Parametric equations, find speed and direction

AI Thread Summary
To find the speed and direction of the object at t = 3 seconds, the derivatives dx/dt and dy/dt must be calculated first, yielding vx = 25 and vy = 20 - 10t. Substituting t = 3 gives vy = -10. The speed can be determined using the formula v = √(vx^2 + vy^2), resulting in v = √(25^2 + (-10)^2). The direction can be found using the arctangent of vy/vx to determine the angle of movement.
Mushroom79
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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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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
\frac{dx}{dt} \text{ and } \frac{dy}{dt}
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.
 
Redbelly98 said:
What you are calling dx and dy are really
\frac{dx}{dt} \text{ and } \frac{dy}{dt}
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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