Velocity and Acceleration

Position refers to the displacement of an object relative to its origin. It can be expressed algebraically as:

\(P(t)\)

Velocity refers to the speed and direction of an object. It represents the first derivative of a function used to represent the position of an object.

It can be algebraically expressed as:

\(V(t) = P'(t)\)

If \(V(t) > 0\), this indicates that the position is increasing (or moving right) with respect to time
If \(V(t) < 0\), this indicates that the position is decreasing (or moving left) with respect to time
If \(V(t) = 0\), this indicates that the position isn't changing at all

Acceleration refers to rate of change of velocity per unit time. It represents the second derivative of a function used to represent the position of an object. It can also represent the first derivative of a function used to represent the object's velocity.

It can be algebraially expressed as:

If \(A(t) > 0\), this indicates that the velocity is increasing with respect to time
If \(A(t) < 0\), this indicates that the velocity is decreasing with respect to time
If \(A(t) = 0\), this indicates that the velocity isn't changing at all

Example

For the position function \(y = \cfrac{1}{3}x^3 -x^2 -3x+4\):

Find the first and second derivatives. What do the derivatives represent?
Graph the original function and its derivatives.
Explain what the different values (\(y\), \(x\), slope, concavity) indicate for each function.

i. We can find the first derivative by differentiating the original equation:

\(y' = x^2 -2x -3\)

We can find the second derivative by differentiating the first derivative equation:

\(y'' = 2x -2\)

Therefore, we can determine that the first and seond derivatives of the original function are \(y' = x^2 -2x -3\) and \(y'' = 2x -2\) respectively. The first and second derivatives represent the velocity and acceleration functions respectively.

ii. Now that we have our 3 equations, we can graph each as such:

Graph of position, velocity, and acceleration functions graphed on the same axis.

iii. \(y = \cfrac{1}{3}x^3 -x^2 -3x+4\)

The \(y\)-value represents the position of the object relative to its origin
The slope represents the velocity
The concavity represents the acceleration
The inflection point represents the turning point of \(y'\)
The turning point represents the zero(s) of \(y'\)

\(y' = 2x^2 -2x -3\)

The \(y\)-value represents the velocity
The slope represents the acceleration
The turning point represents the zero of \(y''\)

\(y'' = 2x -2\)

The \(y\)-value represents the acceleration

Example

A construction worker accidentally drops a hammer from a height of \(90\;[\text{m}]\) while working on the roof of a new apartment building. The height of the hammer \(s\) in metres, after \(t\) seconds, is modelled by the function \(s(t) = 90 - 4.9t^2, t \geq0\)

Determine the average velocity of the hammer between \(1\;[\text{s}]\) and \(4\;[\text{s}]\)
Determine the velocity of the hammer at \(1\;[\text{s}]\) and \(4\;[\text{s}]\)
When will the hammer hit the ground?
Determine the impact velocity of the hammer.
Determine the acceleration function. What do you notice?

i. First, we can determine the respective positions from the ground at each time.

First, we will find the position from the ground at \(t = 1\):

\(s(1) = 90 - 4.9(1)^2\)

\(s(1) = 90 - 4.9\)

\(\textcolor{blue}{s(1) = 85.1}\)

Next, we will find the position from the ground at \(t = 4\):

\(s(4) = 90 - 4.9(4)^2\)

\(s(4) = 90 - 4.9(16)\)

\(s(4) = 90 - 78.4\)

\(\textcolor{red}{s(4) = 11.6}\)

In order to determine the average velocity of the hammer between these times, we need to use the Average Rate of Change (or \(a_{roc}\)) formula. We can substitute \(1\) and \(4\) for \(a\) and \(b\) respectively:

\(a_{roc} = \cfrac{\textcolor{red}{f(b)} - \textcolor{blue}{f(a)}}{\textcolor{red}{b} - \textcolor{blue}{a}}\)

\(a_{roc} = \cfrac{\textcolor{red}{11.6} - \textcolor{blue}{85.1}}{\textcolor{red}{4} - \textcolor{blue}{1}}\)

\(a_{roc} = \cfrac{-73.5}{3}\)

\(a_{roc} = -24.5\;[\text{m/s}]\)

Therefore, we can determine that the average velocity of the hammer between \(1\;[\text{s}]\) and \(4\;[\text{s}]\) is \(-24.5\;[\text{m/s}]\).

ii. In order to determine the velocities of the hammer at their respective times, we can determine the first derivative of the original function:

\(s(t) = 90 - 4.9t^2\)

\(s'(t) = 0 - 2(4.9t)\)

\(s'(t) = -9.8t\)

We can now plug in the respective times into the new equation to determine their velocities:

\(s'(1) = -9.8(1)\)
\(s'(1) = -9.8\;[\text{m/s}]\)

\(s'(4) = -9.8(4)\)
\(s'(4) = -39.2\;[\text{m/s}]\)

