Newtonian mechanics

Introduction

SUVAT

Introduction

For a constant acceleration environment we want to find equations to link:

  • Initial speed: \(v_{t_0}\)

  • End speed: \(v_{t_1}\)

  • Time: \(t_1-t_0\)

  • Acceleration: \(a\)

  • Displacement \(s_{t_1}-s_{t_0}\)

The SUVAT equations

These are the following, and are derived below.

  • \(v_{t_1}=a(t_1-t_0)+v_{t_0}\)

  • \((s_{t_1}-s_{t_0})=v_{t_0}(t_1-t_0)+\dfrac{1}{2}a(t_1-t_0)^2\)

  • \((s_{t_1}-s_{t_0})=v_{t_1}(t_1-t_0)-\dfrac{1}{2}a(t_1-t_0)^2\)

  • \(v_{t_1}^2= v_{t_0}^2+2a(s_{t_1}-s_{t_0})\)

  • \((s_{t_1}-s_{t_0})=(t_1-t_0)\dfrac{v_{t_1}+v_{t_1}}{2}\)

Equation 1: No displacement

This equation is:

\(v_{t_1}=a(t_1-t_0)+v_{t_0}\)

To derive this start with:

\(v_t:=\dfrac{\delta s_t}{\delta t}\)

\(a:=\dfrac{\delta v_t}{\delta t}\)

If acceleration is constant, then

\(\dfrac{\delta v_t}{\delta t}=a\)

\(v_t=\int a dt +v_0\)

\(v_t=at+v_0\)

Equation 2: No end velocity

This equation is:

\((s_{t_1}-s_{t_0})=v_{t_0}(t_1-t_0)+\dfrac{1}{2}a(t_1-t_0)^2\)

To derive this start with:

\(v:=\dfrac{\delta s_t}{\delta t}\)

Then:

\(\dfrac{\delta s_t}{\delta t}=at+v_0\)

\(s_t=\dfrac{1}{2}at^2+v_0t+s_0\)

\((s_t-s_0)=v_0t+\dfrac{1}{2}at^2 \)

Equation 3: No start velocity

This equation is:

\((s_{t_1}-s_{t_0})=v_{t_1}(t_1-t_0)-\dfrac{1}{2}a(t_1-t_0)^2\)

To derive this start with:

\(v_t=at+v_0\)

\((s_t-s_0)=t\dfrac{v_t+v_0}{2}\)

So:

\(v_0=v_t-at\)

\(v_0=\dfrac{2}{t}(s_t-s_0)- v_t\)

\(v_t-at=\dfrac{2}{t}(s_t-s_0)- v_t\)

\((s_t-s_0)=v_tt-\dfrac{1}{2}at^2\)

Equation 4: No time

This equation is:

\(v_{t_1}^2= v_{t_0}^2+2a(s_{t_1}-s_{t_0})\)

To derive this start with:

\(v_t=at+v_0\)

\((s_t-s_0)=t\dfrac{v_t+v_0}{2}\)

So:

\(t=\dfrac{v_t-v_0}{a}\)

\(t=2\dfrac{s_t-s_0}{v_t+v_0}\)

\(\dfrac{v_t-v_0}{a}=2\dfrac{s_t-s_0}{v_t+v_0}\)

\((v_t-v_0)(v_t+v_0)=2a(s_t-s_0)\)

\(v^2_t= v^2_0+2a(s_t-s_0)\)

Equation 5: No acceleration

This equation is:

\((s_{t_1}-s_{t_0})=(t_1-t_0)\dfrac{v_{t_1}+v_{t_1}}{2}\)

To derive this start with:

\(v_t=at+v_0\)

\(s_t-s_0=\dfrac{1}{2}at^2+v_0t\)

So:

\(a=\dfrac{v_t-v_0}{t}\)

\(a=\dfrac{2[(s_t-s_0)-v_0t]}{t^2}\)

\(\dfrac{v_t-v_0}{t}=\dfrac{2[(s_t-s_0)-v_0t]}{t^2}\)

\(t(v_t-v_0)=2[(s_t-s_0)-v_0t]\)

\(t(v_t+v_0)=2(s_t-s_0)\)

\((s_t-s_0)=t\dfrac{v_t+v_0}{2}\)