Have you ever thought about the physics and mechanics of skydiving? As you fall your acceleration changes with the speed you are falling and then levels out at a certain point- why is this? The answer is terminal velocity, and this is because your acceleration would vary with your velocity.
This means the increase of speed as you fall is governed by a function for acceleration that is driven by your changing velocity and at a certain point, you will have hit your terminal velocity- the rate of your fall that you can't exceed.
Read more for further explanations of this principle.
We first start with the relationship between acceleration and velocity.
Acceleration and Velocity relationship
One big question around acceleration is how does it relate to velocity? The answer to this is that acceleration is the derivative of velocity- this means that acceleration is the rate of change of velocity.
Conversely, if you integrate an expression for acceleration you will arrive at the expression for velocity.
Acceleration and velocity are both vector quantities meaning they have both size and direction. This means when considering the values of acceleration and velocity they have sign values attached to them to indicate direction.
For example, if you had a negative acceleration value, the object you are examining is decelerating to the velocity is decreasing.
But, what are the differences between acceleration and velocity then?
Acceleration and Velocity differences
While we've looked at the relationship and similarities between velocity and acceleration, the fact remains they are different variables and as such, the question arises of how are they different.
Acceleration tracks the change in velocity; as such, they will hold different values but can also be of different signs.
For example, if you were examining the acceleration of a car slowing down, the acceleration would be negative for deceleration as the car slows but the velocity would still be positive as it is moving forwards, just at a decreasing rate.
Now, that we know how acceleration and velocity are related and how they are different, we move next to acceleration and velocity derivatives.
Acceleration and Velocity derivatives
We know that acceleration is the rate of change of velocity but we also have the relationship between velocity and displacement: velocity is the rate of change of displacement. That means that acceleration is the second derivative of displacement.
This relationship then works the other way as well - if you have an expression for acceleration and integrate it, you will have an expression for velocity, and if you have an expression for velocity and integrate it, you will have an expression for displacement. The graphic below demonstrates these relationships.
Fig. 1. Graphic to show relations between displacement, velocity and acceleration.
Examining the graphic we see how displacement, velocity, and acceleration are related but how do we write this mathematically?
If we have an expression for the position of an object given as \(r,\) we can see that the velocity will be how this position changes with time,\[v=\frac{dr}{dt}.\] We also know that acceleration is measured by how much the velocity changes with time so is given by:\[a=\frac{dv}{dt}=\frac{d^2r}{dt^2}.\]These are the derivative relationships we use to assess velocity and acceleration. As seen in the graphic above, if we wanted to work in the other direction, we would just integrate,\[v=\int a\quad dt\]\[r=\int v\quad dt.\]
We move now to interpret the graphs of acceleration and velocity.
Acceleration and Velocity graph
We've now seen how displacement, velocity, and acceleration are related to each other, we can further visualize this by examining velocity/acceleration time graphs.
Fig. 2. Graphic showing acceleration and velocity-time graphs for comparison.
Working from what we saw previously with the relationship between velocity and acceleration, we can visualize this on the graphs above.
When we differentiate we are finding the gradient of the velocity line. Examining the figure above you can see this - look between \(4\) and \(5\) seconds on the top graph and you will see the velocity climbing from \(1 m/s\) to \(4 m/s\) over one second. This gives a gradient of \(3\) which you can see at the corresponding point on the acceleration time graph.
Inversely, if we look between \(0\) and \(1\) seconds on the acceleration time graph and calculate the area enclosed by the line we will return our velocity profile. This is the integration technique - if we integrate the acceleration (the area above the line in this case) we will result in the velocity. The answer of the integral is \(-3\) as it's on the underside of the axes and if we examine the velocity graph we can see over this time period the velocity decreased by \(3 m/s.\)
These relations should help you understand how we can differentiate and integrate between displacement, velocity, and acceleration as you've seen previously.
Acceleration and Velocity formula
When considering acceleration that varies with the velocity, we would usually be given an expression using a function of velocity.
We know our formula for acceleration and velocity is given by, \[a=\frac{dv}{dt},\] and if we are given a function of velocity to solve we would have a question in the form, \[\frac{dv}{dt}=f(v).\] This type of question is stating that the acceleration for the question is governed by a function of the velocity. So how would we solve this?
The question has a \(f(v)\) term which is equal to our acceleration \((a)\). Hence, we can expect to solve this via the separation of variables method. This means we collect like terms on either side of the equals sign, integrate and then solve for \(v\) to gain an expression for velocity \((v)\) in terms of time \((t)\) and any constants. Let's look at the logic behind this.
We know that acceleration is the rate of change of velocity,\[\therefore a=\frac{dv}{dt}.\]We also know that we have a function \(f(v)\) that is equal to the acceleration. This leads to the expression, \[\frac{dv}{dt}=f(v).\]From here, we would collect out \(v\) terms with the \(dv\) term and collect any \(t\) terms with our \(dt\) term on the other side of the equals sign.
For example if \(f(v)=\frac{t}{v}\) where \(t\) is a time variable, our separation of variables would look like, \[\begin{align}a&=f(v)\\ \frac{dv}{dt}&=\frac{t}{v} \\ v dv&= tdt. \end{align}\] From here we would integrate and gain an expression of how velocity varies with time from an expression for acceleration. Let's look at what this might look like in an example.
