A kayaker needs to paddle north across a 110 m wide harbor.?
If you study to get a pilot's license you learn to do problems like this, called 'wind triangles'. You have to calculate a 'vector sum' of the distance the kayaker paddles and the distance the current moves him. Think about it. He's paddling 110 at 3 m/s. So that would take 36.66 seconds. But during that time he will also be moved to the right at 1.7 m/s, which in 36.66 sec would move him 62.33 m off course. So he will land 62.33 m to the east of where he was aiming for. So instead he modifies his course. He aims for a point 62.33 to the WEST of where he wants to hit the opposite shore. Then the combination of his paddling and the current will take him right where he wants to be. You can think of this as a right triangle. The base of the triangle is his chosen heading (the direction he's pointing, the direction on his compass). The hypotenuse is his actual course (actual direction of travel). The height is the amount the wind moves him. Do you know how to figure out the angles of a triangle with trig? The tangent of the angle is the ratio between the base leg and the height. I don't have my calculator here 8^< But you want to divide 67.33 by 110 and that is the tangent of the correction angle, to the east, from North (i.e. that angle subtracted from 360).
A kayaker needs to paddle north across a 85 m wide harbor. The tide is going out, creating a tidal current tha?
The kayaker has to travel a greater distance than the width to cross the harbour, than when attempting to paddle straight across. (85/3.4) = 25 secs. to travel straight across. In that time, he would have ended up (25 x 2) = 50m. downstream. So now you can sketch a right triangle, 85m. north/south, and 50m. "upstream" on the far bank. a) Angle to row = atn. (50/85), = 29.17 deg., west of north. b) Distance he actually paddles = sqrt. (85^2 + 50^2), = 98.6 metres. Time taken = (98.6/3.4), = 29 secs.
A kayaker needs to paddle north accross a 100m wide harbor.?
The kayaker can paddle 3.0 m/s, which must be 2.0 m/s to the west (to offset the eastward flow of the river) plus "v" speed north. What is v? well, we know that the total speed is 3.00 and we can make a triangle with the west side 2.0, north side v, and hypotyeneuse 3.0 so 2.02 + v2 = 3.02 if we solve we get v = √5 = 2.23 m/s northward The direction then should be tanθ = 2.236/2.0 or θ = 48.19 deg north of west Now, for the time to cross the harbor... we know his northward speed is 2.236 m/s so time = dist / speed = 100 m / 2.236 m/s = 44.72 seconds
A kayaker needs to paddle north across a 90 m wide harbor.?
The current DOES affect your north speed since you have to aim "upstream" in order to actually travel north. You need to build a western component into your direction: 2.8m/s * sinΘ = 1.8m/s Θ = 40.0º Then your northern speed is v = 2.8m/s * cos40.0º = 2.14 m/s and the crossing time is t = d / v = 90m / 2.14m/s = 42 seconds If you find this helpful, please award Best Answer!
Physics help with vectors!!!!!!!?
A kayaker needs to paddle north across a 115 m wide harbor. The tide is going out, creating a tidal current that flows to the east at 2.2 m/s. The kayaker can paddle with a speed of 3.0 m/s. a.) In which direction should he paddle in order to travel straight across the harbor? ____degrees west of north b.) How long will it take him to cross? ____seconds
Mastering Physics, HELP!!!?
Assume 90° = North We know the kayaker must have a western 2m/s component that cancels the easterly 2m/s so we can write |Vk| = √(2² + Vky²) = 3 where Vky is the north component. Vk² = 2² + Vky² = 3² -----> Vky² = 3² - 2² = 5 Thus Vkx = -2 or 2m/s at 180° or 2m/s West The direction = arcos(2/3) = 48.2° above west = 180-48.2 = 131.8° OR 41.8° west of north. <----- A His northern component = 3*cos41.8 = √5 = 2.236m/s 100m/2.236m/s = 44.7s <----- B