A man can row a boat at a speed of 4.40 km/h in still water. If he is crossing a river where the current is 2.45 km/h, in what direction should his boat be headed if he wants to reach a point directly opposite his starting point?
I've tried 56.16 degrees and 29.1 degrees and neither is right any other ideas???
Yes
To determine the correct direction in which the man should steer the boat, we need to understand the concept of vector addition.
Let's consider the velocity of the boat in still water as one vector and the velocity of the river current as another vector. The boat's velocity in still water is 4.40 km/h, and the current is flowing at 2.45 km/h.
Now, imagine these two vectors being added together to obtain the resultant velocity, which will give us the actual direction and speed of the boat when moving across the river.
To find the direction of the resultant vector, we can use the law of cosines. The resultant vector can be calculated using the magnitudes and angle between the two vectors.
Let's assume the angle between the boat's velocity in still water and the resultant velocity is θ. We can apply the law of cosines as follows:
(resultant velocity)^2 = (velocity of the boat in still water)^2 + (velocity of the river current)^2 - 2 * (velocity of the boat in still water) * (velocity of the river current) * cos(θ)
Plugging in the values:
(resultant velocity)^2 = (4.40 km/h)^2 + (2.45 km/h)^2 - 2 * (4.40 km/h) * (2.45 km/h) * cos(θ)
Since we want the resultant velocity to be opposite to the river's flow, the angle θ should be 180 degrees.
Now we can solve for the angle θ:
(resultant velocity) = √((4.40 km/h)^2 + (2.45 km/h)^2 - 2 * (4.40 km/h) * (2.45 km/h) * cos(180))
Simplifying further:
(resultant velocity) = √(19.36 km^2/h^2 + 6.0025 km^2/h^2 + 21.56 km^2/h^2)
(resultant velocity) = √(46.9225 km^2/h^2)
(resultant velocity) = 6.848 km/h (rounded to three decimal places)
Therefore, the boat should be headed at an angle of 180 degrees, directly opposite the river's current, to reach a point directly opposite the starting point.