Integrating with respect to time

Expert reviewed • 21 July 2024 • 4 minute read

HSC Maths Advanced Syllabus

use any of the functions covered in the scope of this syllabus and their derivatives to solve practical and abstract problems
solve optimisation problems for any of the functions covered in the scope of this syllabus, in a wide variety of contexts including displacement, velocity, acceleration, area, volume, business, finance and growth and decay

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Integration and Motion

Not only can we derive equations involving an objects displacement, to determine its velocity and acceleration, but we can also integrate velocity and acceleration to find displacement.

\int a.\frac{da}{dt}=v\\\int v.\frac{dv}{dt}=x

Practice Question 1

Find the equation for the displacement of a particle at any time $(t)$ , when $v=5$ and the displacement of the particle is 10 at $t=0$ .

To determine the equation for displacement, we must integrate the velocity with respect to time:

x=\int v\frac{dv}{dt}\\=\int5\frac{dv}{dt}\\=5t+C

Now that we know the equation for $x$ we must find the value of $C$ . We do this by substituting in values $x=10$ when $t=0$ , (information given in the question):

x=5t+C\\10=5(0)+C\\C=10

$\therefore$ The equation to find the displacement of the particle at any time is $x=5t+10$

Acceleration due to Gravity

Before integrating acceleration, it is important to know that the value of the acceleration of gravity on Earth is its own constant: $g\approx 9.8\;ms^{-2}$

Now depending on the direction of positive and negative values, the sign of gravity changes when integrating. If upwards is taken as the positive direction, $a=-g$ must be integrated. However, if downwards is taken as positive, $a$ is integrated using $g$ .

It is also important to note that this constant can only be used if a question specifies that air resistance is to be ignored.

Velocity and Acceleration as Derivatives

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Review of Related Rates

Return to Module 7: Motion and Rates