Therefore, we can determine that the respective velocities for \(t=1\) and \(t=4\) are \(-9.8\;[\text{m/s}]\) and \(-39.2\;[\text{m/s}]\).

iii. In order to determine when the hammer will hit the ground, we need to set the original equation to \(0\). We can then solve for \(t\):

\(0 = 90 - 4.9t^2\)

\(\cfrac{4.9t^2}{4.9} = \cfrac{90}{4.9}\)

\(\sqrt{t^2} = \sqrt{\cfrac{90}{4.9}}\)

\(t = \pm 4.29 = 4.29 \;[\text{s}]\)

Therefore, we can determine that the hammer will hit the floor at \(4.29\;[\text{s}]\).

iv. In order to determine the impact velocity of the hammer, we can substitute \(4.29\) for \(t\) in the derivative function:

\(s'(4.29) = -9.8(4.29)\)

\(s'(4.29) = -42\;[\text{m/s}]\)

Therefore, we can determine that the impact velocity of the hammer is \(-42\;[\text{m/s}]\).

v. In order to determine the accerleration function, we need to find the second derivative of the original function:

\(s''(t) = -9.8\;[\text{m/s}^2]\)

We can determine that the acceleration is \(-9.8\;[\text{m/s}^2]\). We can also identify this value as the value of gravity.

The position of a particle moving along a straight line is represented by the function \(s(t) = t^3 -12t^2 +36t\) where distance \((s)\) is in metres, time \((t)\) is in seconds and \(t \geq 0\).

At what time(s) is the object at rest?
When is the object moving in a positive direction?
When does the object return to its original position?
Separately graph the position-time graph, the velocity-time graph, and the acceleration-time graph.
Describe the position of the particle using information from above.

Show Answer

i. In order to determine when the object is at rest, we first need to determine the derivative of the original function. This will give us the function representing velocity:

\(s(t) = t^3 -12t^2 +36t\)

\(s'(t) = 3t^2 -24t +36\)

Next, we can set the velocity equation to 0. We can then factor out a common term:

\(0 = 3t^2 -24t +36\)

\(0 = 3(t^2 -8t +12)\)

Finally, we can factor the inside of the function into two pairs. This will help determine our times:

\(0 = 3(t^2 -2t-6t +12)\)

\(0 = 3(t(t-2)-6(t-2))\)

\(0 = 3((t-2)(t-6))\)

Therefore, we can determine that the object is at rest at \(2\;[\text{s}]\) and \(6\;[\text{s}]\).

ii. In order to determine when the object is moving in a positive direction, we can first determine the Axis of Symmetry. We can do this by dividing the sum of the \(x\)-intercepts (or \(0\)'s) by \(2\):

\(\text{AOS} = \cfrac{2+6}{2}\)

\(\text{AOS} = \cfrac{8}{2}\)

\(\text{AOS} = 4\)

We can use the AOS to determine the Optimal Point (OP) by substituting the value into the original function. If the OP is negative (or a minimum), then all non-negative values before \(2\) and after \(6\) will indicate the object moving in a positive direction. If negative, then all values in between \(2\) and \(6\) will indicate the positive direction:

\(\text{OP} = 3(4)^2 -24(4) +36\)

\(\text{OP} = 3(16) -96 +36\)

\(\text{OP} = 48 -96 +36\)

\(\text{OP} = -12\)

Since the OP is negative this indicates that the object moves in a positive direction before \(t=2\) and after \(t=6\).

Therefore, we can determine that the object moves in a positive direction for \(tϵ(0,2)\) and \(tϵ(6,∞)\).

iii. In order to determine when the object returns to its original position, we can set the original function to \(0\). We can then factor the function to find its \(0\)'s:

\(0 = t^3 -12t^2 +36t\)

\(0 = t(t^2 -12t +36)\)

\(0 = t(t^2 -6t - 6t +36)\)

\(0 = t(t(t-6) -6(t-6))\)

\(0 = t((t-6)^2)\)

Therefore, we can determine that the object will return to its original position at \(0\;[\text{s}]\) and \(6\;[\text{s}]\).

iv. First, we need to determine the acceleration formula. We can do this by calculating the second derivative of the original function:

\(s''(t) = 6t -24\)

We can now draw our respective graphs as such:

Position-Time Graph showing the position of a particle in meters in relation to the time in seconds.

Velocity-Time Graph showing the velocity of the particle in meteres/second in relation to the time in seconds.

Acceleration Graph showing the acceleration of the particle in meters/second² in relation to the time in seconds.

v. At \(t=0\), the object is going away and gradually slowing down.

At \(t = 2\), the object is turning back and gradually speeding up until \(t=4\). It continues moving in the same direction but continues to slow down until \(t = 6\).

At \(t=6\), the object returns to the origin and turns away and continues to speed up.

Velocity and Acceleration

Review these lessons:

Try these questions: