Question

Two ships, 1 and 2, move with constant velocities 3 m/s and
4 m/s along two mutually perpendicular straight tracks toward
the intersection point 0. At the moment 1 = 0 the ships were
located at the distances 120 m and 200 m from the point. How
soon will the distance between the ships become the shortest
and what is it equal to ?​

Answer 1

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Let the distance between then be the least after a time $\sf{t}$ seconds of their journey.

During this time $\sf{t}$ seconds, ship 1 travels a distance $\sf{3t}$ metres and ship 2 travels a distance $\sf{4t}$ metres.

Now ship 1 is at a distance $\sf{120-3t}$ metres and ship 2 is at a distance $\sf{200-4t}$ metres.

Now the distance between the two ships will be,

$\longrightarrow\sf{d=\sqrt{(120-3t)^2+(200-4t)^2}}$

$\longrightarrow\sf{d=\sqrt{14400-720t+9t^2+40000-1600t+16t^2}}$

$\longrightarrow\sf{d=\sqrt{54400-2320t+25t^2}}$

For minimum distance, its derivative with respect to time should be zero.

$\longrightarrow\sf{d'=0}$

$\longrightarrow\sf{\dfrac{d}{dt}\left[\sqrt{54400-2320t+25t^2}\right]=0}$

$\longrightarrow\sf{\dfrac{50t-2320}{2\sqrt{54400-2320t+25t^2}}=0}$

$\longrightarrow\sf{50t-2320=0}$

$\longrightarrow\sf{t=\dfrac{2320}{50}}$

$\longrightarrow\sf{\underline{\underline{t=46.4\ s}}}$

Thus the distance becomes the shortest after a time $\bf{46.4\ seconds.}$

And the distance is,

$\longrightarrow\sf{d=\sqrt{54400-2320(46.4)+25(46.4)^2}}$

$\longrightarrow\sf{d=\sqrt{576}}$

$\longrightarrow\sf{\underline{\underline{d=24\ m}}}$

Hence the least distance between them is $\bf{24\ m.}$