It’s the tail end of summer, and your parents suggest one last family beach day. While driving down, you’re not paying much attention as you listen to music and play on your phone. However, you suddenly notice the car begins to slow down. When you pick your head up, you see why, the dreaded “traffic.” Now, you may not realize it, but the action your parents just performed is a classic example of physics, specifically involving the concepts of average velocity and average acceleration. When you hit the brakes, your car's velocity starts to drop over a certain distance, and the car now has acceleration due to the change in velocity. Therefore, let this article define average velocity and acceleration as well as explain how one can compute average velocity and average acceleration based on what kinematic equations one has been given.
Difference Between Average Velocity and Average Acceleration
Average velocity and average acceleration are not the same things. Although both velocity and acceleration are vectors with magnitude and direction each describes a different aspect of motion. Average velocity describes an object's change in position with respect to time while average acceleration describes an object's change in velocity with respect to time. Moreover, an object is accelerating if either the magnitude or direction of the object's velocity is changing.
Average quantities refer to quantities that are calculated only considering the initial and final values of that quantity.
Definition of Average Velocity and Average Acceleration
We will define average velocity and acceleration as well as discuss their corresponding mathematical formulas.
Average Velocity
Average velocity is a vector quantity that relies on the final and initial position of an object.
Average velocity is an object's change in position with respect to time.
The mathematical formula corresponding to this definition is $$v_{\text{avg}}=\frac{\Delta{x}}{\Delta{t}}$$
where \( \Delta{x} \) represents the change in position and \( \Delta{t} \) represents the change in time.
The SI unit for velocity is \( \mathrm{\frac{m}{s}} \).
One can also calculate average velocity using the initial and final values of velocity.
$$v_{\text{avg}}=\frac{v_o + v}{2}$$
where \( v_o \) is initial velocity and \( v \) is final velocity.
This equation is derivable from the kinematic equation for average distance as follows:
$$\begin{aligned}\Delta{x}=& \frac{v_o+v}{2}(t) \\ \frac{\Delta{x}}{t}= & \frac{v_o+v}{2} \\ v_{\text{avg}}= & \frac{v_o+v}{2}. \\ \end{aligned}$$
Note from the above that \( \frac{\Delta{x}}{t} \) is the definition of average velocity.
Since we've defined the average velocity and discussed two corresponding formulas we can use to determine its value, let's solve a simple example to help us understand this before moving on.
For exercise, an individual walks \( 3200\,\mathrm{m} \) every day. If it takes \( 650\,\mathrm{s} \) to complete this, what is the average velocity of the individual?
Walking is an example of determining average velocity and average acceleration.CC-iStock
Based on the problem, we are given the following:
As a result, we can identify and use the equation,
\( v_{\text{avg}}=\frac{\Delta{x}}{\Delta{t}} \) to solve this problem. Therefore, our calculations are:
$$\begin{aligned}v_{\text{avg}} &=\frac{\Delta{x}}{\Delta{t}} \\ v_{\text{avg}}&=\frac{3200\,\mathrm{m}}{650\,\mathrm{s}} \\ v_{\text{avg}}&=4.92\,\mathrm{\frac{m}{s}}. \\\end{aligned}$$
The average velocity of the individual is \( 4.92\,\mathrm{\frac{m}{s}}. \)
Average Acceleration
Average acceleration is a vector quantity that relies on the final and initial velocities of an object.
Average acceleration is an object's change in velocity with respect to time.
The mathematical formula corresponding to this definition varies depending on different quantities such as velocity and time or velocity and distance.
We'll introduce the formula in another section. But first, we'll discuss two ways to calculate average velocity given kinematic variables.
Calculating Average Velocity from Acceleration and Time Variables
Above we saw that the definition of average velocity does not depend on intermediate values of velocity over a time interval. This means that we only need the values of the initial and final velocity of an object if we want to calculate its average velocity. But what happens if, instead of knowing the initial and final velocity, we only know the initial velocity and the acceleration? Can we still determine the average velocity? Yes! But, to do so, we have to use the kinematic equations.
What is kinematics? Well, kinematics is a field in physics that focuses on the motion of an object without reference to the forces that cause it. The study of kinematics focuses on four variables: velocity, acceleration, displacement, and time. Note that velocity, acceleration, and displacement are all vectors, which means they have magnitude and direction. Therefore, the relationship between these variables is described by the three kinematic equations.
These are the linear kinematic equation,
$$v=v_o + at;$$
the quadratic kinematic equation,
$$\Delta{x}=v_o{t} + \frac{1}{2}at^2;$$
and the time-independent kinematic equation,
$$v^2= {v_o}^2 + 2a\Delta{x}.$$
Here \( v \) is final velocity, \( v_o \) is initial velocity, \( a \) is acceleration, \( t \) is time, and \( \Delta{x} \) is displacement.
These kinematic equations only apply when acceleration is constant.
