Calculating the time to cross a river

[gnits]

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.

 

[PeroK]

The N-S component of the velocity is variable. You have to integrate to get the distance, surely?

 

[BvU]

(faulty?) reasoning

faulty indeed. You can check yourself that this does not satisfy the given equation.

Cause: there is no motion with constant acceleration in the y direction. The river doesn't 'push' with a constant force !

You will have to solve that differential equation to obtain he given answer, which should not be $1/6$ but something with the dimension of length.

LaTeX isn't working for me

That is strange. For most others it works just fine. Do you enclose things in ## for in-line math and in $$ for displayed math ?
e.g. $$ \vec v = V \left ( {\bf\hat\imath } + {x(a-x)\over a^2 } {\bf\hat\jmath } \right ) $$ gives
$$ \vec v = V \left ( {\bf\hat\imath } + {x(a-x)\over a^2 } {\bf\hat\jmath } \right ) $$ (which is correct).

 

[Delta2]

Your equation 1 is wrong. The i-component is correct but the j-component is wrong. To find the j-component correctly you have to integrate the velocity $u$, first replace $x=Vt$ in the expression of $u$ and then integrate with respect to the time $t$.

EDIT:I edited this post, integration with respect to x was a bad idea.
EDIT2: Well because there is linear relationship between x and t it turns out that the integral $$\int_0^a\frac{x(a-x)}{a^2}dx=\int_0^{\frac{a}{V}}\frac{Vt(a-Vt)}{a^2}Vdt=AC$$

 

[gnits]

Thanks all for your help. I see my error now. Thanks very much for the quick help.