| LoFi version for PDAs |
Help
Search
Members
Calendar
|
| Welcome Guest ( Log In | Register ) | Resend Validation Email |
Add reply · Start new topic · |
| bobby77 |
Posted: Oct 2 2006, 08:29 PM
|
|
Newbie ![]() Group: Members Posts: 10 Joined: 1-September 06 Positive Feedback: 0% Feedback Score: 0 |
A man can row a boat through still water at 4km/h. How long will it take him to cross an 18 m wide river flow at 2 km/h (a ) by rowing at right angles to the current; (B ) by rowing in a direction such that he lands directly opposite his starting point?
This post has been edited by bobby77 on Oct 2 2006, 08:30 PM |
|
Send PM · Send email ·
|
| mr_homm |
Posted: Oct 3 2006, 03:14 PM
|
|
Advanced Member ![]() ![]() ![]() ![]() ![]() Group: Power Member Posts: 881 Joined: 31-March 06 Positive Feedback: 96.83% Feedback Score: 143 |
This is a relative velocity problem. To find his total velocity, you should add his rowing velocity (which is relative to the water) to the river flow velocity. Basically the rule is, for three objects A, B, and C, V_AB + V_BC = V_AC. You can always check that your equation is correct by using the following trick:
Delete the "V"s, and insert parentheses around the subscripts and minus signs between the letters, like (A-B) + (B-C) = (A-C). Now pretend that A B and C are the variables in this equation. If this equation comes out to be true, then your original velocity equation is correct. For your problem, rowing is something happening between the boat and the water, so the rowing velocity is V_boat,water = V_BW. The flow of the river is the water moving over the riverbed, which is part of the ground, so the river flow velocity is V_water,ground = V_WG. The motion of the boat from one bank to the other is between the boat and the riverbank, which is part of the ground, so this velocity is V_boat,ground = V_BG. Using the trick, you can see that the correct equation for your problem is V_BW + V_WG = V_BG. Now notice that any velocity component of V_BG that is parallel to the banks of the river will not help get the boat across, so what you want to know is the velocity component of V_BG that points straight across the river. In the first problem, you are rowing straight across (let's say that's the X direction) and the river is flowing in the Y direction. Therefore in your equation V_BG = V_BW + V_WG, ALL of the X component comes from V_BW, which is just 4km/hr. Therefore the time to cross is D/V = (18m)/(4km/hr). Don't forget to convert V to m/s. For the second part, if the boat must go straight across the river, then V_BG must have NO Y component, only X component. Therefore, since V_BF = V_BW + V_WG, the Y components of V_BW and B_WG must cancel out to zero. Since the river is flowing 2km/hr in the Y direction, V_BW must contribute 2km/hr in the NEGATIVE Y direction to cancel it out. Now you know that V_BW has a Y component of -2 and a total speed of 4, which means that the boat is turned 30degrees upstream (4sin(30) = 2), and by the Pythagorean Theorem, (or by trig, whichever you like better), the X component of V_BW must by 2*sqrt(3). Since this is the ONLY contribution to the X component of V_BG (remember the river flows in the Y direction, so V_WG has no X component), that means V_BG has an X component of 2*sqrt(3). Therefore, the time to cross the river is now D/V = 18m/(2*sqrt(3)km/hr). Again, remember to convert V to m/s. Hope this helps! --Stuart Anderson -------------------- A hallmark of intelligence is the ability to give precise answers to vague questions.
|
|
Add reply · Start new topic · |