A particle moves along the x axis according to the equation x=2.04 + 2.98t - 1.00t^2, where x is in meters and t is in seconds. Find its acceleration at t=2.90 s.

Question

anonymous · Answer

I found that the position of the particle at t=2.90 s is 2.27 m and the velocity at t=2.90 s is 1.56 m/s

anonymous · Answer

@Kkutie7 Can you maybe help me with this?

kkutie7 · Answer

that equation is for the position. You need to take the first and second derivative. as for the first one you need to find x for the first equation.

kkutie7 · Answer

Would you like me to walk you through and explain... I can do that.

kkutie7 · Answer

Oh I didn't read it right. you need to find x" or the second derivative. do you know how to do this?

kkutie7 · Answer

$$x=2.04 + 2.98t - 1.00t^2$$
$$x=-1.00t^2 +2.98t+2.04$$
$$x'=2.00t+2.98$$
$$x"=2.00$$
So it looks like there is a constant acceleration of 2.00
look when t= 2.90
you should be able to get the answer from here =)

kkutie7 · Answer

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anonymous · Answer

I understand that I have to find the second derivative, I guess I just don't know how to go about it. And im still confused on what you plug t=2.90 s into to get the answer?

kkutie7 · Answer

you dont... look at the graph no matter what t value you have its (a) is always 2

anonymous · Answer

So the answer is 2?

kkutie7 · Answer

yes I believe so

anonymous · Answer

Because when I try to submit my answer it says its incorrect, that's why I wasn't sure

anonymous · Answer

Okay it's negative 2 because the second derivative of x was actually -2.00t + 2.98

anonymous · Answer

so x"=-2

kkutie7 · Answer

yeah you are right