A child walks due east on the deck of a ship at 5 kilometers per hour.

The ship is moving north at a speed of 30 kilometers per hour.

Find the speed and direction of the child relative to the surface of the water.

(a) Speed = __?___ km/h

(b) The angle of the direction is __?__radians.

Note : Traditionally north is 0 degrees so 0 rad, east is 90 degrees so (pi2) rad, south is 180 degrees so pi rad and west is 270 degrees or ((3pi)/2) rad.

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The child is walking east, and the ship is moving north. The relative motion of the child to the water is found using vector addition. This means that the speed of the child to the water is

`v=sqrt{3^2+30^2}`

`=sqrt{9+900}`

`approx 30.15` km/h

The angle of direction is found using trigonometry, where 0 radians is given as north and east is `pi/2` radians. This means that the angle is:

`theta=tan^{-1}(30/3)=tan^{-1}10 approx 1.471`

But this is the angle north of east, and we want the angle east of north, so subtract it from `pi/2` to get:

`alpha=pi/2-theta=0.0997`

**The speed of the child relative to the water is 30.15 km/h and the angle of direction is 0.0997 radians.**

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