Kinematics using Integration

Distances from Velocity Graphs

Suppose a car travels at a constant positive velocity $80 \text{ km h}^{-1}$ for $2$ hours.
Distances from Velocity Graphs $$ \begin{align} \displaystyle \text{distance travelled} &= \text{speed} \times \text{time} \\ &= 80 \text{ km h}^{-1} \times 2 \text{ h} \\ &= 160 \text{ km} \end{align} $$ We we sketch the graph velocity against time, the graph is a horizontal line, and we can see that the distance travelled is the area shaded in green.
Therefore the distance travelled is found by the definite integral: $$ \begin{align} \displaystyle \int_{0}^{2}{80}dt &= \big[80t\big]_{0}^{2} \\ &= 80 \times 2 - 80 \times 0 \\ &= 160 \text{ km} \end{align} $$ Now the speed decreased at a constant rate so taht the car, initially travelling at $80 \text{ km h}^{-1}$, stops in 2 hours. In this case, the average speed is $40 \text{ km h}^{-1}$. $$ \begin{align} \displaystyle \text{distance travelled} &= \text{speed} \times \text{time} \\ &= 40 \text{ km h}^{-1} \times 2 \text{ h} \\ &= 80 \text{ km} \end{align} $$ Distances from Velocity Graphs $$ \begin{align} \displaystyle \text{area of the triangle} &= \dfrac{1}{2} \times \text{based} \times \text{height} \\ &= \dfrac{1}{2} \times 2 \times 80 \\ &= 80 \text{ km} \end{align} $$ Again, the area shaded in green is the distance travelled, and we can find it using the difinite integral: $$\displaystyle \begin{align} \int_{0}^{2}{(80-40t)}dt &= \big[80t-20t^2\big]_{0}^{2} \\ &= (80 \times 2 - 20 \times 2^2) - (80 \times 0 - 20 \times 0^2) \\ &= 80 \text{ km} \end{align} $$ $$\displaystyle \therefore \text{distance travelled} = \int_{t_1}^{t_2}{v(t)}dt$$

Example 1

The velocity-time graph for a car journey is shown in the graph. Find the total distance travelled by the car, where $t$ is in hours and $v$ is in $ \text{ km h}^{-1}$. Velocity Time Graph

Displacement and Velocity Functions

For some displacement $s(t)$, the velocity function is $v(t)=s'(t)$.
So, given a velocity function it is determined that the displacement function by the integral: $$s(t) = \int{v(t)}dt$$ The constant of integration determines where on the line the object begins, called the initial potition. Using the displacement function we can determine the change in displacement in a time interval $t_1 \le t \le t_2$.
$$ \displaystyle \begin{align} \text{Displacement} &= s(t_2) - s(t_1) \\ &= \int_{t_1}^{t_2}{v(t)}dt \end{align} $$ To find the total distance travelled given a velocity function $v(t)=s'(t)$ on $t_1 \le t \le t_2$:
  • Draw an accurate sign diagram for $v(t)$ to determine any changes of direction
  • Find $s(t)$ by integration, including a constant of integration
  • Find $s(t)$ at each time the direction chagnes.
  • Draw a motion diagram.
  • Calculate the total distance travelled from the motion diagram.

Example 2

Find the distance travelled by a car with a velocity $v(t) = 2t+5$, where $t$ is in hours and $v$ is in $\text{km h}^{-1}$ for the first $5$ hours.

Displacement and Velocity Functions

The acceleration functin is the derivative of the velocity function, so $a(t) = v'(t)$. Given an acceleration function, we can determine the velocity function by integration.:
$$v(t) = \int{a(t)}dt$$

Example 3

A particle moves in a straight line with velocity $v(t)=6t^2-18t+12 \text{ m s}^{-1}$. Find the total distance travelled that the particle moves in the first $3$ seconds of motion.