Trajectory parametric equations

In summary, the conversation discusses a problem related to the position, velocity, and acceleration of a particle at different points in time. The equations used for this problem involve finding the average velocity and displacement, and then manipulating them to solve for the unknown variables. The book's solution involves using simultaneous equations to find the values of velocity and acceleration.
  • #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 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.
 
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  • #2
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
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?
 

1. What are trajectory parametric equations?

Trajectory parametric equations are mathematical equations used to describe the path of an object in two or three-dimensional space. They use parametric variables, typically denoted as t, to represent time and calculate the position of the object at any given time.

2. How are trajectory parametric equations different from standard equations?

Trajectory parametric equations use parametric variables to describe the position of an object, while standard equations use only variables of x and y. This allows for a more flexible and accurate description of the object's path, especially in cases where the object's velocity or acceleration may vary.

3. What are the benefits of using trajectory parametric equations?

Using trajectory parametric equations allows for a more precise and detailed analysis of an object's motion. They can accurately describe complex paths and account for variations in velocity and acceleration. They are also useful for predicting future positions of the object.

4. How are trajectory parametric equations used in real life?

Trajectory parametric equations have many practical applications, such as in predicting the path of projectiles in physics or the trajectory of a rocket in aerospace engineering. They are also used in computer graphics to create realistic motion for animated objects.

5. What is an example of a trajectory parametric equation?

A simple example of a trajectory parametric equation is x = 2t, y = t^2, where t represents time. This equation describes a parabolic path, with the object's horizontal position increasing linearly with time and its vertical position following a quadratic function.

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