A particle moves with velocity vector v(t) = ti - 2j + t^2 k At time t = 0 the particle is at the point (0, −1, 1). What is the position of the particle at time t?

Question

asked Sep 6, 2020 153k views

1 Answer

OfficeYA · Answer 1 · 2020-09-12T05:39:08+0000

Answer:

$\vec{s(t)}=(1)/(2)t^2\hat{i}-(2t+1)\hat{j}+((1)/(3)t^3+1)\hat{k}$

Explanation:

We are given that velocity vector of a particle

$\vec{v(t)}=t\hat{i}-2\hat{j}+t^2\hat{k}$

When t=0 then the particle is at the point (0,-1,1).

We have to find the position of particle at time t.

We know that

Velocity =

Therefore,
$\vec{v}=\frac{\vec{ds}}{dt}$

$\int{ds}=\int (t\hat{i}-2\hat{j}+t^2\hat{k})dt$

Integrate on both sides then we get

$\vec{s(t)}=(1)/(2)t^2\hat{i}-2t\hat{j}+(1)/(3)t^3\hat{k}+C$

$\int x^n dx=(x^(n+1))/(n+1)+C$

Substitute the value of point at time t=0 then we get

C=
$-\hat{j}+\hat{k}$

Substitute the value of C then we get

$\vec{s(t)}=(1)/(2)t^2\hat{i}-2t\hat{j}+(1)/(3)t^3\hat{k}-\hat{j}+\hat{k}$

Therefore, the position of particle at time t

$\vec{s(t)}=(1)/(2)t^2\hat{i}-(2t+1)\hat{j}+((1)/(3)t^3+1)\hat{k}$