Answer:
[tex]r(t)=<\frac{7}{2}t^2+\frac{56}{3\sqrt{11}}t+2,\frac{7}{2}t^2+\frac{56}{3\sqrt{11}}t+1,-\frac{1}{2}t^2-\frac{8}{3\sqrt{11}}t+6>[/tex]
Step-by-step explanation:
We are given that
Initial position of particle =[tex]r_0=r(0)=<2,1,6>[/tex]
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=[tex]\frac{dv}{dt}[/tex]
[tex]dv=adt[/tex]
[tex]dv=(7i+7j-k)dt[/tex]
Integrating on both sides then, we get
[tex]v(t)=<7t,7t-t>+v_0[/tex]
Where [tex]v_0=,v_1,v_2,v_3>[/tex]
[tex]v_0=v(0)=8[/tex]
[tex]\mid <v_1,v_2,v_3>\mid=8[/tex]
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
[tex]v_1=7s, v_2=7s, v_3=-s[/tex]
[tex]\sqrt{v^2_1+v^2_2+v^2_3}=8[/tex]
[tex]\sqrt{49s^2+49s^2+s^2}=8[/tex]
[tex]\sqrt{99s^2}=8[/tex]
[tex]99s^2=64[/tex]
[tex]s^2=\frac{64}{99}[/tex]
[tex]s=\sqrt{\frac{64}{99}}=\frac{8}{3\sqrt{11}}[/tex]
[tex]v_0=<\frac{56}{3\sqrt{11}},\frac{56}{3\sqrt{11}},-\frac{8}{3\sqrt{11}}>[/tex]
[tex]v(t)=<7t,7t,-t>+<\frac{56}{3\sqrt{11}},\frac{56}{3\sqrt{11}},-\frac{8}{3\sqrt{11}}>[/tex]
[tex]v(t)=<7t+\frac{56}{3\sqrt{11}},7t+\frac{56}{3\sqrt{11}},-t-\frac{8}{3\sqrt{11}}>[/tex]
[tex]dr=vdt[/tex]
Integrating on both sides then we get
[tex]r(t)=\int <7t+\frac{56}{3\sqrt{11}},7t+\frac{56}{3\sqrt{11}},-t-\frac{8}{3\sqrt{11}}> dt[/tex]
[tex]r(t)= <\frac{7}{2}t^2+\frac{56}{3\sqrt{11}}t,\frac{7}{2}t^2+\frac{56}{3\sqrt{11}}t,-\frac{t^2}{2}-\frac{8}{3\sqrt{11}}t>+r_0[/tex]
Where [tex]r_0=r(0)=<r_1,r_2,r_3>[/tex]=Some constant vector
[tex]r_0=<2,1,6>[/tex]
Substitute the value
[tex]r(t)=<\frac{7}{2}t^2+\frac{56}{3\sqrt{11}}t,\frac{7}{2}t^2+\frac{56}{3\sqrt{11}}t,-\frac{1}{2}t^2-\frac{8}{3\sqrt{11}}t>+<2,1,6>[/tex]
[tex]r(t)=<\frac{7}{2}t^2+\frac{56}{3\sqrt{11}}t+2,\frac{7}{2}t^2+\frac{56}{3\sqrt{11}}t+1,-\frac{1}{2}t^2-\frac{8}{3\sqrt{11}}t+6>[/tex]
This is required equation of the position vector r(t) of the particle at time t.
