A boat begins at point A and heads straight across a 60-meter wide river with a speed of 4 m/s (relative to the water). The river water flows north at a speed of 3 m/s (relative to the shore). The boat reaches the opposite shore at point C. Which would cause the boat to reach the opposite shore at a location SOUTH of C?

Now to answer this, look on the picture attached. As you can see point C is in a diagonal direction from A, therefore, the boat is not crossing the river in a straight line, because of the speed of the river flow.

The direction is heading is like a triangle (See picture), so, if we know the speed of the boat in a straight line which is 4 m/s and the speed of river which is 3 m/s, we can know the speed in the diagonal direction, using the following expression:

V = √Vx² + Vy²

Now, we want a location south of C, but it's the same, because looking the picture, you can actually see that it's not going to poinnt B, just a little south from C. We still have the x and y components, which is the speed of the boat and the river, therefore replacing in the above equation:

V = √4² + 3²

V = 5 m/s

This is the speed which the boat needs to cross the river to reach a point south of C.

Now, the river is exerting something here too. All we need to do, is see the difference of the 2 speeds, and then, substract than from the speed of the flow. This has logic, because is the speed of the river is lower and the speed of boat is higher, he can reach a point at C in the south still so:

V = 4 -3 = 1 m/s

Vriver = 3 - 1 = 2 m/s

obasola1 obasola1 · Answer 2 · 2019-12-13T15:29:28+01:00

Answer:

river flows north at 2 m/s and boat heads across with 5 m/s

Explanation:

from pythagoras formula which states that the square of the hypotenuse of a right angle triangle is equal to the square of the opposite plus the square of the adjacent

(4.0 m/s)2 + (3.0 m/s)2 = R2

16 m2/s2 + 9 m2/s2 = R2

25 m2/s2 = R2

SQRT (25 m2/s2) = R

5.0 m/s = R

This is the speed which the boat needs to cross the river to reach a point south of C

tan (theta) = (opposite/adjacent)

tan (theta) = (3/4)

theta - invtan (3/4)

theta = 36.9 degrees

Now, the difference in the two speed an be derived as thus. note that the boat s speed is greater than that of the river and are contrary to each other

V = 4 -3 = 1 m/s

V of the river = 3 - 1 = 2 m/s