Question

A river flows due south with a speed of 2.0 m/s .You steer a motorboat across the river; your velocity relative to the water is 4.8 m/s due east. The river is 600 m wide.
a) What is the magnitude of your velocity relative to the earth
b) What is the direction of his velocityrelative to the earth?
c) How much time is required for the man tocross the river?
d) How far south of his starting point willhe reach the opposite bank?

Answer 1

Answer:

a) $v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

b) $\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

c)$t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

d) $Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South

Explanation:

Part a

For this case the figure attached shows the illustration for the problem.

We know that $v_y = 2 m/s$ represent the velocity of the river to the south.

We have the velocity of the motorboard relative to the water and on this case is $V_x= 4.8 m/s$

And we want to find the velocity of the motord board relative to the Earth $v_m$

And we can find this velocity from the Pythagorean Theorem.

$v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

Part b

We can find the direction with the following formula:

$\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

Part c

For this case we can use the following definition

$D = Vt$

The distance would be D = w = 600 m and the velocity V = 4.8m/s and if we solve for t we got:

$t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

Part d

For this case we can use the same definition but now using the y compnent we have:

$Y = v_y t$

And replacing we got:

$Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South