Parametric equations, find speed and direction

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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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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