Question

A motorboat sets out for open sea. From home it travels 5 km due East, then heads [N35E] at 2 m/s for 90 minutes. After a break, it travels South for 6.5 km. The captain is lost at sea. Where is the boat relative to home?

Answer 1

The distance of the motorboat relative to the home, which travels in various directions, is 3.07 km.

What is relative velocity?

Relative velocity is a velocity of an object which is in respect to another observer, or it is a ratio of change in the position of a point to time with respect to another observer.

Given: Travel in East direction ($d_{E}$)= 5 km,

velocity in N35E(v) = 2 m / s, time taken(t)  = 90 min = 5400 seconds,
Displacement in south ($D_S}$) = 6.5 km

Firstly, find the displacement  in N35E so

Displacement   = Velocity × time,

$d_{NE}$ = 2 × 5400 = 10800 meters or 10.8 km

Now by using Pythagoras' theorem
We can calculate the return displacement in the south direction($d_{S}$),

$d_{S}$ = $\sqrt{d^{2} _{NE}- d^{2} _{E } }$
$d_{S}$ = $\sqrt{10.8^{2}-5^{2} }$

$d_{S}$ = 9.57 km

Therefore, the distance of the boat from the home (D) = $d_S}$ - $D_S}$

D = 9.57 - 6.5;

D = 3.07 Km

To know more about Relative velocity:

brainly.com/question/19260269

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