Kinematic Equations (Calculus Method)

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Note: The calculus method will only derive 3 of the 4 kinematic equations. $LaTeX: \Delta x = {1\over{2}}(v_i+v_f)t$ $\Delta x = {1\over{2}}(v_i+v_f)t$ must be obtained through the algebraic method. For the derivation of this equation, please see Kinematic Equations (Algebraic Method). (Make link)

Core concepts

$LaTeX: {dv\over{dt}}=a$ ${dv\over{dt}}=a$

$LaTeX: {dx\over{dt}}=v$ ${dx\over{dt}}=v$

a is a constant value

Equation #1: LaTeX: v_f = v_i +at $v_f = v_i +at$

$LaTeX: {dv\over{dt}}=a$ ${dv\over{dt}}=a$ Apply separation of variables

$LaTeX: dv=a\space dt$ $dv=a\space dt$ Integrate

$LaTeX: \displaystyle\int dv = \displaystyle\int a \space dt$ $\displaystyle\int dv = \displaystyle\int a \space dt$ Apply initial and final limits. Let LaTeX: t_i = 0 $t_i = 0$ and LaTeX: t_f=t $t_f=t$ .

$LaTeX: \displaystyle\int\limits_{v_i}^{v_f}dv= \displaystyle\int\limits_0^t a \space dt$ $\displaystyle\int\limits_{v_i}^{v_f}dv= \displaystyle\int\limits_0^t a \space dt$

$LaTeX: v \bigg|_{v_i}^{v_f} = at \bigg|_0^t$ $v \bigg|_{v_i}^{v_f} = at \bigg|_0^t$

$LaTeX: v_f-v_i = at-\color{red}\cancelto{0}{\color{black}a(0)}$ $v_f-v_i = at-\color{red}\cancelto{0}{\color{black}a(0)}$

LaTeX: v_f = v_i +at $v_f = v_i +at$

Equation #2: $LaTeX: \Delta x = v_i t + {1\over 2}at^2$ $\Delta x = v_i t + {1\over 2}at^2$

$LaTeX: {dx \over dt}=v$ ${dx \over dt}=v$ Apply separation of variables

$LaTeX: v \space dt = dx$ $v \space dt = dx$

Because v is a function of time, and not a constant (like acceleration), v dt cannot be directly integrated. An equation needs to be substituted for v.

According to eq #1, the velocity at any point in time is LaTeX: v_f = v_i + at $v_f = v_i + at$ where v_i and a are constants. Eq #1 represents the velocity at any point as a function of time.

LaTeX: (v_i + at)dt = dx $(v_i + at)dt = dx$

$LaTeX: \displaystyle\int\limits_0^t (v_i + at)dt =\displaystyle \int\limits_{x_i}^{x_f}dx$ $\displaystyle\int\limits_0^t (v_i + at)dt =\displaystyle \int\limits_{x_i}^{x_f}dx$

$LaTeX: \left( v_{i}t + {1 \over 2}at^2 \right) \Bigg|_0^t = x \Bigg|_{x_i}^{x_f}$ $\left( v_{i}t + {1 \over 2}at^2 \right) \Bigg|_0^t = x \Bigg|_{x_i}^{x_f}$

$LaTeX: v_i t + {1 \over 2}at^2 - z \color{red}\cancelto{0}{\color{black}v_i(0)} \color{black}- \color{red}\cancelto{\color{red}0}{\color{black}{1 \over 2} a (0)^2} \color{black} = x_f - x_i = \Delta x$ $v_i t + {1 \over 2}at^2 - z \color{red}\cancelto{0}{\color{black}v_i(0)} \color{black}- \color{red}\cancelto{\color{red}0}{\color{black}{1 \over 2} a (0)^2} \color{black} = x_f - x_i = \Delta x$

$LaTeX: \Delta x = v_i t + {1 \over 2} a t^2$ $\Delta x = v_i t + {1 \over 2} a t^2$

Equation #3: $LaTeX: v_f^2 = v_i^2 +2a \Delta x$ $v_f^2 = v_i^2 +2a \Delta x$

To derive this equation, a time-independent derivative is required.

$LaTeX: { dx \over dv }= \space ?$ ${ dx \over dv }= \space ?$

$LaTeX: { dx \over dt } = \space v$ ${ dx \over dt } = \space v$ (given)

$LaTeX: { dv \over dt } = \space a$ ${ dv \over dt } = \space a$ Find the reciprocal.

$LaTeX: { dt \over dv } = \space {1 \over a }$ ${ dt \over dv } = \space {1 \over a }$

$LaTeX: { dx \over \color{red}\cancel{\color{black}dt} } \cdot { \color{red}\cancel{\color{black}dt} \over dv } = \space v \cdot { 1 \over a }$ ${ dx \over \color{red}\cancel{\color{black}dt} } \cdot { \color{red}\cancel{\color{black}dt} \over dv } = \space v \cdot { 1 \over a }$

$LaTeX: { dx \over dv } = \space { v \over a }$ ${ dx \over dv } = \space { v \over a }$ Separation of variables.

$LaTeX: v \space dv = a \space dx$ $v \space dv = a \space dx$

$LaTeX: \displaystyle\int\limits_{v_i}^{v_f} v \space dv = \displaystyle\int\limits_{x_i}^{x_f} a \space dx$ $\displaystyle\int\limits_{v_i}^{v_f} v \space dv = \displaystyle\int\limits_{x_i}^{x_f} a \space dx$

$LaTeX: { 1 \over 2 } v^2 \bigg|_{v_i}^{v_f} = ax \bigg|_{x_i}^{x_f}$ ${ 1 \over 2 } v^2 \bigg|_{v_i}^{v_f} = ax \bigg|_{x_i}^{x_f}$

$LaTeX: { 1 \over 2 } \left(v_f^2 - v_i^2 \right) = a \left(x_f - x_i\right)$ ${ 1 \over 2 } \left(v_f^2 - v_i^2 \right) = a \left(x_f - x_i\right)$ Multiply both sides by 2.

$LaTeX: \color{red}\cancel{\color{black}\left(2\right)} \color{black} {1 \over \color{red}\cancel{\color{black}2} } \left( v_f^2 -v_i^2 \right) = \left( 2 \right) a \left( x_f - x_i \right)$ $\color{red}\cancel{\color{black}\left(2\right)} \color{black} {1 \over \color{red}\cancel{\color{black}2} } \left( v_f^2 -v_i^2 \right) = \left( 2 \right) a \left( x_f - x_i \right)$

$LaTeX: v_f^2 - v_i^2 = 2 a \left( x_f = x_i \right)$ $v_f^2 - v_i^2 = 2 a \left( x_f = x_i \right)$ Substitute $LaTeX: x_f - x_i = \Delta x$ $x_f - x_i = \Delta x$ and rearrange.

$LaTeX: v_f^2 = v_i^2 + 2 a \Delta x$ $v_f^2 = v_i^2 + 2 a \Delta x$