Asked by John
A boat is capable of 20 km/h in still water. You wish to cross the river downstream so that the angle the boat's push makes with the bank is 50 degrees. At what angle to the bank should the boat steer if the current is 8 km/h? How long will it take to cross if the river is 1.7 km/h?
Answers
Answered by
Henry
A. Vr = Vb + Vc.
20[50o] = Vb - 8i,
Vb = 20[50]+8i,
Vb = 20*Cos50+20*sin50 + 8i,
Vb = 12.9 + 15.3i + 8i = 12.9 + 23.3i.
Tan A = Y/X =23.3/12.9 = 1.80620,
A = 61o.
20[50o] = Vb - 8i,
Vb = 20[50]+8i,
Vb = 20*Cos50+20*sin50 + 8i,
Vb = 12.9 + 15.3i + 8i = 12.9 + 23.3i.
Tan A = Y/X =23.3/12.9 = 1.80620,
A = 61o.
Answered by
Henry
Your last question doesn't make sense.
Answered by
John
1.7 km wide.
Srry
Srry
Answered by
Henry
B. sin50 = 1.7/d. d = 2.22 km.
Answered by
Henry
50o W. of N. = 140o CCW.
Vr = Vb + Vc.
20[140] = Vb + (-8i),
Vb = 20[140] + 8i,
Vb = 20*Cos140+20*sin140 + 8i.
Vb = -15.3 + 12.9i + 8i = -15.3 + 20.9i
Tan A = Y/X = 20.9/(-15.3) = -1.36601. A = -53.8o = 53.8o N. of W. = 126.2o CCW = 36.2O W of N.
Vr = Vb + Vc.
20[140] = Vb + (-8i),
Vb = 20[140] + 8i,
Vb = 20*Cos140+20*sin140 + 8i.
Vb = -15.3 + 12.9i + 8i = -15.3 + 20.9i
Tan A = Y/X = 20.9/(-15.3) = -1.36601. A = -53.8o = 53.8o N. of W. = 126.2o CCW = 36.2O W of N.