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Trajectory parametric equations

  • Thread starter krisrai
  • Start date
  • #1
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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 dont 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 in to 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 dont understand how they subtract3s(a)
I would really appreciate it.
 

Answers and Replies

  • #2
Shooting Star
Homework Helper
1,977
4
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.)
 
  • #3
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.
 
  • #4
alphysicist
Homework Helper
2,238
1
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:

[tex]
\begin{align}
\vec v &= \vec v_0 +\vec a t \\
(5\hat i - 5 \hat j) &= \vec v_0 + 3\vec a
\end{align}
[/tex]

and the displacement equation is:

[tex]
\begin{align}
\vec r &= \vec r_0 + \vec v_0 t + \frac{1}{2}\vec a t^2\\
(8 \hat i - 2 \hat j) &= (2\hat i + 4\hat j) + 3\vec v_0 + \frac{1}{2}3^2 \vec a
\end{align}
[/tex]

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

So the displacement equation becomes:

combining [itex]\vec r[/itex] and [itex]\vec r_0[/itex]:
[tex]
(6 \hat i - 6 \hat j) &=3\vec v_0 + \frac{1}{2}3^2 \vec a
[/tex]

dividing by 3:
[tex]
(2 \hat i - 2 \hat j) &=\vec v_0 + \frac{1}{2}3 \vec a
[/tex]

and finally multiplying by 2:
[tex]
(4 \hat i - 4 \hat j) &=2 \vec v_0 + 3 \vec a
[/tex]
to get the result from your book.

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

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