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# Mechanics watch

1. hey could you give me a hand with this please....Im rather stuck

a boat sails across a straight river, of uniform width W, starting from point O
on one bank of the river. the velocity of the river at a distance Y from the bank is u(y)=ay(W-y)
where a is a positive constant, the boat travels at a constant speed V relative to the current and steers a course set at a constant angle S where (0<s<pi) to the downstream direction.

a) show that the velocity of the boat relative to a cartesian coordiante system with origin at O and i pointing in the downward direction and j pointing across
(u+vcosS)i+(vsinS)j

b) at what time does the boat reach the other bank
c) show what when the boat has reached the other bank the downsteram distance it has travelled is equal to
aW³ ÷ 6vsinS + WcotS

thanks!
2. number one, sketch a picture that really makes the difference.
part a) is resolving into components, in the i direction there is the added speed of the water.

b) time = distance/speed, just got to consider the component of velocity

c) once you find time t, you can find downstream distance by using distance = speed*time (same time that you calculated), the speed is the component of velocity in that direction. It means you get an expression almost at the final answer except there is a u(y)
it is found by integration with respect to y, the limits are from y = 0 to y = , I'll let you consider the upper limit.

hope it works out ok.

(correction, there seems like an extra factor of W in the integral term when you work it out, I'm not entirely sure about it, may be in doing the integration you got to multiply by a factor of 1/W so that you are essentially taking the average speed, if you don't multiply by 1/W I'm not sure if it works, it may not be the right method but it's a way)
3. (Original post by apple tree)
number one, sketch a picture that really makes the difference.
part a) is resolving into components, in the i direction there is the added speed of the water.

b) time = distance/speed, just got to consider the component of velocity

c) once you find time t, you can find downstream distance by using distance = speed*time (same time that you calculated), the speed is the component of velocity in that direction. It means you get an expression almost at the final answer except there is a u(y)
it is found by integration with respect to y, the limits are from y = 0 to y = , I'll let you consider the upper limit.

hope it works out ok.
Hey I'm not entirely sure if I follow your method. How do I resolve into components?
(correction, there seems like an extra factor of W in the integral term when you work it out, I'm not entirely sure about it, may be in doing the integration you got to multiply by a factor of 1/W so that you are essentially taking the average speed, if you don't multiply by 1/W I'm not sure if it works, it may not be the right method but it's a way)
I'm not entirely sure if i follow your method, how would you resolve into components?

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Updated: January 27, 2006
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