A boat, which has a speed of 9.00 m/s in still water, would like to reach a point on the shore directly across a river. The river has a current of 3.50 m/s and the river is 550 meters wide.
a. In what direction should the boat be aimed in order for the boat to head directly across
the river?
b. How long will it take the boat to reach the opposite shore?
It has to travel upstream at an angle θ (where straight across is θ=0) such that
sinθ = 3.50/9.00
the distance traveled will be 550/cosθ
and of course, time = distance/speed, so ...
a. Vr = Vb - 3.5i = 9.0 = resultant Velocity,
Vb = 9 + 3.5i = 9.66m/s[21.25o] CCW.
Direction = 21.25o CCW = 21.25o N. of E.
b. d = V * t = 550,
9*t = 550,
t = 61 s.
To determine the direction in which the boat should be aimed in order to head directly across the river, we can use the concept of vector addition.
a. To reach a point directly across the river, the boat needs to counteract the effect of the river current. The boat should be aimed slightly upstream from the point directly across.
The velocity of the boat in still water is 9.00 m/s, and the velocity of the river current is 3.50 m/s. The resultant velocity needed to counteract the river current and head directly across the river can be found by subtracting the river current velocity from the velocity of the boat in still water.
Resultant velocity = Boat velocity - River current velocity
= 9.00 m/s - 3.50 m/s
= 5.50 m/s
Therefore, the boat should be aimed in a direction that results in a velocity of 5.50 m/s.
b. To calculate the time it will take for the boat to reach the opposite shore, we can use the equation:
time = distance / velocity
The distance the boat needs to cover is the width of the river, which is given as 550 meters. The velocity of the boat needed to cross the river is the resultant velocity we calculated earlier, which is 5.50 m/s.
Substituting these values into the equation, we have:
time = 550 m / 5.50 m/s
= 100 seconds
Therefore, it will take the boat 100 seconds to reach the opposite shore.