Yeh, +1 on that. You don't know the hypotenuse as that's the total distance travelled. I'm still trying to think about what the correct answer is for the direction the boat should be steered. I think it's 0 degrees, which is North. The angle of the vector the boat travels in isn't 0, but the question doesn't ask for that. I'm sure they word these questions in the most difficult way possible just to try and confuse you, thats why I only got C in my A Level maths, but then managed to get a 1st degree in Games Programming, which was nearly all the same sort of maths,
Your post says the shortest distance is required. That would require a track of 360°. At 3m/s for 300m gives a time of 100secs. In that time, due to the current of 5m/s from west to east, the water would flow 500m from west to east. So he must point the boat 59° to the west to overcome the current and allow for the drift. 500/300=tan drift angle = 1.666, = 59° Therefore he needs to head 301° to make good a track of 360° to cross from south to north say from pier to pier straight across the river. It would take him 194 secs. distance=500/sin59 =583m time en route=583/3 =194secs
I know what you mean. The first part of the question was asking for the fastest way to travel across and you instinct is the correct solution to that one. By steering a little into the current you reduce the drift at the cost of increasing the time taken to travel across. The further into the current you steer, the more you reduce drift...up to a point...which is where the optimal angle comes in. You're trading drift off for a more travel time, which counteracts the reduced drift past a certain angle. Thus, there is an optimal solution.
I'm still not sure how you can travel 500m in 125s, which is 4ms when pointing upstream (which the question says the boat can't do (max 3ms)). If the fastest the rower can go is 3ms, then pointing the boat directly into the current would result in a speed of 2ms going downstream, any sort of angle to the current (still pointing upstream) is going to reduce that 2ms further as some of the boats velocity is used to travel north. Don't headings start from 0 pointing north and then rotate clockwise? I get confused with all the different coord systems,
The initial setup should look like this. The river flows parallel to the bank. Both the velocity of the boat and the velocity of the boat relative to the river must point across the river but for the shortest possible distance they must be perpendicular to one another:
See it now too, , this is obviously why I only got a D or something in A-level maths, . I think it was the bit that says the rower can row at 3ms in still water that got me, I didn't think he'd be able to row upstream at 3ms as it wasn't still water. Think I was also going about adding the river speed vector wrong to things. I'll go sit in the corner with the dunce hat on now,
Half the battle is the comprehension. If you're ever looking for more practice/fun try the OCR-MEI syllabus.
Thank you all for taking an interest in the thread and contributing, chewing the fat of a question with other people really helps me think it through from angles I'd not considered. I was really worried this would just sink, unnoticed. It's really heartening to see so many with a passion for a good problem. I've got another thread coming soon, change of subject on this as I've been launching rockets with a pupil as part of his extended project qualification and I'd like to share the video footage when I finally get around to uploading it (later this week, homework to mark and reports to write first). Once the videos are up I'll be happy to answer any questions on rocketry in the UK.
Your post says the shortest distance is required. That would require a track of 360°. At 3m/s for 300m gives a time of 100secs. In that time, due to the current of 5m/s from west to east, the water would flow 500m from west to east. So he must point the boat 59° to the west to overcome the current and allow for the drift. 500/300=tan drift angle = 1.666, = 59° Therefore he needs to head 301° to make good a track of 360° to cross from south to north say from pier to pier straight across the river. It would take him 194 secs. distance=500/sin59 =583m time en route=583/3 =194secs [/URL][/IMG] Not sure what happened. My post on the 21st never appeared.
You've mixed up the initial conditions from your first scenario with your second scenario (the change in angle of attack to 59 degrees). The speed of 3m/s is now pointed at 59 degrees, changing the shape of the triangle. It no longer takes 100s to cross the river, it'll take longer but drift will be reduced: Time to cross the river = 300/3cos(59) = 194s Distance travelled across the river (perpendicular to the bank)=300m Distance travelled down the river (parallel to the bank) = (5-3sin(59))x194 = 471m (to 3sf) Total distance travelled = sqrt(300^2+471^2)=559m (to 3sf) This is further than my solution of 500m. To find the shortest distance, velocity of the boat must be perpendicular to the velocity of the boat relative to the river:
Is the reason that the two velocities need to be at a right angle so that the added vector is pointing sort of straight across the river? Obviously with the different speeds it's not direct straight across, I'd assume with a river flow of 3ms the angle would be 45 degrees?
