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

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

t = 3 sec

v = √(dy/dt)^2 / (dx/dt)^2

Pythagoras theorem

x'(t)=25
y('t)=20-10t

How do I put these in by using the two formulas above to get dy/dt and dx/dt?
I need to find the speed and direction
mathman
x'(t) = dx/dt, y'(t) = dy/dt (it is simply notation).

v=√{(dy/dt)^2 + (dx/dt)^2} (not ratio)

Direction is arctan((dy/dt)/(dx/dt))
Monocerus
√(25^2+(20-10t)^2)

Okay, but I still can't get an understanding of what I need to do.
I've been trying to figure it out for 2 days now.
mathman
QUOTE (Monocerus+Aug 8 2012, 01:51 PM)
√(25^2+(20-10t)^2)

Okay, but I still can't get an understanding of what I need to do.
I've been trying to figure it out for 2 days now.

x' is constant. Since the question is for t = 3, evaluate y' for t=3. Next compute v. x'/v isw the horizontal component and y'/v is the vertical component. The speed is v, while direction comes from the components.
Monocerus
QUOTE (mathman+Aug 8 2012, 09:59 PM)
x' is constant.  Since the question is for t = 3, evaluate y' for t=3.  Next compute v.  x'/v isw the horizontal component and y'/v is the vertical component.  The speed is v, while direction comes from the components.

Ah!

So, just making sure I've understand correctly

x'=25
y' = 20-(10*3) = -10

√(25^2+(-10)^2)=

√725 = speed

"Direction is arctan((dy/dt)/(dx/dt))" = -21.8 degrees
mathman
I didn't check the angle, but it looks right (neg. sign, mag. < 45 deg).
Monocerus
Thank you very much for the help, it cleared things up
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