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vlada-n [284]
3 years ago
14

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

4t i + sin t j + cos 2t k, v(0) = i, r(0) = j
Physics
1 answer:
kondor19780726 [428]3 years ago
4 0

Answer:

Explanation:

Given

Acceleration a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}[/tex]

and v(0)=\hat{i}

r(0)=\hat{j}

we know a=\frac{\mathrm{d} v}{\mathrm{d} t}

\int dv=\int adt

v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt

v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c

at t=0

v(0)=0-1\cdot \hat{j}+0+c

c=\hat{i}+\hat{j}

v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}

and \frac{\mathrm{d} r}{\mathrm{d} t}=v(t)

\int dr=\int vdt

r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt

r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2

at t=0

r(0)=\hat{j}

r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}

       

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