Submitted by jtedds on Thu, 04/27/2017 - 10:27

When forming my expression for R in part (i), I understand that the vertical part (k) is $5\sqrt{5}t -5t^2$, the north part (j) is $5\sqrt{15}t$ but I don't understand how in the solution they've obtained $50 - 15\sqrt{5}t$ for i.

The horizontal part of the $25\ms^{-1}$ is $25\cos(arctan(\frac{1}{2})) = \frac{2}{\sqrt{5}} * 25$ but multiplying through then by $cos(60)$ obtains $5\sqrt{5}$ not $15\sqrt{5}$

## typo!

Thanks for your post - it was a typo and should indeed read $5\sqrt5$. I have updated it accordingly, sorry for not catching it sooner.

## Part (iii)

I have a query about the last part of (iii). I agree with the horizontal and vertical distances given in the solutions, but how to show that the total distance is approx 3? Is this approach reasonable?

$5\sqrt{\frac{5}{64}+\frac{15}{64}+\left(\frac{1}{64}\right)^2}$

$=\frac{5}{8}\sqrt{\frac{1281}{64}}$

$\approx\frac{5}{8}\sqrt{\frac{1296}{64}}$

$\approx\frac{5}{8}\times\frac{36}{8}$

$\approx\frac{5}{8}\times\frac{9}{2}$

$\approx\frac{45}{16}$

$\approx\frac{48}{16}$

$\approx3$

The calculator gives $2.796$, which is not that close to 3, which makes me wonder why we've been asked to show it's approximately 3 and that perhaps there's a better way than I've found.

## Yep

That seems reasonable. Alternatively you could argue that the distance is

\[

\sqrt{\frac{25 \times (64\times(5+15)+1)} {64}}\\

=\sqrt{\frac{25 \times 1281}{64^2}}\\

=\sqrt{\frac {32025}{64^2}}\\

\approx \sqrt{\frac{32768}{2^{12}}}\\

=\sqrt{\frac{2^{15}}{2^{12}}}\\

=\sqrt{8}\\

\approx 3

\]

I think I prefer your way though!