A boy can swim with a speed of 26metre per minute in still water. Be wants to swim across a 150metere wide river from point A to point B directly opposite on the other side of the River.The river flows at 10metre per minute (a)if he always swims in the direction parallel toAB,find how far he lands downstream of B. (b) in what direction relative to the bankmust the boy swim so as to cross directly from A to B
2 answers
1)10/26=X/150 X=1500/26=57.7m 2)costheta=10/26 theta=67.3degrees
a. Tan A = Y/X = 10/26 = 0.38462
A = 21.04o E of N.
Tan A = d/150
d = 150*Tan 21.04 = 57.7 m = Distance
downstream.
b. To offset the affect of the current,
the swimmer must head 21.04o W. of N.
Vsc = Vs + Vc = 26i
Vs + 10 = 26i
Vs = -10 + 26i
Tan A = 26/-10 = -2.600
X = -68.96 = 68.96o CW from -x axis
180-68.96 = 111.04o CCW. = 21.04o W. of
N.
A = 21.04o E of N.
Tan A = d/150
d = 150*Tan 21.04 = 57.7 m = Distance
downstream.
b. To offset the affect of the current,
the swimmer must head 21.04o W. of N.
Vsc = Vs + Vc = 26i
Vs + 10 = 26i
Vs = -10 + 26i
Tan A = 26/-10 = -2.600
X = -68.96 = 68.96o CW from -x axis
180-68.96 = 111.04o CCW. = 21.04o W. of
N.
2 answers