Trajectory parametric equations

[krisrai]

Homework Statement

A particle is located at r=(2i+4j)m at t=0s.
At t=3s it is at r=(8i-2j)m and has velocity v=(5i-5j)m/s
a)what is the particles acceleration vector a?

Homework Equations

r1=r0+v0(t1-t0)+1/2a(t1-t0)^2
v1=v0+at

The Attempt at a Solution

v1=v0+at
(5i-5j)m/s=v0+a(3s-0s)
v0=(5i-5j)m/s-3a

this is where i have trouble:
I understand how my text uses this equation i just don't know were they got these numbers:
r1=r0+v0(t1-t0)+1/2a(t1-t0)^2
(4i-4j)m=(3s^2)a+(2s)vo

--> did they assume that r0=0? and how did they get r1?
and where did they get 2sv0 because from the information given i thought it was all 3seconds



okay well anyway i understand they would take the v0 found in the first part and sub it into get this:
(3s^2)a+ (2s)[(5i+5j)m/s - 3s(a)] =(4i+4j)m
and a=(2i+2j)m/s^2

Please can someone explain to me where they got the 2 from
and also i don't understand how they subtract3s(a)
I would really appreciate it.

 

[Shooting Star]

In relevant equations, you have written down the formulae for constant accn, which need not be the case here.

It is not mentioned explicitly, but presumably the velocity is 0 at t=0. If so, then,

v_avg = (r2-r1)/Δt = Δr/Δt. You know Δr.

Finding the avg accn is just one step from here.

(v, r2, r1, Δr are all vectors.)

 

[Phlogistonian]

The equations in the text's answer don't make sense to me either. They look like they're full of typos. I'd ignore them.

 

[alphysicist]

Hi krisrai,

The numbers from the text look right to me; they are just showing one step in solving the problem.

The velocity equation, with the numbers from the problem plugged in, are:

<br /> \begin{align}<br /> \vec v &amp;= \vec v_0 +\vec a t \\<br /> (5\hat i - 5 \hat j) &amp;= \vec v_0 + 3\vec a <br /> \end{align}<br />

and the displacement equation is:

<br /> \begin{align}<br /> \vec r &amp;= \vec r_0 + \vec v_0 t + \frac{1}{2}\vec a t^2\\<br /> (8 \hat i - 2 \hat j) &amp;= (2\hat i + 4\hat j) + 3\vec v_0 + \frac{1}{2}3^2 \vec a<br /> \end{align}<br />

So you have two equations, with two unknown (\vec v_0 and \vec a) and you have to solve for them. What the book is doing, is, since there is a 3\vec a term in the velocity equation, is to manipulate the displacement equation to get a term of 3\vec a; that way you can easily solve the simultaneous equations.

So the displacement equation becomes:

combining \vec r and \vec r_0:
<br /> (6 \hat i - 6 \hat j) &amp;=3\vec v_0 + \frac{1}{2}3^2 \vec a<br />

dividing by 3:
<br /> (2 \hat i - 2 \hat j) &amp;=\vec v_0 + \frac{1}{2}3 \vec a<br />

and finally multiplying by 2:
<br /> (4 \hat i - 4 \hat j) &amp;=2 \vec v_0 + 3 \vec a<br />
to get the result from your book.

So you can solve this last equation for 3 \vec a, and plug it directly into the 3 \vec a term in the velocity equation. Then you have an equation with only one unknown \vec v_0 and you can find it's \hat i and \hat j values. Once you have that, you can plug it back into the velocity equation and find the acceleration \vec a. What do you get?