Question : A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat.
Doubt by Faizal
Solution :
Here
Let magnitude of velocity of motorboat (Vb) = 25 km/h
Magnitude of velocity of water current (Vc) = 10 km/h
Clearly, angle between Vb and Vc
= (90°+30°)
= 120°
Let magnitude of velocity of motorboat (Vb) = 25 km/h
Magnitude of velocity of water current (Vc) = 10 km/h
Clearly, angle between Vb and Vc
= (90°+30°)
= 120°
Using
R = √(A²+B²+2ABcosθ)
Magnitude of resultant velocity
V = √(Vb²+Vc²+2VbVccosθ)
V = √[(25)²+(10)²+2(25)(10)cos120°]
R = √(A²+B²+2ABcosθ)
Magnitude of resultant velocity
V = √(Vb²+Vc²+2VbVccosθ)
V = √[(25)²+(10)²+2(25)(10)cos120°]
V = √[625+100+500(-1/2)] [∵cos120° = -1/2]
V = √[725-250]
V = √(475)
V = 21.7025
V ⋍ 21.7 km/h
Direction of resultant velocity
tanβ=Bsinθ/(A+Bcosθ)
tanβ=Vcsinθ/(Vb+Vccosθ)
tanβ=(10)sin120°/(25+10cos120°)
tanβ=(10)(√3/2)/(25+10(-1/2)) [∵sin120° =√3/2]
tanβ=(10√3/2)/(25-5)
tanβ=5√3/20
tanβ=√3/4
tanβ=1.732/4
tanβ=0.433
β = tan-1(0.433)
β=23.41
β=23.4°
V = √[725-250]
V = √(475)
V = 21.7025
V ⋍ 21.7 km/h
Direction of resultant velocity
tanβ=Bsinθ/(A+Bcosθ)
tanβ=Vcsinθ/(Vb+Vccosθ)
tanβ=(10)sin120°/(25+10cos120°)
tanβ=(10)(√3/2)/(25+10(-1/2)) [∵sin120° =√3/2]
tanβ=(10√3/2)/(25-5)
tanβ=5√3/20
tanβ=√3/4
tanβ=1.732/4
tanβ=0.433
β = tan-1(0.433)
β=23.41
β=23.4°
Hence, the resultant velocity of the boat is 21.7 km/h and it is making angle of 23.4° with the North direction.