Motion and integration

Finding velocity and displacement by integrating acceleration and velocity, using initial conditions, and motion under gravity.

Working backwards

$v = \displaystyle\int a\,dt$ and $x = \displaystyle\int v\,dt$. Each integration brings a constant, found from what is known at one moment, usually $t = 0$.

Example 1●●

A particle starts from rest at the origin with $a = 6t - 4$. When does it return to the origin?

$v = 3t^2 - 4t + C$, and $v(0) = 0$ gives $C = 0$.

$x = t^3 - 2t^2 + D$, and $x(0) = 0$ gives $D = 0$.

$x = t^2(t - 2) = 0$ at $t = 2$ seconds.

Example 2●●

A particle has $v = 4 - 2t$ and starts at $x = 3$. Find its greatest displacement.

$x = 4t - t^2 + 3$. It turns around when $v = 0$, at $t = 2$.

$x(2) = 8 - 4 + 3 = 7$ m.

Motion under gravity

Near the Earth, a thrown object has constant acceleration downwards, about $10$ m/s² (more precisely $9.8$). With upwards as positive, $\ddot{x} = -10$.

Example 3●●●

A stone is thrown upwards at $20$ m/s from the top of a $25$ m cliff. Find its greatest height, when it hits the ground below, and its impact speed.

Take the base of the cliff as the origin. $v = -10t + 20$ and $x = -5t^2 + 20t + 25$.

Integrate

$v = 0$ at $t = 2$, so $x = -20 + 40 + 25 = 45$ m.

Greatest height

$-5t^2 + 20t + 25 = 0$ gives $t^2 - 4t - 5 = 0$, so $(t - 5)(t + 1) = 0$ and $t = 5$ s.

Hits the ground

$v(5) = -30$, so the speed is $30$ m/s.

Impact speed

The stone's height against time

It rises for 2 seconds, then falls and hits the ground at t = 5.

Flashcards

$v$ from $a$

$v = \displaystyle\int a\,dt$, then use $v(0)$.

$x$ from $v$

$x = \displaystyle\int v\,dt$, then use $x(0)$.

Greatest height

Where $v = 0$.

Practice

Starting from rest at the origin with $a = 6t - 4$, when does the particle return to the origin?
$v = 4 - 2t$ and $x(0) = 3$. Find the greatest displacement.
In Example 3, find the stone's greatest height above the ground.
When does the stone hit the ground?
Find the stone's impact speed in m/s.
$a = -10$ and $v(0) = 15$. Find $v$ at $t = 2$.
$v = 3t^2$ and $x(0) = 3$. Find $x$ at $t = 2$.
A particle has $v = 2\cos t$ and starts at the origin. Its displacement always satisfies:
The constant of integration is found from: