Finding Acceleration in Two-Dimensional Kinematics

  • Thread starter indietro
  • Start date
  • Tags
    Kinematics
In summary, a flying saucer with a constant acceleration is observed with positions and velocities shown in a diagram. The saucer's acceleration can be found using the equation a = ∆v/t, but since the time is not given, the chain rule can be used to find the acceleration. The diagram shows a point at (0,0) with a vertical velocity of 200j m/s and another point at (2000,1000) with a south-east velocity of 200i-100j m/s. The position vector, r, only tells the position and the direction of the tangent of the trajectory shows the direction of velocity. The question provides the information in the form of vx and vy graphs, but they can also
  • #1
indietro
46
0

Homework Statement


a flying saucer manueverign with a constant acceleration is observed with the positions and velocities shown below. what is the saucer's acceleration?

Homework Equations


a = [tex]\Delta[/tex]v / t

The Attempt at a Solution


so a time is not given.. is there another way to find acceleration?
** actually if someone knows a site that gives a clear and concise tutorial on two-dimensional kinematics that would be great :). I know there is one attached to this site.. but i do better with actual examples and more visual.
 

Attachments

  • 7.jpg
    7.jpg
    3.1 KB · Views: 371
Physics news on Phys.org
  • #2
indietro said:
a flying saucer manueverign with a constant acceleration is observed with the positions and velocities shown below. what is the saucer's acceleration?

so a time is not given.. is there another way to find acceleration?

Hi indietro! :smile:

(I can't see the picture yet, but …)

The standard trick for finding dv/dt from v and s without involving t is to use the chain rule …

dv/dt = dv/dx dx/dt = v dv/dx :wink:
 
  • #3
oo ok for the picture it is a x-y graph that shows a point at (0.0) with a vertical vector (200[tex]\hat{j}[/tex] m/s) and a second point at (2000, 1000) with a south-east vector (200[tex]\hat{i}[/tex] - 100[tex]\hat{j}[/tex] m/s)

for the chain rule: is the v the final velocity?
 
  • #4
v = dx/dt
 
  • #5
so i have a question: for an x-y graph showing the trajectory, what does [tex]\vec{r}[/tex] (starts at origin and goes to a point on the trajectory) tell me? the velocity at that point? or does it only tell me the direction of velocity at that point?
 
  • #6
sorry but I am really confused as how to relate an xy-graph, a vx graph, a vy graph and acceleration. Like what does each tell me, how can i find position after a certain time, how can i find acceleration?
 
  • #7
Hi indietro! :smile:

(just got up :zzz: …)
indietro said:
so i have a question: for an x-y graph showing the trajectory, what does [tex]\vec{r}[/tex] (starts at origin and goes to a point on the trajectory) tell me? the velocity at that point? or does it only tell me the direction of velocity at that point?

r only tells you the position, and the direction of the tangent of the trajectory tells you the direction of the velocity.
indietro said:
sorry but I am really confused as how to relate an xy-graph, a vx graph, a vy graph and acceleration. Like what does each tell me, how can i find position after a certain time, how can i find acceleration?

Why are you using a graph? Does the question tell you to?

If it doesn't, then forget graphs, and just use equations. :smile:
 
  • #8
yes the question gives me all the information in the form of an vx graph and vy graph ...:(
 
  • #9
indietro said:
yes the question gives me all the information in the form of an vx graph and vy graph ...:(

hmm … I still can't see your picture.

I'll have to wait until I can see it.

(I've reported it, so hopefully it'll come up soon :redface:)
 
  • #10
oooh, I see it now! :tongue2:​

ok, that isn't a graph (a graph would be a continuous curve) …

it's just a diagram, defining the two velocities in a picture instead of in words.

So you don't have to use a "graph method" …

just write the velocities as 200j and 200i - 100j, and carry on from there. :smile:
 

What is two dimensional kinematics?

Two dimensional kinematics is the study of motion in two dimensions, typically represented on a coordinate plane. It involves analyzing the position, velocity, and acceleration of an object in both the x and y directions.

How is two dimensional kinematics different from one dimensional kinematics?

The main difference between two dimensional kinematics and one dimensional kinematics is that two dimensional kinematics takes into account motion in both the x and y directions, while one dimensional kinematics only considers motion in a single direction. This allows for a more comprehensive analysis of an object's motion.

What equations are used in two dimensional kinematics?

The equations used in two dimensional kinematics are the same as those used in one dimensional kinematics, such as the equations for velocity and acceleration. However, in two dimensions, these equations are applied separately to the x and y components of an object's motion.

How can two dimensional kinematics be applied in real-world situations?

Two dimensional kinematics is commonly applied in fields such as physics, engineering, and sports. For example, in physics, it is used to analyze the motion of objects in projectile motion problems. In engineering, it is used to design structures and machines that move in two dimensions. In sports, it can be used to analyze the trajectory of a ball in motion.

What are some common misconceptions about two dimensional kinematics?

One common misconception is that acceleration must always be constant in two dimensional motion. In reality, acceleration can vary in both the x and y directions. Another misconception is that an object's motion in one direction does not affect its motion in the other direction. In fact, the two directions are often interdependent and must be analyzed together.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
228
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
848
  • Introductory Physics Homework Help
Replies
8
Views
2K
Replies
5
Views
936
  • Introductory Physics Homework Help
Replies
6
Views
967
  • Introductory Physics Homework Help
Replies
2
Views
927
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
14K
  • Introductory Physics Homework Help
Replies
33
Views
3K
Back
Top