A 0.40-km wide river has a uniform flow speed of 7.1 m/s toward the east. It takes 39 s for a boat to cross the river to a point directly north of its departure point on the south bank. In what direction must the boat be pointed in order to accomplish this?
plzz...help me solving this!!
thanks..
2006-09-02 21:47:07 · 3 answers · asked by steve 2 in Science & Mathematics ➔ Physics
i mean give the angle also!!
2006-09-02 21:51:44 · update #1
Suppose velocity vector of the boat (without considering the stream of water) is:
Vb=Ai+Bj
When we add the velocity vector of the river stream (Vr=7.1i) to the above vector the result should be a straight northward vector which its magnitude is 400/39= 10.26 m/s (the velocity of boat considering the stream of water).
(Ai+Bj)+(7.1i)=10.26j
A+7.1=0
B=10.26
A=-7.1
B=10.26
The tangent of the direction of the boat (teta) with reference to the stream direction is -10.26/7.1= -1.445
teta=124.68 degree
2006-09-02 22:09:41 · answer #1 · answered by Farshad 2 · 0⤊ 0⤋
In 39 s the river will flow through a distance of 39s x 7.1 m/s =276.9 m from West to East.
The boat must be pointed to a point ‘W’ which is 276.9 m to the West of the exact North point (N). ‘S’ is the starting point.
tan ( WSN) = 276.9/ 400 = 0.69225.
Angle ( WSN) = 34 degree 41 minutes 34 second (From North to West).
The boat’s speed is distance/ time.
Distance traveled is square root of the sum of the squares of 400 and 276.9 = 486.5m
The boat’s speed is 486.5m / 39s = 12.47 m/s.
The resultant speed of both the boat and river is Sq. root of 12.47^2 - 7.1^2 = 10.25m/s
10.25 x 39 = 400 m the width of the river.
2006-09-03 07:15:52 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
to south
2006-09-03 04:49:29 · answer #3 · answered by x_squared 4 · 0⤊ 0⤋