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Parametric equations, find speed and direction

 
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Aug6-12, 06:35 PM   #1
 

Parametric equations, find speed and direction


1. The problem statement, all variables and given/known data

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


2. Relevant equations

v = √(dy/dt)^2 / (dx/dt)^2

Pythagoras theorem

3. 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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Aug6-12, 08:00 PM   #2
 
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Quote by Mushroom79 View Post
1. The problem statement, all variables and given/known data

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


2. Relevant equations

v = √(dy/dt)^2 / (dx/dt)^2

Pythagoras theorem

3. 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.
 
Aug7-12, 03:38 PM   #3
 
Quote by Redbelly98 View Post
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
 
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