1.3) \ \

Speed of Vehicle V  v_v = 125 km/h in north-east direction. \ \

Speed of Vehicle W  v_w = 125 km/h in east direction. \ \

Consider that two vehicles are starts from same point. \ \

\ \

By using vector analysis \ \

OA = 125 km/h \ \

OB = 125 km/h \ \

OA = BC \ \

OB = AC \ \

AB = CD \ \

OD = OB+BD \ \

OC = Relative velocity magnitude \ \

∠OCD = Relative velocity angle \ \

From above figure \ \

Angle ∠NOE = 90° \ \

Angle ∠AOB is half in ∠NOE \ \

Angle ∠AOB = ∠CBD = θ = 45° \ \

From CBD right angled triangle \ \

sinθ = BC/CD \ \

BC = CD sin 45° \ \

BC = 125 sin 45° \ \

BC = 88.388 km/h \ \

From CBD right angled triangle \ \

cosθ = BD/CB \ \

BD = CB cos 45° \ \

BD = 125 cos 45° \ \

BD = 88.388 km/h \ \

OD = OB+BD = 125+88.388 = 213.388 km/h \ \

From OAB right angled triangle \ \

sinθ = AB/OA \ \

AB = OA sin45 \ \

AB = 125 sin45 \ \

AB = 88.388 km/h \ \

AB = CD = 88.388 km/h \ \

From OCD right angled triangle \ \

OC² = OD² + CD² \ \

OC² = 213.388² + 88.388² \ \

OC = 230.969 km/h \ \

∠OCD = tan^-1(88.388/213.388) \ \

∠OCD = 22.3° \ \

Solution : \ \

Magnitude of the velocity of vehicle W relative to the velocity of vehicle V is 230.969 km/h \ \

Direction of the velocity of vehicle W relative to the velocity of vehicle V is 22.3° upside inclined to east \ \