Have you ever wondered what the feeling is in a lift as it accelerates away and then when it stops? This jerky feeling is due to the acceleration of the lift and the fact it isn't constant. Over your time in the lift, the acceleration of the lift will change and this change in acceleration with time is what leads to that jerky feeling. Read further for explanations and examples of acceleration varying with time.
Relationship between Acceleration and Time
Acceleration is the rate of change in velocity per unit of time and has the units of \(\text{m/s}^2.\)
Acceleration is directly related to time in that it is the rate of change in velocity over a change in time. This rate of change could be a linear or quadratic relationship.
If you are examining a linear relationship with constant acceleration with given values for a combination of displacement, initial velocity, final velocity, acceleration and time you would apply the SUVAT equations and solve for your unknown variable in your solution. You can see a further explanation of this below.
You can apply SUVAT to multiple dimensions- for example in projectile motion you would consider the motion in the \(x\) and \(y\) planes and resolve each plane separately using SUVAT and trigonometry. For more information see Further Projectiles.
If you were to be presented with a function that leads to a non-constant acceleration - this may be quadratic- you would apply the differential or integral technique as the relationship is different.
Most questions that assess a solution via the differential or integral method will be with respect to \(\mathrm{d} t\) meaning the functions are tracking the change in a variable with respect to time. These variables could be position, velocity or acceleration. The below graphic shows how this would work.
Fig. 1. Graphic to show relations between displacement, velocity and acceleration.
Acceleration and Time Equation
Examining the graphic above you see how displacement, velocity and acceleration are related but how do you write this mathematically?
If you have an expression for the position of an object given as \(r,\) you can see that the velocity will be how this position changes with time:
\[v=\frac{\mathrm{d} r}{\mathrm{d} t}.\]
You also know that acceleration is measured by how much the velocity changes with time so is given by:
\[a=\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} ^2r}{\mathrm{d} t^2}.\]
These are the derivative relationships you use to assess velocity and acceleration. As seen in the graphic if you wanted to work in the other direction, you would just integrate:
\[v=\int a\, \mathrm{d} t\]\[r=\int v\, \mathrm{d} t.\]
As you can see in the above formulas, all are in relation to \(dt\) which highlights the relationship between time and these other variables- they change with time.
Acceleration and Time Graph
The acceleration-time graph can be used to find the velocity change over a time period. As you saw above, the method for moving from acceleration to velocity is via integration with respect to time. In graph format, this means finding the area between the line and the \(x\) axis. Let's look at the graphic below to contextualize this.
Fig. 2. Example of an acceleration-time graph.
You've seen how finding the area enclosed by the line and axis gives the change in velocity and how this is due to integration, but let's have a look at how this works.
First, examine \(0\) to \(1\) seconds. You see a \(y\) axis value of \(-3\) which would indicate the velocity is decreasing and as this time interval is one second, you can understand that in the interval the velocity dropped by \(3\text{ m/s}\). This fits as the area between the line and axis in this interval is \(-3\cdot 1\).
Let's now look at the interval directly after between \(1\) and \(3\) seconds. Here the line is on the x-axis indicating there is no change in velocity over the time period- there is no acceleration.
Let's now move on to look between \(5\) and \(7\) seconds- here the area between the line and axis is governed by:
\[a \Delta t= -2\cdot 2 = -4\text{ m/s}.\]
This means the velocity was decreasing at \(-2\text{ m/s}\) over \(2\) seconds so the velocity decreased by a total of \(4\text{ m/s}.\)
Let's now have a look at an example question using this graphic.
In regards to the acceleration-time graph above, what is the change in velocity values at \(3\) and \(7\) seconds?
Solution
Looking at the graph you can see there are two changes in acceleration over the time period. To tackle this you will therefore find the area under each of the acceleration changes separately and add them to find the total change in velocity. You will then solve the same situation via verbal reasoning to see if your answers line up.
