Is anyone able to point me in the right direction to answering these questions please.
Surely you can solve b). Point a) can be solved with a simple application of the Pythagoras theorem.
In general, let P' be the projection of P on OR. If you know angle POR, then you an find OP' and PP' using trigonometric functions. Knowing PP', you can find P'Q using the Pythagoras theorem. Therefore, you can find OQ.
1. What you have calculated is the distance $\displaystyle |\overline{OQ}|$ if P is placed on the left end of the diameter of the circle.
2. As far as I understand the question you are asked to determine the length of the way the point Q is traveling on the line OR if the point P describes a semicircle. And that must be a complete diameter, that means the answer is 1.5 m.
Thank you once again earboth, with your help I was able to understand the question. Hopefully I have answered them correctly (although you have already confirmed b)
a) Initially Q is @ 3.2 m after P rotates 90 degress using the pythagoras theorem we find
$\displaystyle c^2=b^2-a^2$
$\displaystyle sqrt(2.5^2-0.75^2)= 2.38 m$
Distance the end Q travelled = 3.2 m - 2.38 m = 0.82 m
b) Distance the end Q travelled (half a turn) 2 x .75 = 1.5 m
c) using the sine rule we find sin(30) c = 3.12 m
using the sine rule again we find sin(170) c =1.76 m
Distance the end Q travelled 1.36 m
If I have made any mistakes please let me know. Thank you.
It's 3.25 m, so the traveled distance in a) is 0.87 m, not 0.82 m.
I don't understand this. First, what is c? 3.12 / sin(30) is not the same as 1.76 / sin(170). Second, sin(30) determines the vertical position of P, not the horizontal position of Q.
As I said, one way is to find PP' and OP' (where P' is the projection of P) and then use Pythagoras theorem to find P'Q; then OQ = OP' + P'Q.
1. I'll take emakarov's suggestions. I only use some different labels. (see attachment. I've modified your image a little bit)
2.
- $\displaystyle |\overline{OF}| = p = 0.75 \cdot \cos(\theta)$
- $\displaystyle |\overline{PF}| = 0.75 \cdot \sin(\theta)$
- $\displaystyle |\overline{FQ}| = q = \sqrt{2.5^2-(0.75 \cdot \sin(\theta))^2}$
The traveled distance w:
$\displaystyle w = 3.25 - (p+q)$
3. The advantage of emakarov's method is that the traveled distance w depends only on the angle of rotation $\displaystyle \theta$. So you only have to plug in the angle to get the distance from the start-point.