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

Answer:

$r(t)=$

Step-by-step explanation:

We are given that

Initial position of particle =$r_0=r(0)=$

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=$\frac{dv}{dt}$

$dv=adt$

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

Integrating on both sides then, we get

$v(t)=+v_0$

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

$v_0=v(0)=8$

$\mid \mid=8$

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

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

$\sqrt{v^2_1+v^2_2+v^2_3}=8$

$\sqrt{49s^2+49s^2+s^2}=8$

$\sqrt{99s^2}=8$

$99s^2=64$

$s^2=\frac{64}{99}$

$s=\sqrt{\frac{64}{99}}=\frac{8}{3\sqrt{11}}$

$v_0=$

$v(t)=+$

$v(t)=$

$dr=vdt$

Integrating on both sides then we get

$r(t)=\int dt$

$r(t)= +r_0$

Where $r_0=r(0)=$=Some constant vector

$r_0=$

Substitute the value

$r(t)=+$

$r(t)=$

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