Question

Ships A and B leave port together. For the next two hours, ship A travels at 20.0 mph in a direction 70.0 degrees west of north while the ship B travels 80.0 degrees east of north at 30.0 mph

A) What is the distance between the two ships two hours after they depart?

B) What is the speed of ship A as seen by ship B?

Answer 1

A)
Distance travelled by ship A in two hours = 20 x 2 = 40 mi
Location of ship A = (-40 sin(70), 40 cos(70))
= (- 37.58, 13.68)

Distance travelled by ship B in two hours = 30 x 2 = 60 mi
Location of ship B = (60 sin(80), 60 cos(80))
= (59.09, 10.42)

Distance between the ships = SQRT[(59.09 - (-37.58))² + (10.42 - 13.68)²]
= 96.73 mi

B)
The velocity of ship A in component form can be written as,
Va = -20 sin(70) + 20 cos(70)
Similarly velocity of ship B can be written as
Vb = 30 sin(80) + 30 cos(80)

Velocity of ship A with respect to ship B = Va - Vb
= [-20 sin(70) + 20 cos(70) ] + [30 sin(80) + 30 cos(80) ]
= 10.75 + 12.05
Magnitude of velocity = SQRT[(10.75)² + (12.05)²]
= 16.15 mph