Easy-acceleration with unit vectors

In summary, the conversation discusses kinematic equations for components of displacement, velocity, and acceleration. The speaker mentions two sets of kinematic equations for each component and asks for the saucer's acceleration in unit vectors and in terms of magnitude and direction. The other person clarifies that the given equations are valid for each component and asks for the results of the calculations.
  • #1
PEZenfuego
48
0

Homework Statement



untitled.jpg


r1=(0m)
r2=(2000i + 1000j)m
v1=(200j)m/s
v2=(200i-100j)m/s

Homework Equations


tantheta=vy/vx

v2^2=v1^2+2ax
v2=v1+at
x2=x1+v1t+1/2at^2

The Attempt at a Solution



IMG_20120909_150513.jpg


Thank you for any help
 
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  • #2
So what do we have to find??
 
  • #3
find the saucer's acceleration in unit vectors and in terms of magnitude and direction.
 
  • #4
PEZenfuego said:

Homework Statement



View attachment 50618

r1=(0m)
r2=(2000i + 1000j)m
v1=(200j)m/s
v2=(200i-100j)m/s

Homework Equations


tantheta=vy/vx

v2^2=v1^2+2ax
v2=v1+at
x2=x1+v1t+1/2at^2

The Attempt at a Solution



View attachment 50617

Thank you for any help
Are those kinematic equations for components of vectors, or for vectors themselves?
 
  • #5
SammyS said:
Are those kinematic equations for components of vectors, or for vectors themselves?

If you click the first picture, you will have an understanding equal to mine. I'm sorry.
 
  • #6
PEZenfuego said:
If you click the first picture, you will have an understanding equal to mine. I'm sorry.
Sorry for the slow response.

I should have been more direct with my earlier response.

The set of kinematic equations you have
(v2)2=(v1)2+2ax
(v2)=(v1)+at
x2=x1+(v1)t+1/2at2
are valid for each component of displacement, velocity, and acceleration.

So you have two such sets of kinematic equations:
(v2)x2 = (v1)x2 + 2axx
(v2)x = (v1)x + axt
x2 = x1 + (v1)xt + 1/2axt2

(v2)y2 = (v1)y2 + 2ayy
(v2)y = (v1)y + ayt
y2 = y1 + (v1)yt + 1/2ayt2
 
  • #7
SammyS said:
Sorry for the slow response.

I should have been more direct with my earlier response.

The set of kinematic equations you have
(v2)2=(v1)2+2ax
(v2)=(v1)+at
x2=x1+(v1)t+1/2at2
are valid for each component of displacement, velocity, and acceleration.

So you have two such sets of kinematic equations:
(v2)x2 = (v1)x2 + 2axx
(v2)x = (v1)x + axt
x2 = x1 + (v1)xt + 1/2axt2

(v2)y2 = (v1)y2 + 2ayy
(v2)y = (v1)y + ayt
y2 = y1 + (v1)yt + 1/2ayt2

It makes sense now. thank you.
 
  • #8
PEZenfuego said:
It makes sense now. thank you.
What did you get for answers?
 

1. What are unit vectors?

Unit vectors are vectors that have a magnitude of 1 and are used to indicate the direction of a vector. They are commonly denoted by the symbol "i", "j", or "k" in three-dimensional coordinate systems.

2. How are unit vectors used in easy-acceleration?

Unit vectors are used in easy-acceleration to break down a vector into its individual components in order to simplify the calculations involved in acceleration. By using unit vectors, the acceleration of an object can be easily determined in each direction.

3. What is the formula for easy-acceleration with unit vectors?

The formula for easy-acceleration with unit vectors is a = axi + ayj + azk, where a is the acceleration vector and ax, ay, and az are the components of acceleration in the x, y, and z directions, respectively.

4. How do you find the components of acceleration using unit vectors?

To find the components of acceleration using unit vectors, you first need to determine the direction of the acceleration vector. Then, using trigonometric functions, you can calculate the magnitude and direction of the components of acceleration in each direction.

5. What are some real-world applications of easy-acceleration with unit vectors?

Easy-acceleration with unit vectors is commonly used in fields such as engineering, physics, and mechanics. It can be applied to various real-world scenarios, such as calculating the acceleration of a car on a curved road or determining the forces acting on an airplane during takeoff and landing.

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