Is There a Calculus Relationship Between These Kinematics Equations?

[tahayassen]

{ y }_{ f }={ y }_{ i }+{ v }_{ yi }t+\frac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }\\ { v }_{ yf }={ v }_{ yi }+{ a }_{ y }t

It almost looks like the second equation is the derivative of the first equation with respect to time.

 

[Pengwuino]

Exactly. The velocity of an object is simply the time-derivative of its position function.

 

[tahayassen]

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?

0={ v }_{ yi }+{ a }_{ y }t

 

[jtbell]

$v_{yf}$ 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]

$v_{yf}$ 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]

Any way I can derive the below equation from the position and velocity equations?

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

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