Recent content by Charge

Yes, both of you guys were incredibly helpful and I appreciate the time you guys took to help me out with this problem.

Ahh okay this makes a bit more sense. Thank you, you've been a tremendous help!

See this is the bit I don't understand, how is this calculated? What I've tried to do is add the Vx to the Vy, but I get the same velocity as if he were not counteracting the current. I'm sure I must be missing some key element to this? I've found a resource online that claims the velocity is...

So if the x component is zero doesn't that mean the y velocity is still 4 m/s? That wouldn't make sense to me though. I've been trying to add the Vx of -1 to the Vy of 4 m/s but I get the same velocity as if he was under the effect of the current. Any tips?

vx would be -1, I'm thinking because that would cancel out with the velocity of the current. I would also think that the x component of his velocity relative to the water and the ground would be the same because there is no velocity in the y direction from the water. The thing that confuses me...

But why would they be anything other than right angles? From my understanding the water is the velocity in the x direction and the boats velocity is in the y direction. For finding the resultant velocity into account it was easy to just find the resultant vector of those two vectors. However not...

That sounds like how to solve part 1 I think? Part 1 the cat just goes forward and doesn't account for the current. But when Phil goes the problem says he "takes the water into account so that the path of his boat relative to the ground is due north", which I'm not exactly sure what that means...

Problem: Dr. L and his cat Kepler are coming home from fishing. They ended their trip on the south bank of a river, directly across a 1200 m wide river, from where Caroline was going to pick them up on the north bank. They are in identical boats that can travel at 4 m/s (Vb). The river is...