What Time Does the Particle Reach the Y-Axis?

[Feldoh]

Homework Statement

A particle moves in the xy-plane so that its position at any time t, 0 =< t =< pi, is given by:
x(t) = \frac{t^2}{2}-ln(1+t)
y(t) = 3sint

-- At that time is the particle on the y-axis on the interval? Find the speed and acceleration vector of the particle at this time.

Homework Equations

None?

The Attempt at a Solution

I'm actually lost -- I believe that when x(t) = 0 it will be on the y-axis but I must be doing the algebra wrong I can't seem to get an exact solution...

 

[Dick]

You can't solve it exactly algebraically. A well reasoned guess for a value of t such that x(t)=0 will give you the answer though.

 

[Feldoh]

You can't solve it exactly algebraically. A well reasoned guess for a value of t such that x(t)=0 will give you the answer though.

So my method is right up to the point I was at? I should mention I needed to find a t value such that 0 < t < pi, so I don't think 0 is the answer, I should have added that, sorry. Eye-balling the graph it's something like t=1.25 but my teacher claims that it is possible to get an exact solution, but not that you've confirmed it I really don't think there is...

Oh and, velocity is <x'(t), y'(t)> or (x'(t))i+(y'(t))j and acceleration is <x''(t), y''(t)> or (x''(t))i+(y''(t))j

 

[Dick]

What's wrong with t=0?

 

[Feldoh]

What's wrong with t=0?

Nothing I just forgot to add that the problem asks for a value of t > 0. I know it works but the problem isn't asking for it I don't think. My mistake I should have clarified:(

 

[Dick]

Your post does say t>=0. You've eyeballed the t>0 root correctly. But there's no way you can solve for that one in terms of elementary functions.