Position, Velocity, and Acceleration Position, Velocity, and Acceleration

Introduction

Position:

s(t)

Velocity:
v(t)

Acceleration:
a(t)
v(t) = \frac{ds}{dt} \\ \text{} \\ a(t) = \frac{dv}{dt} = \frac{d^2 s}{dt^2}
Acceleration is the derivative of velocity.
Velocity is the derivative of position.

Therefore,
Position is the antiderivative of velocity.
Velocity is the antiderivative of acceleration.
v(t) = \int a(t) \, dt + C_1 \\ \text{} \\ s(t) = \int v(t) \, dt + C_2
C_1
is the initial velocity.
C_2
is the initial position.

Examples

Example 1: An object in free fall is accelerating at

-9.8 \, \text[m/s²]
.
Find its position function if
v(0) = 0
and
s(0) = 20 \, \text{m}
.
\begin{align*} a(t) &= -9.8 \\\\ v(t) &= \int (-9.8) \, dt + C_1 \\\\ &= -9.8t + C_1 \\\\ v(0) &= 0 \\\\ C_1 &= 0 \\\\ s(t) &= \int (-9.8t) \, dt + C_2 \\\\ &= -4.9t^2 + C_2 \\\\ s(0) &= 20, \, C_2 = 20 \\\\ s(t) &= -4.9t^2 + 20 \end{align*}


Example 2: A particle accelerates at the rate:
a(t) = 2t + 1
.
v(0) = 2
and
s(0) = 5
.
Find
s(t)
.
\begin{align*} a(t) &= 2t + 1 \\\\ v(t) &= \int (2t + 1) \, dt + C_1 \\\\ &= t^2 + t + C_1 \\\\ v(0) &= 2 \\\\ C_1 &= 2 \\\\ s(t) &= \int (t^2 + t + 2) \, dt + C_2 \\\\ &= \frac{t^3}{3} + \frac{t^2}{2} + 2t + C_2 \\\\ s(0) &= 5 \\\\ C_2 &= 5 \\\\ s(t) &= \frac{t^3}{3} + \frac{t^2}{2} + 2t + 5 \end{align*}