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
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))
v=√{(dy/dt)^2 + (dx/dt)^2} (not ratio)
Direction is arctan((dy/dt)/(dx/dt))
√(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.
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.
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.
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.
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
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
I didn't check the angle, but it looks right (neg. sign, mag. < 45 deg).
Thank you very much for the help, it cleared things up 
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