# Calculus/Kinematics

 ← Centre of mass Calculus Parametric and Polar Equations → Kinematics

## Introduction

Kinematics or the study of motion is a very relevant topic in calculus.

If ${\displaystyle x}$ is the position of some moving object, and ${\displaystyle t}$ is time, this section uses the following conventions:

• ${\displaystyle x(t)}$ is its position function
• ${\displaystyle v(t)=x'(t)}$ is its velocity function
• ${\displaystyle a(t)=x''(t)}$ is its acceleration function

## Differentiation

### Average Velocity and Acceleration

Average velocity and acceleration problems use the algebraic definitions of velocity and acceleration.

• ${\displaystyle v_{\rm {avg}}={\frac {\Delta x}{\Delta t}}}$
• ${\displaystyle a_{\rm {avg}}={\frac {\Delta v}{\Delta t}}}$

#### Examples

Example 1:

A particle's position is defined by the equation ${\displaystyle x(t)=t^{3}-2t^{2}+t}$ . Find the
average velocity over the interval ${\displaystyle [2,7]}$ .

• Find the average velocity over the interval ${\displaystyle [2,7]}$ :
 ${\displaystyle v_{\rm {avg}}}$ ${\displaystyle ={\frac {x(7)-x(2)}{7-2}}}$ ${\displaystyle ={\frac {252-2}{5}}}$ ${\displaystyle =50}$
Answer: ${\displaystyle v_{\rm {avg}}=50}$ .


### Instantaneous Velocity and Acceleration

Instantaneous velocity and acceleration problems use the derivative definitions of velocity and acceleration.

• ${\displaystyle v(t)={\frac {dx}{dt}}}$
• ${\displaystyle a(t)={\frac {dv}{dt}}}$

#### Examples

Example 2:

A particle moves along a path with a position that can be determined by the function ${\displaystyle x(t)=4t^{3}+e^{t}}$ .
Determine the acceleration when ${\displaystyle t=3}$ .

• Find ${\displaystyle v(t)={\frac {ds}{dt}}}$ .
${\displaystyle {\frac {ds}{dt}}=12t^{2}+e^{t}}$
• Find ${\displaystyle a(t)={\frac {dv}{dt}}={\frac {d^{2}s}{dt^{2}}}}$ .
${\displaystyle {\frac {d^{2}s}{dt^{2}}}=24t+e^{t}}$
• Find ${\displaystyle a(3)={\frac {d^{2}s}{dt^{2}}}{\bigg |}_{t=3}}$
 ${\displaystyle {\frac {d^{2}s}{dt^{2}}}{\bigg |}_{t=3}}$ ${\displaystyle =24(3)+e^{3}}$ ${\displaystyle =72+e^{3}}$ ${\displaystyle =92.08553692\dots }$
Answer: ${\displaystyle a(3)=92.08553692\dots }$


## Integration

• ${\displaystyle x_{2}-x_{1}=\int \limits _{t_{1}}^{t_{2}}v(t)dt}$
• ${\displaystyle v_{2}-v_{1}=\int \limits _{t_{1}}^{t_{2}}a(t)dt}$