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Is there some sort of calculus relationship between these two kinematics equations?

by tahayassen
Tags: calculus, equations, kinematics, relationship, sort
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tahayassen
#1
Dec25-12, 08:11 PM
P: 273
[tex]{ y }_{ f }={ y }_{ i }+{ v }_{ yi }t+\frac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }\\ { v }_{ yf }={ v }_{ yi }+{ a }_{ y }t[/tex]

It almost looks like the second equation is the derivative of the first equation with respect to time.
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Pengwuino
#2
Dec25-12, 08:14 PM
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Exactly. The velocity of an object is simply the time-derivative of its position function.
tahayassen
#3
Dec25-12, 08:24 PM
P: 273
Quote Quote by Pengwuino View Post
Exactly. The velocity of an object is simply the time-derivative of its position function.
Maybe I'm incredibly rusty on my calculus, but isn't the time-derivative of the first equation the following?

[tex]0={ v }_{ yi }+{ a }_{ y }t[/tex]

jtbell
#4
Dec25-12, 08:47 PM
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Is there some sort of calculus relationship between these two kinematics equations?

##v_{yf}## isn't a constant, it's a variable, more specifically the dependent variable, a function of t. Written as functions, your two equations are

$$y(t) = y_i + v_{yi} t + \frac{1}{2}a_y t^2 \\ v_y(t) = v_{yi} + a_y t$$
tahayassen
#5
Dec25-12, 08:48 PM
P: 273
Quote Quote by jtbell View Post
##v_{yf}## isn't a constant, it's a variable, more specifically the dependent variable, a function of t. Written as functions, your two equations are

$$y(t) = y_i + v_{yi} t + \frac{1}{2}a_y t^2$$

$$v_y(t) = v_{yi} + a_y t$$
Thank you!
tahayassen
#6
Dec25-12, 08:51 PM
P: 273
Any way I can derive the below equation from the position and velocity equations?

[tex]{ { v }_{ yf } }^{ 2 }={ { v }_{ yi } }^{ 2 }+2{ a }_{ y }({ y }_{ f }-{ y }_{ i })[/tex]

Edit: Never mind. http://www.physicsforums.com/showthread.php?t=660863


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