Parametric equations, find speed and direction

In summary: I'm not sure how to simplify this or where to go next.In summary, the object's speed and direction can be determined by using the formula v = √(dy/dt)^2 / (dx/dt)^2 and the Pythagorean theorem. The object's coordinates, x = 25t and y = 20t-5t^2, can be used to find the components of the velocity vector (vx and vy). By plugging in the time value of t = 3 sec, the values of vx and vy can be calculated. These values can then be used to find the object's speed and direction at 3 sec. However, further simplification is needed to find the final result.
  • #1
Mushroom79
26
0

Homework Statement



An object moves so it's coordinates at the time t is given by the relationships

x = 25t
y = 20t-5t^2

What is the object's speed and direction at 3 sec?

t = 3 sec


Homework Equations



v = √(dy/dt)^2 / (dx/dt)^2

Pythagoras theorem

The Attempt at a Solution



dx=25
dy=20-10t

I'm not sure how I should use this by combining the two formulas above.
 
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  • #2
Mushroom79 said:

Homework Statement



An object moves so it's coordinates at the time t is given by the relationships

x = 25t
y = 20t-5t^2

What is the object's speed and direction at 3 sec?

t = 3 sec


Homework Equations



v = √(dy/dt)^2 / (dx/dt)^2

Pythagoras theorem

The Attempt at a Solution



dx=25
dy=20-10t

I'm not sure how I should use this by combining the two formulas above.
What you are calling dx and dy are really
[tex]\frac{dx}{dt} \text{ and } \frac{dy}{dt}[/tex]
which is the same as vx and vy. So you have found the two components of the velocity vector v -- or at least you'll have them once you plug the time value into your expression. So as a start, figure out what the values of vx and vy are, using the expressions you got.
 
  • #3
Redbelly98 said:
What you are calling dx and dy are really
[tex]\frac{dx}{dt} \text{ and } \frac{dy}{dt}[/tex]
which is the same as vx and vy. So you have found the two components of the velocity vector v -- or at least you'll have them once you plug the time value into your expression. So as a start, figure out what the values of vx and vy are, using the expressions you got.

Ah, but that is the part I get stuck on,

√25^2+(20-10t)^2
 

1. What are parametric equations?

Parametric equations are a type of mathematical representation that involves expressing the variables in a system in terms of one or more independent parameters. In other words, the equations describe the relationship between different variables in terms of a common parameter.

2. How do you find the speed in parametric equations?

To find the speed in parametric equations, you can use the formula: speed = √(dx/dt)^2 + (dy/dt)^2, where dx/dt and dy/dt represent the derivatives of the x and y components, respectively.

3. How do you find the direction in parametric equations?

The direction in parametric equations is given by the tangent vector, which can be found by taking the derivative of the parametric equations with respect to the parameter. The angle of the tangent vector with the horizontal axis can then be used to determine the direction.

4. Can parametric equations be used to model motion?

Yes, parametric equations can be used to model motion as they can describe the position, speed, and direction of an object at any given time. This makes them useful for studying the motion of objects in physics and engineering.

5. Are there any real-world applications of parametric equations?

Yes, parametric equations have many real-world applications, such as in physics, engineering, and computer graphics. They are used to model the motion of objects, design curves and surfaces, and create animations and special effects in movies and video games.

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