Question

A river flows south with a speed of 2 m/s. A motorboat crosses the river due east with a velocity of 4.2 m/s relative to the water. The river is 500 m wide. a) What is the magnitude and direction of the boats velocity relative to the earth? b) How much time is required for the boat to cross the river? c) How far south of the boats starting position will the boat reach the opposite bank of the river? d) What direction should the boat head to reach a point on the opposite bank directly east from the boats starting position? Assume the boats speed is still 4.2 m/s. e) What would be the boats velocity relative to the earth using the heading in part (d)? f) How much time is required to cross the river using the heading in part (d)?

Answer 1

Answer:

a) Vb = 4.65m/s at 25.46° due south of east

b) t = 119s

c) d = 238m

d) 28.43° due north of east.

e) Vb = 3.69m/s due east

f) t = 135.5s

Explanation:

Velocity relative to the earth is:

$V_b = V_{b/w} + V_w = [4.2,0]+[0,-2]=[4.2,-2]m/s$

Vb = 4.65m/s < -25.46°

Since the distance to travel is 500m:

$t = X / V_{b-x} = 500/4.2 =119s$

The distance on the y-axis is given by:

$Y = V_{b-y}*t=2*119=238m$

If the final position is directly east from the starting position:

$V_b = V_{b/w}+V_w$

$V_b= [V_{b-x}, 0] = [4.2*cos\beta ,4.2*sin\beta ]+[0,-2]$

From the y-components:

$0=4.2*sin\beta -2$ Solving for β:

β=28.43°

With this angle, the velocity would be:

Vb = 4.2*cos(28.43°) = 3.69m/s

And the time it would take it to cross:

t = 500/3.69 = 135.5s