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

The boat's velocity relative to the earth is 4.5 m/s, heading 26.6° south of east. It takes 119.0 s for the boat to cross the river. The boat will reach a distance of 238.0 m south of its starting position on the opposite bank of the river.

Step-by-step explanation:

a) The magnitude of the boat's velocity relative to the earth can be found using the Pythagorean theorem. The magnitude is the square root of the sum of the squares of the boat's velocity relative to the water and the velocity of the river. In this case, the magnitude is 4.2 m/s2 + 2 m/s2 = 4.5 m/s. The direction of the boat's velocity relative to the earth can be found using the inverse tangent function. In this case, the direction is tan-1(2 m/s / 4.2 m/s) = 26.6° south of east.

b) The time required for the boat to cross the river can be found by dividing the width of the river by the component of the boat's velocity that is perpendicular to the river. In this case, the width of the river is 500 m and the component of the boat's velocity perpendicular to the river is 4.2 m/s. So, the time required is 500 m / 4.2 m/s = 119.0 s.

c) The distance south of the boat's starting position that the boat will reach the opposite bank of the river can be found by multiplying the time required to cross the river by the component of the river's velocity that is perpendicular to the river. In this case, the time required is 119.0 s and the component of the river's velocity perpendicular to the river is 2 m/s. So, the distance is 119.0 s * 2 m/s = 238.0 m.

d) The boat should head in a direction directly east in order to reach a point on the opposite bank directly east from the boat's starting position.

e) To calculate the boat's velocity relative to the earth using the heading in part (d), we can use the same method as in part (a). The magnitude is still 4.5 m/s, but the direction is now tan-1(0 m/s / 4.5 m/s) = 0°.

f) The time required to cross the river using the heading in part (d) is the same as in part (b), which is 119.0 s.

Answer 2

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

Step-by-step 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