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Question: A motorboat is racing towards north at 25 km<sup>-1</sup> and the water current in that region is 10...

A motorboat is racing towards north at 25 km-1 and the water current in that region is 10 km h-1 in the direction of 60o east of south. The resultant velocity of the boat is

A

11kmh−111 \mathrm { kmh } ^ { - 1 }

B

C

33kmh−133 \mathrm { kmh } ^ { - 1 }

D

44kmh−144 \mathrm { kmh } ^ { - 1 }

Answer

Explanation

Solution

The velocity of the motorboat and the velocity of the water current are represented by vectors and V→c\overrightarrow { \mathrm { V } } _ { \mathrm { c } } as shown in the figure.

Here,

∴\thereforeAccording to parallelogram law of vector additions, the magnitude of the resultant velocity of the boat is

(25)2+(10)2+2(25)(10)(−12)\sqrt { ( 25 ) ^ { 2 } + ( 10 ) ^ { 2 } + 2 ( 25 ) ( 10 ) \left( \frac { - 1 } { 2 } \right) } =625+100−250=25kmh−1= \sqrt { 625 + 100 - 250 } = 25 \mathrm { kmh } ^ { - 1 }