Step one: Calculate the area under the graph
\[\begin{align} \Delta v &= \mbox{area of segment one} + \mbox{area of segment two} +\mbox{area of segment two} \\ &=(a_1\cdot \Delta t_1)+(a_2\cdot \Delta t_2)+(a_3\cdot \Delta t_3) \\&= (1\cdot 1)+(3\cdot 1)+(-2\cdot 2) \\&=1+3-4\\ &=0\text{ m/s}. \end{align}\]
Step two: Verbal reasoning
Section one is between \(3\) and \(4\) seconds and the velocity increases by \(1\text{ m/s}^2\). Section two then has an increase in velocity of \(3\text{ m/s}^2\) over an interval of one second. This means the velocity has increased by \(3\text{ m/s}\) and so far the total increase in your velocity is \(4\text{ m/s}\). Between \(5\) and \(7\) seconds, you then have a decrease of \(2\text{ m/s}^2\) for \(2\) seconds. This leads to a decrease of \(4\text{ m/s}\). Therefore the change in velocity values at \(3\) and \(7\) seconds is \(0\text{ m/s}\).
Acceleration and Time to Velocity
You've now seen the general principle to move between displacement, acceleration and time when you have a function defining the motion in terms of time via the methods of integration and differentiation. Let's now have a look at an example that puts this principle into practice.
A particle has an acceleration that varies with time:
\[a=3\sin(2t).\]
If it starts from rest, find its velocity after 30 seconds.
Solution
Your methodology here is as simple as a question of this type can be. You have one transformation step - you integrate acceleration to get velocity- you apply initial conditions and then you solve for time being 30 seconds.
Step one: Integrate
\[\begin{align} v&=\int a\, \mathrm{d}t\\&=\int 3\sin(2t)\, \mathrm{d} t\\&=3\cdot \int \sin(2t)\, \mathrm{d} t\\&=3\left( -\frac{1}{2}\right) \cos(2t)+C\\&=- \frac{3}{2}\cos(2t)+C \end{align}\]
Step two: Apply initial conditions and find the constant of integration
You know the particle starts at rest so therefore: \(v=0\) and \(t=0\).
\[\begin{align} v&=- \frac{3}{2}\cos(2t)+C\\ 0&=- \frac{3}{2}\cos(2\cdot 0)+C \\ 0&=\left[- \frac{3}{2}\cdot 1 \right] +C\\ \frac{3}{2}&=C. \end{align}\]
Therefore
\[ v=- \frac{3}{2}\cos(2t)+ \frac{3}{2} . \]
Step three: Calculate the velocity at \(t=30\)
Now you take the expression for velocity in terms of time and sub in your time value.
\[\begin{align} v&=- \frac{3}{2}\cos(2t)+ \frac{3}{2} \\&=- \frac{3}{2}\cos(2\cdot 30)+ \frac{3}{2} \\&=0.75\text{ m/s}\end{align}\]
Acceleration and Time to Distance
You now have a good understanding of how to find the solution to a question where acceleration varies with time- you can integrate it twice to gain distance and you've graphically examined how this theory works when changing to velocity.
You've also had a look at an example that moves once from acceleration to velocity. Let's now have a look at an example that moves from an acceleration function in terms of time to distance traveled.
An object has an acceleration that varies with time and this acceleration is defined as:
\[a=\frac{4t-2}{t^2}.\]
First find the velocity of the object in terms of time \(t\) and then find its position after \(7\) seconds. After \(1\) second, the object has a velocity of \(6\text{ m/s}\) and has travelled \(2.4\text{ m}\) after \(1.5\) seconds.
Solution
Step one: Calculate velocity in terms of time
You first apply the integration technique to move from acceleration to velocity.
\[\begin{align}v&=\int a\, \mathrm{d} t\\&=\int \frac{4t-2}{t^2} \, \mathrm{d} t\\&= \int \frac{4}{t} -\frac{2}{t^2}\, \mathrm{d} t\\&=\int \frac{4}{t}\, \mathrm{d} t - \int \frac{2}{t^2}\, \mathrm{d} t\\&=4\ln|t|+C-\left[-\frac{2}{t}\right]\\&= 4\ln|t|+\frac{2}{t} +C. \end{align}\]
You know the object has a velocity of \(6\) m/s after \(1\) second so you can apply these as initial conditions: \(v=6\) and \(t=1.\)
\[\begin{align}v&= 4\ln|t|+\frac{2}{t} +C \\ 6&=4\ln|1|+ \frac{2}{1}+C \\ 6&= 0+2+C\\4&=C. \end{align} \]
Therefore
\[ v= 4\ln|t|+\frac{2}{t} +4.\]
Step two: Calculate position after 7 seconds
To get the expression for position you will need to integrate again- this time from velocity to acceleration.
