Skip to main content
Ch 03: Motion in Two or Three Dimensions

Chapter 3, Problem 3

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 4.2 m/s due east. The river is 500 m wide. (b) How much time is required to cross the river?

Verified Solution
Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
638
views
1
rank
Was this helpful?

Video transcript

Welcome back everybody. We have a river that is flowing north with a velocity relative to the earth of 1.8 m/s. We are told that a boatman named john is crossing the river and he has a velocity relative to the river of seven m per second and we have to figure out how much time it takes for john to cross the river. Well, just based off this little diagram that I drew here, we know that john's velocity relative to the earth is going to be the hypotenuse of this little triangle right here. Now, this is important to know because in order to figure out the time we're going to use the fact that distance is equal to velocity times time. But is this velocity john's velocity relative to the earth? Well, we are told that the river is 3000 m long. So we know that that's going to be our distance. But we truly just need his velocity going to the left, but still relative to the earth, A. K. A. We are going to have the X component of this velocity. Well, as you might be able to see the X component of his velocity relative to the earth and his velocity relative to the river are the same thing. Let me just write that out here, just so we can visualize that the X component of john's velocity relative to the earth is equal to john's velocity relative to the river, Which is of course equal to 7m/s and this is the velocity that we are going to use to find our time. So we have our values now. So let's go ahead and plug those in. We're told that our distance is equal to velocity times time and we're trying to find time. So I'm gonna divide both sides by our velocity here. This yields the equation that time is equal to distance over velocity. So let's plug in our values. We have the entire distance of the river that he needs to cross, having a width of 3000 m. And we know that the velocity that we're looking for is his X velocity. Let me go ahead and fix that. His X velocity relative to the earth of seven m per second. And when you plug this into your calculator, you get 428. seconds. Which when you divide by 60 seconds gives us 7. minutes corresponding to our answer choice of a thank you guys so much for watching. Hope this video helped and we will see you all in the next one.
Related Practice
Textbook Question
An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is 320.0 km/h (about 200 mi/h), in which direction should the pilot head?
1770
views
Textbook Question
A canoe has a velocity of 0.40 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.
2176
views
Textbook 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 4.2 m/s due east. The river is 500 m wide. (a) What is your velocity (magnitude and direction) relative to the earth?
735
views
Textbook Question
A 'moving sidewalk' in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5 m/s relative to the moving sidewalk, how much time does it take her to reach the opposite end if she walks (a) in the same direction the sidewalk is moving?
451
views
Textbook Question
A 'moving sidewalk' in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5 m/s relative to the moving sidewalk, how much time does it take her to reach the opposite end if she walks (b) In the opposite direction?
581
views
Textbook Question
A dog running in an open field has components of velocity υx = 2.6 m/s and υy = −1.8 m/s at t1 = 10.0 s. For the time interval from t1 = 10.0 s to t2 = 20.0 s, the average acceleration of the dog has magnitude 0.45 m/s2 and direction 31.0° measured from the +x–axis toward the +y–axis. At t2 = 20.0 s, (a) what are the x- and y-components of the dog's velocity?
2275
views