Please see the previous thread for background to this question.
This is the follow on from the question posted previously.
I've managed to get a little bit further with this one than the last. For ease i'll post the whole question again.
Rower A is stationary at the point A adjacent to the river bank when he spots an object floating down the river. He estimates that the object is 50m away and heading south at a speed of 3m/s (the same speed as the current) see diagram below.
Assuming rower A can sustain a rowing speed of 4m/s
1 Draw a diagram of the velocity vectors involved if rower a is to intercept the object
2 calcuate the direction with respect to the river bank he needs to row (in degrees) in order to achieve his goal.
3 Rower A miscalculated the distance and the angle the object was from him, in fact the object was 60m away at 21 degrees to the bank. Assuming he stayed on the ame heading and speed as part 2, determine the closest point of the object to Rower A
With th ehelp of the forum i finished the first two, its the third that i'm having trouble with.
For this question I had it that I would have to work out the actual direction of travel not just the direction he rows in. So I tried to work out the rowers angle of travel
This is what I worked out
A position = (0,0)
B position = (50sin20,50cos20)
I had the velocity vector of the rower (4sin(20),4cos(20)-3)
Therefore the angle for the rower was Arctan (4sin(20))/ (4cos(20)-3)
Which equalled an angle of 60.99 degrees
The velocity vector of the object I had as (0,-3cos(20))
Now because I need to know the positions of the rower and object I was trying to find the positional vectors, so that I could then work out where each was at any given time. This is where Iím really getting shaky on what Iím doing. Because ei know that on the original 20 degrees and 50 m they would intersect I used that to try and work out the positional vectors. My working out is below (and I donít think its that good.
To calculate the vector position of A
Which becomes (4t*sin(20),4t*cos(20)-3) (not sure if Iím right here)
To calculate the vector position of B
At time t B is at (50sin20,50Cos20)+t(0,-3cos(20))
Which becomes (50sin20,50Cos20)+(0,-3t*cos(20))
I then made the objects vector positions equal to each other
(4t*sin(20),4t*cos(20)-3) = (50sin20,(50cos20)-3t*cos(20))
4t*sin(20) = (50Sin20) 4t*cos(20)-3 = (50cos20) Ė 3t*cos(20)
4t=50Sin20/sin20 4t*cos(20)-3 + 3t*cos20 = 50 Cos 20
4t = 50
t = 12.5
Any help with this is really appreciated, (please use baby steps with me)
Would I be right in saying that the shortest distance between the rower and the object can only occur when either the x co-ordinates of both are the same or when the y co-ordinates of both are the same?