Velocity and Acceleration as Derivatives

Expert reviewed • 22 November 2024 • 7 minute read

HSC Maths Advanced Syllabus

  • define and interpret the concept of the second derivative as the rate of change of the first derivative function in a variety of contexts, for example recognise acceleration as the second derivative of displacement with respect to time

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Instantaneous Velocity and Speed

Unlike average velocity and average speed, instantaneous velocity and speed are measurements of an objects movement, at a specific point in time. As discussed before, velocity is a vector quantity and speed is a scalar quantity. This means that a measurement of instantaneous velocity will refer to an objects magnitude and direction, while a measure of instantaneous speed will only provide a magnitude. As such instantaneous speed can never be a negative value.

To find the formula for instantaneous velocity we must derive the displacement of the object. This means that the the velocity of an object can also be measured by the gradient of the tangent of a displacement-time graph. Thus, we use the following formulas:

v=dxdt=x˙v=\frac{dx}{dt}=\dot{x}

Practice Question 1

Find the velocity of an object, given it’s displacement is equal to: x=20t−5t3x=20t-5t^3, at any time (t)(t):

Using knowledge learnt in previous modules, we must derive the equation for xx with respect to tt to determine the object’s velocity.

v=dxdt=1×20−3×5t2=20−15t2v=\frac{dx}{dt}\\=1\times 20-3\times5t^2\\=20-15t^2

∴\therefore We see that the velocity of the particle at any given time is equal to, v=20−15t2v=20-15t^2

Finding Acceleration

Following on from velocity as the derivative of the displacement of an object, the acceleration of an object is found by deriving its velocity.

a=dvdt=v˙a=\frac{dv}{dt}=\dot{v}

This means that acceleration is also the double derivate of xx.

Practice Question 2

Find the acceleration of an object at t=3t=3, where velocity at any given time is: v=20−15t2v=20-15t^2. Note that time is in seconds and velocity is in meters per second.

First we must derive the equation for velocity, to determine an equation for the object’s acceleration.

a=dvdt=0×20−2×15t=−30ta=\frac{dv}{dt}\\=0\times20-2\times15t\\=-30t

Now that we know that at any given time, a=−30ta=-30t, we must find the acceleration of the object when t=3t=3

a=−30(3)=−90a=-30(3)\\=-90

∴\therefore At t=3t=3, we can see that the objects acceleration is −90 ms−2-90\:ms^{-2}

What are Stationary Particles?

A stationary particle or stationary point (on a graph, which is discussed in module 3) refers to any instantaneous point, where the velocity of a particle is equal to zero. Thus, to determine any stationary particles, simply find vv as an equation and solve for tt

The Properties of Acceleration

When the acceleration of a particle is equal to zero, it has a unique affect on the velocity of the particle. A result of a=0a=0, will mean that the particle has a constant velocity. This means that a velocity-time graph, will display a straight horizontal line, representing that the particle will be travelling at the same speed, as long as a=0a=0.

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Average Velocity and Speed

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Integrating with respect to Time

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