Question

A river flows due south with a speed of 3.50 m/s. A man steers a motorboat across the river; his velocity relative to the water is 3.60 m/s due east. The river is 700 m wide.

Part A.) What is the magnitude of his velocity relative to the earth?

Part B.) What is the direction of his velocity relative to the earth?

Part C.) How much time is required for the man to cross the river?

Part D.) How far south of his starting point will he reach the opposite bank?

Answer 1

A

velocity relative to Earth = V = Vb + Vr = Vector velocity of boat+vector Velocity of river

Magnitude of V = Sqrt ( magnitude Vb^2 + magnitude Vr^2) = sqrt[( 3.60 m/s)^2+(3.50m/s)^2]

= 5.02 m/s

B

If a be angle made by Resultant velocity with respect to River flow

Tan (a) = 3.60 m/s/3.50 m/s angle (a) = 45.8 deg

C

time required for boat to cross river = Distance between two banks of river / Velocity to East of boat = 700m/3.60 m/s =194 sec

D

distance traveled south along the river bank = velocity x time = 3.5 m/s x 194s = 679 m