A rowboat crosses a river with a velocity of 3.98 mi/h at an angle 62.5° north o

Question

A rowboat crosses a river with a velocity of 3.98 mi/h at an angle 62.5° north of west relativeto the water. The river is 0.595 mi wideand carries an eastward current of 1.25 mi/h. How far upstream isthe boat when it reaches the opposite shore?
                                ft
A rowboat crosses a river with a velocity of 3.98 mi/h at an angle 62.5° north of west relativeto the water. The river is 0.595 mi wideand carries an eastward current of 1.25 mi/h. How far upstream isthe boat when it reaches the opposite shore?
                                ft

Explanation / Answer

Therefore VRG,x = 1.25mi/h andVRG,y = 0mi/h
Velocity of the rowboat relative to river isVBR = 3.98mi/h at 62.5o north ofwest Therefore VBR,x =-(3.98mi/n)cos62.5o  = -1.839mi/h
And VBR,y =(3.98mi/n)sin62.5o  = 3.539mi/h Then the x-component of velocity of the rowboat relative toground is VBG,x = VBR,x + VRG,x    =-1.839mi/h + 1.25mi/h
      = -0.589mi/h And  VBG,y = VBR,y + VRG,y       = 3.539mi/h + 0mi/h       = 3.539mi/h
Now time taken for the boat to travel 0.595mi in they-direction is t =(0.595mi)/(3.539mi/h) = 0.16813h Then the distance reached by the boat in the upstream is(0.589mi/h)(0.16813h) = 0.09902857 mi                      =(0.09902857 mi)(5280ft/mi)                      =522.87ft                        ˜ 523ft    =-1.839mi/h + 1.25mi/h
      = -0.589mi/h And  VBG,y = VBR,y + VRG,y       = 3.539mi/h + 0mi/h       = 3.539mi/h
Now time taken for the boat to travel 0.595mi in they-direction is t =(0.595mi)/(3.539mi/h) = 0.16813h Then the distance reached by the boat in the upstream is(0.589mi/h)(0.16813h) = 0.09902857 mi                      =(0.09902857 mi)(5280ft/mi)                      =522.87ft                        ˜ 523ft       = 3.539mi/h + 0mi/h       = 3.539mi/h
Now time taken for the boat to travel 0.595mi in they-direction is t =(0.595mi)/(3.539mi/h) = 0.16813h Then the distance reached by the boat in the upstream is(0.589mi/h)(0.16813h) = 0.09902857 mi                      =(0.09902857 mi)(5280ft/mi)                      =522.87ft                        ˜ 523ft

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