- #1
gnits
- 137
- 46
- Homework Statement
- To calculate time to cross a river
- Relevant Equations
- d = st
Could I please ask for help on the last part of this question:
So, part b, I get the right time but not the right distance.
Book answers are: distance = 1/6 and time = a/V.
Here's my (faulty?) reasoning (LaTeX isn't working for me):
The boat is steered due east and so would have a velocity of V east and 0 north but for the current of speed u = x(a-x)V / a^2 which will push it north, so velocity vector of boat is:
V i + ( x(a-x)V/a^2 ) j
Therefore the position vector of the boat at time t will be:
V t i + ( x(a-x)V t /a^2 ) j - Call this "Equation 1"
We can see that the j component is equal to x(a-x)/a^2 times the i component and so the gradient of this line is a(a-x)/a^2 as required.
So boat will have crossed river when Vt = a and so t = a/V as required.
Finally, to calculate the distance AC, I put this time into the y component of Equation 1 and this gives me y = x(a-x) / a
Not 1 / 6.
Thanks for any help,
Mitch.
So, part b, I get the right time but not the right distance.
Book answers are: distance = 1/6 and time = a/V.
Here's my (faulty?) reasoning (LaTeX isn't working for me):
The boat is steered due east and so would have a velocity of V east and 0 north but for the current of speed u = x(a-x)V / a^2 which will push it north, so velocity vector of boat is:
V i + ( x(a-x)V/a^2 ) j
Therefore the position vector of the boat at time t will be:
V t i + ( x(a-x)V t /a^2 ) j - Call this "Equation 1"
We can see that the j component is equal to x(a-x)/a^2 times the i component and so the gradient of this line is a(a-x)/a^2 as required.
So boat will have crossed river when Vt = a and so t = a/V as required.
Finally, to calculate the distance AC, I put this time into the y component of Equation 1 and this gives me y = x(a-x) / a
Not 1 / 6.
Thanks for any help,
Mitch.