the distance of a particle from a fixed point O is given by s= 3cos2t + 4sin2t
show that the velocity v and the acceleration a
are given by v^2 + 4s^2 =100, a+4s=0.
Please give me a hint someone!!

Question

the distance of a particle from a fixed point O is given by s= 3cos2t + 4sin2t 
show that the velocity v and the acceleration a 
are given  by v^2 + 4s^2 =100, a+4s=0.
Please give me a hint someone!!

anonymous · Answer

Velocity is the derivative of the position and acceleration is the second derivative of position and first derivative of velocity!

anonymous · Answer

also @stamp has a weak pellet

anonymous · Answer

I know that but what does the v^2 + 4s^2 =100 mean?

anonymous · Answer

cause when I diffrentiate I get trig ratios.

anonymous · Answer

I don't have v , I only have s !!

anonymous · Answer

no read the question ,

anonymous · Answer

it says

anonymous · Answer

he distance of a particle from a fixed point O is given by s= 3cos2t + 4sin2t show that the velocity v and the acceleration a are given by v^2 + 4s^2 =100, a+4s=0.

anonymous · Answer

Task 1, find the velocity equation.  Can you do it?

anonymous · Answer

Can you take the derivative?

anonymous · Answer

yes is it -6sint+8cos2t?

anonymous · Answer

Okay, so then substitute that in for $v$ in $v^2+4s^2=100$

anonymous · Answer

$$
v^2+4s^2=(-6\sin t+8\cos2t)^2+4(3\cos2t + 4\sin2t)^2=\dots
$$

anonymous · Answer

oh

anonymous · Answer

Hint:  $\sin^2(t)+\cos^2(t)=1$

anonymous · Answer

Also actually you got velocity wrong I think, should have a $2t$ rather than just $t$.

anonymous · Answer

oh yes !  sorry