The acceleration of a particle varies with its velocity and is defined by the equation, \[a=3v-4.\]Find the velocity of the particle after 0.5 seconds if the particle is moving at \(20\text{ m/s}\) at \(t=0.\)
Solution
Step 1. Separation of variables
\[\begin{align} a&=3v-4\\ \frac{dv}{dt}&=3v-4\\ \frac{1}{3v-4}dv&=1dt\end{align}\]
Step 2. Integrate
\[\begin{align} \int \frac{1}{3v-4}dv&=\int 1dt \\ \frac{\ln(|3v-4|)}{3}&=t+C\end{align}\]
Step 3. Rearrange for \(v\)
\[\begin{align} \frac{\ln(|3v-4|)}{3}&=t+C \\ \ln(|3v-4|) &= 3t+C\\ 3v-4&= e^{3t+C} \\ v&=\frac{ e^{3t+C} +4}{3}\end{align}\]
Step 4. Apply initial conditions
We know at \(t=0\text{ s}\) the velocity is \(20\text{ m/s}\) so, \(v(0)=20.\) Subbing into our equation, we get,
\[\begin{align}v&=\frac{ e^{3t+C} +4}{3}\\ 20&=\frac{ e^{3(0)+C} +4}{3} \\ 60&=e^C+4\\ \ln(56)&=C\end{align}\]
Therefore,
\[\begin{align}v=\frac{ e^{3t+\ln(56)} +4}{3}\end{align}\]
Step 5. Find velocity at \(t=0.5s\)
\[\begin{align}v&=\frac{ e^{3t+\ln(56)} +4}{3} \\ &=\frac{ e^{3(0.5)+\ln(56)} +4}{3}\\ &= 84.99\text{ m/s} \end{align}\]
Terminal or limiting velocity
We can also calculate the terminal velocity of an object using the same expression format as seen above.
If we take the terminal velocity to be the maximum velocity the object can reach, this means that the rate of change of velocity will be 0. The object has reached its maximum velocity so the velocity can't rise anymore.
An example of this could be an object in freefall. At some point, the weight of the object will limit the speed at which it is falling and this speed cannot be exceeded. We will still have the acceleration \(a\) expressed as a function of velocity \(v\) which results in the general form,\[\frac{dv}{dt}=f(v)\]\[0=f(v).\]If we then rearrange the function for \(v\) we will have the limiting velocity. Let's have a look at this as an example.
An object is in freefall and its acceleration is given as,\[a=\frac{237-6v^2}{24}.\]Find the object's terminal velocity.
Solution
Step 1. Write the equation in the rate of change form
\[\begin{align}a&=\frac{237-6v^2}{24}\\ \frac{dv}{dt}&=\frac{237-6v^2}{24}.\end{align}\]We know that at the terminal velocity the rate of change of velocity is 0 so,\[0=\frac{237-6v^2}{24}\]
Step 2. Solve for \(v\)
We know the left-hand side of our equation must be equal to 0. As such, the numerator of the fraction must equal 0 so when it's divided by the denominator (24) it will return 0. This means that \(6v^2\) must be equal to 237.
\[\begin{align}6v^2&=237\\ v&=\sqrt{\frac{237}{6}} \\&=6.28\text{ m/s}\end{align}\]
Acceleration and Velocity equations
We also have another methodology we can use when tracking acceleration with varying velocity. We can apply the SUVAT equations to a problem with constant acceleration and varying velocity. In terms of notation we have,\[\begin{align} s&=\mbox{Displacement}\\u&=\mbox{Initial Velocity}\\v&=\mbox{Final Velocity}\\a&=\mbox{Acceleration}\\t&=\mbox{Time}\end{align}\]We then have the five equations for SUVAT calculations: \[\begin{align} &(1)\quad s= ut+\frac{at^2}{2}\\&(2)\quad s= vt-\frac{at^2}{2}\\&(3)\quad s=\frac{v+u}{2}t\\&(4)\quad v^2=u^2+2as\\&(5)\quad v=u+at\end{align}\]
We can see how equations (1), (2), (4) and (5) have an acceleration term in and as such the relevant equations rearranged for acceleration are as follows,\[\begin{align} &(1)\quad a= \frac{2s-2ut}{t^2}\\&(2)\quad a= - \left[\frac{2s-2vt}{t^2}\right]\\&(4)\quad a=\frac{v^2-u^2}{2s}\\&(5)\quad a=\frac{v-u}{t}.\end{align}\]
What is the acceleration experienced by an object if it starts at \(5\text{ m/s} \) after \(10\) seconds having travelled a distance of \(37\text{ m}\)?
Solution
We know the initial velocity, time and distance and want to know the acceleration. That means we can use equation (1) above which is,\[s= ut+\frac{at^2}{2}\]Rearranging for our unknown acceleration and solving:
\[\begin{align}a&= \frac{2s-2ut}{t^2}\\&=\frac{(2\cdot 37)-(2\cdot 5\cdot 10)}{10^2}\\&=-0.26\text{ m/s}^2\end{align}\]
This means the object is decelerating at \(0.26\text{ m/s}^2\) after \(10\) seconds from the start.
Acceleration and Velocity - Key takeaways
- You can move between displacement, velocity, and acceleration by differentiating along that order or integrating backwards.
- Acceleration and velocity graphs are linked in that the gradient of the velocity graph gives your acceleration value and the area under the acceleration graph for a time interval gives you the velocity change.
- When you have an equation that gives acceleration as a function of velocity you can use the separation of variables technique and integration to get an expression for velocity.
- The terminal velocity of an object can be calculated by setting the rate of change of velocity as \(0\) and solving for \(v.\) This takes the general form \(0=f(v).\)
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