2)/(2)-(8)/(3√(11))t>+r_0`

Where
`r_0=r(0)=<r_1,r_2,r_3>`=Some constant vector

`r_0=<2,1,6>`

Substitute the value

`r(t)=<(7)/(2)t^2+(56)/(3√(11))t,(7)/(2)t^2+(56)/(3√(11))t,-(1)/(2)t^2-(8)/(3√(11))t>+<2,1,6>`

`r(t)=<(7)/(2)t^2+(56)/(3√(11))t+2,(7)/(2)t^2+(56)/(3√(11))t+1,-(1)/(2)t^2-(8)/(3√(11))t+6>`

This is required equation of the position vector r(t) of the particle at time t.