1. ## Vector problem

I apologize for the unhelpful title. I don't know how to otherwise classify this problem.

A particle moves along the graph of y = cos(x) so that its x-component of acceleration is always 2. At time t=0, the particle is at the point ($\displaystyle \pi$, -1) and the velocity of the particle is {0,0} (that's the notation my book uses to show a vector).

(a) find the position vector of the particle.
Got that. {$\displaystyle t^2 + \pi , cos (t^2 + \pi)$}

(b). Find the speed for the particle when it is at the point (4, cos(4))
I don't get the question.
Speed is the absolute value of velocity So, it seems like I should set the x part (for lack of a better word) of vector to 4 and the y part of the vector.

However, the answer is some big complicated thing over a square root.
Specifically, $\displaystyle \sqrt((2t)^2 + 2t(-sin(t^2 +\pi))^2$ where t= $\displaystyle \sqrt(4 - \pi)$ .

Basically, how do I do (b)?

Thanks!

2. Originally Posted by Truthbetold
I apologize for the unhelpful title. I don't know how to otherwise classify this problem.

A particle moves along the graph of y = cos(x) so that its x-component of acceleration is always 2. At time t=0, the particle is at the point ($\displaystyle \pi$, -1) and the velocity of the particle is {0,0} (that's the notation my book uses to show a vector).

(a) find the position vector of the particle.
Got that. {$\displaystyle t^2 + \pi , cos (t^2 + \pi)$}

(b). Find the speed for the particle when it is at the point (4, cos(4))
I don't get the question.
Speed is the absolute value of velocity So, it seems like I should set the x part (for lack of a better word) of vector to 4 and the y part of the vector.

However, the answer is some big complicated thing over a square root.
Specifically, $\displaystyle \sqrt((2t)^2 + 2t(-sin(t^2 +\pi))^2$ where t= $\displaystyle \sqrt(4 - \pi)$ .

Basically, how do I do (b)?

Thanks!

$\displaystyle \frac{d^2x}{dt^2} = 2$

$\displaystyle \frac{dx}{dt} = 2t + C$

$\displaystyle C = 0$ since velocity = 0 at t = 0

$\displaystyle \frac{dy}{dt} = -\sin{x} \cdot \frac{dx}{dt} = -2t\sin{x}$

speed = $\displaystyle \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$

speed = $\displaystyle \sqrt{4t^2 + 4t^2\sin^2{x}}$