I've attempted this question but can't seem to get the right answer.
Three identical particles A, B and C are moving in a plane and, at time $t$, their position vectors, $â$, $b̂$ and $ĉ$, with respect to an origin O, are (in metres):
$$â=(2t+1)î+(2t+3)ĵ$$ $$b̂=(10-t)î+(12-t)ĵ$$ $$ĉ=(t^{3}-15t+4)î+(-3t^{2}+2t+1)ĵ$$
The first part of the question was to find the magnitude of the velocity of particle C relative to particle A when $t=2.00s$, which I did correctly. I substituted $t=2$ into the position vectors and then found the relative displacement: ($â-ĉ=23î+14ĵ$). Then I calculated its magnitude using Pythagoras ($\sqrt{23^{2}+14^{2}}=26.9$, after which I divided by $t=2$ to find the relative velocity, which is $13ms^{-1}$.
The second part of the question asks to find the angle which this relative velocity makes with the unit vector $î$ at this time. I drew myself a little diagram and realised that the required angle should be $90°$ plus the angle which the relative velocity vector makes with the vertical. To calculate the angle that the relative velocity makes with the vertical, I used trigonometry ($tan^{-1}(23/14)=58.67°$. Adding the two together, I got an angle of $148.7°$, but this is incorrect. I'm not sure what the correct answer is as I can only confirm this if I actually submit the correct answer.
$V_{(a,o)}=2i +2j$ (differentiate wrt t) and $ V_{(c,o)}=(3t^2-15)i+(-6t+2)j$ where $V_a$ and $V_c$ are velocity relative to origin,
Now $V_{(c,a)}=V_{(c,o)}+V_{(o,a)}=(3t^2-15)i+(-6t+2)j-2i -2j=(3t^2-17)i-6tj$
Now substitute $t=2$ for the velocity of $c$ relatively $a$ which is $-5i-12j$
For the second part take the dot product with $i$ then $\cos (\theta)=\dfrac{-5}{13}$.