Ship A is moving westwards with a speed of 20 | vedclass

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(A) Let the position of ship $A$ at $t=0$ be $(0, 200)$ and ship $B$ be $(0, 0)$.Velocity of ship $A$ is $\vec{v}_A = -20 \hat{i} \ km h^{-1}$.Velocit

Vedclass · Accepted Answer

$4 \ h, 80 \sqrt{5} \ km$

Vedclass · Answer

(A) Let the position of ship $A$ at $t=0$ be $(0, 200)$ and ship $B$ be $(0, 0)$.Velocity of ship $A$ is $\vec{v}_A = -20 \hat{i} \ km h^{-1}$.Velocity of ship $B$ is $\vec{v}_B = 10 \hat{j} \ km h^{-1}$.Relative velocity $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = 20 \hat{i} + 10 \hat{j} \ km h^{-1}$.Relative position $\vec{r}_{BA} = \vec{r}_B - \vec{r}_A = (0 - 0) \hat{i} + (0 - 200) \hat{j} = -200 \hat{j} \ km$.The distance $d$ at time $t$ is given by $|\vec{r}_{BA} + \vec{v}_{BA} t| = |20t \hat{i} + (10t - 200) \hat{j}|$.$d^2 = (20t)^2 + (10t - 200)^2 = 400t^2 + 100t^2 - 4000t + 40000 = 500t^2 - 4000t + 40000$.For minimum distance,$\frac{d(d^2)}{dt} = 1000t - 4000 = 0 \implies t = 4 \ h$.Shortest distance $d = \sqrt{500(4)^2 - 4000(4) + 40000} = \sqrt{8000 - 16000 + 40000} = \sqrt{32000} = 80 \sqrt{5} \ km$.

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