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serious [3.7K]
3 years ago
11

A particle moves with velocity vector

Mathematics
1 answer:
asambeis [7]3 years ago
8 0

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}

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