\[\begin{align} r&=\int v\, \mathrm{d} t \\ &=\int (4\ln|t|+\frac{2}{t} +4) \, \mathrm{d} t\\&=\int 4\ln|t|dt+\int \frac{2}{t} \, \mathrm{d} t+ \int 4 \, \mathrm{d} t\\&= 4t\ln|t|-4t+2\ln|t|+4t+C\\&=4t\ln|t|+2\ln|t|+C\end{align}\]
You know the object has travelled \(2.4\text{ m}\) after \(1.5\) seconds so you can apply the initial conditions of \(r=2.4\) and \(t=1.5\).
\[\begin{align}r&=4t\ln|t|+2\ln|t|+C\\2.4&=4(1.5)\ln|1.5|+2\ln|(1.5)|+C\\ 2.4&=2.43+0.81+C\\-0.84&=C. \end{align} \]
Therefore
\[ r =4t\ln|t|+2\ln|t|-0.84.\]
You then find position after 7 seconds:
\[\begin{align}r&=4t\ln|t|+2\ln|t|-0.84\\&=4(7)\ln|7|+2\ln|7|-0.84\\&=57.54\text{ m}\end{align}\]
You can also go further than acceleration in terms of derivatives of position with respect to time. Jerk is the 3rd derivative of position in respect to time- this is the rate of change of acceleration.
Finding distance with Acceleration and Time
Another method for finding the distance from acceleration and time stems from the SUVAT equations. You can apply the SUVAT equations to a problem with a mix of the variables: displacement, initial velocity, final velocity, time and acceleration. In terms of notation:
\[\begin{align} s&=\mbox{Displacement}\\u&=\mbox{Initial Velocity}\\v&=\mbox{Final Velocity}\\a&=\mbox{Acceleration}\\t&=\mbox{Time}\end{align}\]
You then have the five equations for SUVAT calculations:
\[\begin{align} s&= ut+\frac{at^2}{2}\\ s&= vt-\frac{at^2}{2}\\ s&=\frac{v+u}{2}t\\ v^2&=u^2+2as\\ v&=u+at\end{align}\]
As you can see in the first two equations you can calculate the distance with acceleration and time, but you will also need either the initial or final velocity for this. The exception to this is if you had a starting velocity of \(0\) - if the object starts from rest. This would result in the equation:
\[\begin{align}s&=ut+\frac{at^2}{2}\\&=(0)t+\frac{at^2}{2}\\&=\frac{at^2}{2}.\end{align}\]
Let's have a look at this method in an example.
A car is accelerating at \(2\text{ m/s}^2\). Considering the fact the car's final velocity is \(45\text{ m/s}\) find the distance it travelled over \(5\) seconds.
Solution
You have acceleration, time and final velocity given and as such, you will use equation 2 as seen above to find the distance.
\[ \begin{align}s&= vt-\frac{at^2}{2}\\&=(45\cdot5)-\frac{2\cdot5^2}{2}\\&= 300\text{ m}.\end{align}\]
Acceleration and Time - Key takeaways
- Acceleration is the rate of change in velocity per unit of time and can be expressed as a function of time - the acceleration changes with time.
- You can move between position, velocity and acceleration by differentiating down the list or integrating backwards up the list.\[\begin{align} & a=\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} ^2r}{\mathrm{d} t^2}\\& v=\int a\, \mathrm{d} t \\& r=\int v\, \mathrm{d} t. \end{align}\]
- When starting with an expression for acceleration you will use the given initial conditions to find the constant of integration at each step.
- If acceleration is constant and you have a combination of displacement, initial velocity, final velocity and time, you can use the SUVAT equations to find your solution. The SUVAT equations are: \[\begin{align} s&= ut+\frac{at^2}{2}\\s&= vt-\frac{at^2}{2}\\s&=\frac{v+u}{2}t\\v^2&=u^2+2as\\v&=u+at.\end{align}\]
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