Question

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?

Answer 1

Answer:

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

Step-by-step 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 =$\frac{Displacement }{time}=\frac{ds}{dt}$

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

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

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

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

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

Therefore, the position of particle at time t

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