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