Question

A man is standing on a straight bridge over a river and another man on a boat on the river just below the man on the bridge. If the first man starts walking at the uniform speed of $4m/min$ and the boat moves perpendicularly towards the bridge at the speed of $5m/min$, then at what rate are they separating after $4min$, if the height of the bridge above the boat is $3m$

• A. $\frac{164}{\sqrt{665}}m/min$
• B. $\frac{165}{\sqrt{665}}m/min$
• C. $\frac{163}{\sqrt{665}}m/min$
• D. $\frac{162}{\sqrt{665}}m/min$

Answer 1

The correct option is A $\frac{164}{\sqrt{665}}m/min$

Let the direction of motion of boat be $\hat{i}$, direction of motion of man on the bridge be $\hat{j}$ and height of bridge be in $\hat{k}$

Let the starting point of boat be the origin, then,

Position of boat after $4min$ is $20\hat{i}$

Position of bridge is $3\hat{k}$

Position of man on bridge after $4min$ is $16\hat{j}+3\hat{k}$

Vector joining the positions of the two man$:16\hat{j}+3\hat{k}-20\hat{i}$

Velocity of boat $=5m/min\hat{i}$

Velocity of man on bridge $=4m/min\hat{j}$

Speed of boat along the vector joining the two man: $5\hat{i}.\frac{(16\hat{j}+3\hat{k}-20\hat{i})}{\sqrt{665}}=-\frac{100}{\sqrt{665}}m/min$

Speed of man on bridge along the vector joining the two man: $4\hat{j}.\frac{(16\hat{j}+3\hat{k}-20\hat{i})}{\sqrt{665}}=\frac{64}{\sqrt{665}}m/min$

Subtracting the two speeds, we get,

rate of separation $=\frac{164}{\sqrt{665}}m/min$

So, Option (A)