Trajectory parametric equations

  • Thread starter krisrai
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  • #1

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


The Attempt at a Solution


this is where i have trouble:
I understand how my text uses this equation i just dont know were they got these numbers:

--> 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
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
Homework Helper
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:

\vec v &= \vec v_0 +\vec a t \\
(5\hat i - 5 \hat j) &= \vec v_0 + 3\vec a

and the displacement equation is:

\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

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]:
(6 \hat i - 6 \hat j) &=3\vec v_0 + \frac{1}{2}3^2 \vec a

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

and finally multiplying by 2:
(4 \hat i - 4 \hat j) &=2 \vec v_0 + 3 \vec a
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?