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You want to cross a straight section of a river that has a uniform current of 5.40 m/s and is 126. m wide. You?

  • You want to cross a straight section of a river that has a uniform current of 5.40 m/s and is 126. m wide. Your motorboat has an engine that can generate a speed of 17.0 m/s for your boat. Assume that you reach top speed right away (that is, neglect the time it takes to accelerate the boat to top speed). (a) If you want to go directly across the river with a 90° angle relative to the riverbank, at what angle relative to the riverbank (0° ≤ θ ≤ 90°) should you point your boat? θ = ???° (b) How long will it take to cross the river in this way? (c) In which direction should you aim your boat to achieve minimum crossing time? (0° ≤ θ ≤ 90° relative to the riverbank.) θ = ???° (d) What is the minimum time to cross the river? (e) What is the minimum speed of your boat that will still enable you to cross the river with a 90° angle relative to the riverbank?

  • Answer:

    Hello a) let angle of boat with bank = (angle) cos(angle) = 5.4/17 angle = 71.48° -------- b) the perpendicular velocity of the boat will be v0*sin71.48 m/s = 17*sin71.48 m/s. And t = s/v = 126/(17*sin71.48) t = 7.816 s ------------- c) you should direct your boat at 90° to the river bank. You will cross the river in 126/17 = 7.41 seconds, but you will land 7.41*5,4 m = 40 m downstream from the place directly opposite. -------------- d) the minimum time = 7.41 seconds, as outlined in c) e) the minimum velocity of your boat is equal to the velocity of the river = 5.4 m/s. But in this border case you will need infinitely long. So your velocity should be at least a tiny bit more than 5.4 /s. Regards

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Hello a) let angle of boat with bank = (angle) cos(angle) = 5.4/17 angle = 71.48° -------- b) the perpendicular velocity of the boat will be v0*sin71.48 m/s = 17*sin71.48 m/s. And t = s/v = 126/(17*sin71.48) t = 7.816 s ------------- c) you should direct your boat at 90° to the river bank. You will cross the river in 126/17 = 7.41 seconds, but you will land 7.41*5,4 m = 40 m downstream from the place directly opposite. -------------- d) the minimum time = 7.41 seconds, as outlined in c) e) the minimum velocity of your boat is equal to the velocity of the river = 5.4 m/s. But in this border case you will need infinitely long. So your velocity should be at least a tiny bit more than 5.4 /s. Regards

Ossi G

a) 17 cos θ = 5.40 or θ = cos^-1[5.40/17] = 71.48 b) Time required = 126/{17* sin θ} = 7.82 s c) , d) e) wil be done later

Let'slearntothink

a) 17 cos θ = 5.40 or θ = cos^-1[5.40/17] = 71.48 b) Time required = 126/{17* sin θ} = 7.82 s c) , d) e) wil be done later

Let'slearntothink

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