To calculate average velocity from acceleration and time, we start from the quadratic kinematic equation:
$$\begin{aligned}\Delta{x}&=v_o{t} + \frac{1}{2}at^2 \\ \Delta{x}&= t(v_o + \frac{1}{2}at)\\ \frac{\Delta{x}}{t}&=v_o + \frac{1}{2}at \\v_{\text{avg}}&= v_o + \frac{1}{2}at.\\\end{aligned}$$
Hence, the equation \( v_{\text{avg}}= v_o + \frac{1}{2}at \) can determine the average velocity. Going a step further, we can plug in the definition of acceleration, \( {a=\frac{\Delta{v}}{t}} \), and re-derive the average velocity equation, which includes only its initial and final quantities.
$$\begin{aligned}v_{\text{avg}}&= v_o + \frac{1}{2}at \\ v_{\text{avg}}&= v_o + \frac{1}{2}{\frac{\Delta{v}}{t}}t\\ v_{\text{avg}}&= v_o + \frac{1}{2}\Delta{v} \\v_{\text{avg}}&= \frac{2v_o + (v-v_o)}{2}\\v_{\text{avg}}&= \frac{v_o + v}{2}\\v_{\text{avg}}&= \frac{1}{2}{\left(v_o + v\right)}.\\\end{aligned}$$
By doing this, we've verified that the average velocity indeed depends only on the initial and final velocity. Let's now see how we can calculate the average velocity from a graphical representation.
Calculating Average Velocity from an Acceleration-Time Graph
Another way to calculate average velocity is by means of an acceleration-time graph. When looking at an acceleration-time graph, you can determine the velocity of the object as the area under the acceleration curve is the change in velocity.
$$\text{Area}=\Delta{v}.$$
For example, the acceleration-time graph below represents the function, \( a(t)=0.5t+5 \). Using this, we can show that the change in velocity corresponds to the area under the curve.
The function indicates that as time increases by one second, the acceleration increases by \( 0.5\,\mathrm{\frac{m}{s^2}} \).
Fig. 1 Determining average velocity from an acceleration-time graph.
Using this graph, we can find what the velocity will be after a specific amount of time by understanding that velocity is the integral of acceleration
$$v=\int_{t_1}^{t_2}a(t)$$
where the integral of acceleration is the area under the curve and represents the change in velocity. Therefore,
$$\begin{aligned}v&=\int_{t_1}^{t_2}a(t) \\ v&=\int_{t_1=0}^{t_2=5}(0.5t +5)dt\\ v&=\frac{0.5t^2}{2}+5t \\v&=\left(\frac{0.5(5)^2}{2}+5(5))-(\frac{0.5(0)^2}{2}+5(0)\right)\\v&=31.25\,\mathrm{\frac{m}{s}}.\\\end{aligned}$$
We can double-check this result by calculating the area of two different shapes (a triangle and a rectangle) as the first figure shows.
Start by calculating the area of the blue rectangle:
$$\begin{aligned}\text{Area}&=(\text{height})(\text{width})=hw \\\text{Area}&=(5)(5)\\ \text{Area}&=25.\\\end{aligned}$$
Now calculate the area of the green triangle:
$$\begin{aligned}\text{Area}&=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)=\frac{1}{2}bh \\\text{Area}&=\frac{1}{2}\left(5\right)\left(2.5\right)\\ \text{Area}&=6.25.\\\end{aligned}$$
Now, adding these two together, we retrieve the result for the area under the curve:
$$\begin{aligned}\text{Area}_{\text{(curve)}}&=\text{Area}_{(\text{rec})}+ \text{Area}_{(\text{tri})} \\{Area}_{(\text{curve})}&= 25 + 6.25\\ \text{Area}_{(\text{curve})}&=31.25.\\\end{aligned}$$
The values match clearly, showing that in the acceleration-time graph, the area under the curve represents the change in velocity.
Calculating Average Acceleration Given Velocity and Time
To calculate the average acceleration at a given velocity and time, the appropriate mathematical formula to start with is
$$a_{avg}=\frac{\Delta{v}}{\Delta{t}}$$
where \( \Delta{v} \) represents the change in velocity and \( \Delta{t} \) represents the change in time.
The SI unit for acceleration is \( \mathrm{\frac{m}{s^2}} \).
The following example asks us to use the above equation to find a numerical answer.
A car's velocity increases from \( 20\,\mathrm{\frac{m}{s}} \) to \( 90\,\mathrm{\frac{m}{s}} \) in a span of \( 16\,\mathrm{s} \). What is the average acceleration of the car?
A moving car demonstrating average velocity and average acceleration.CC-Science4fun
Based on the problem, we are given the following:
- initial velocity
- final velocity
- time
As a result, we can identify and use the equation, \( a_{\text{avg}}=\frac{\Delta{v}}{\Delta{t}} \) to solve this problem. Therefore, our calculations are:
$$\begin{aligned}a_{\text{avg}}&=\frac{\Delta{v}}{\Delta{t}} \\a_{\text{avg}}&=\frac{90\,\mathrm{\frac{m}{s}}-20\,\mathrm{\frac{m}{s}}}{16\,\mathrm{s}}\\ a_{\text{avg}}&=\frac{70\,\mathrm{\frac{m}{s}}}{16\,\mathrm{s}}\\a_{\text{avg}}&= 4.375\,\mathrm{\frac{m}{s^2}}.\\\end{aligned}$$
The average acceleration of the car is \( 4.375\,\mathrm{\frac{m}{s^2}}. \)
Next, we'll see how the method to calculate acceleration changes if we've been given the distance instead of the time.
Calculating Average Acceleration with Velocity and Distance
To calculate the average acceleration from the velocity and distance, we have to use the kinematic equations once more. Looking at the list above, note that the first and second equations have an explicit time dependence. This means we have to rule them out and use the third equation instead.
$$\begin{aligned}v^2&={v_o}^2+2a\Delta{x} \\v^2-{v_o}^2&=2a\Delta{x}\\ a&=\frac{v^2-{v_o}^2}{2\Delta{x}}.\\\end{aligned}$$
Recall that the kinematic equations are applicable only in the case of constant acceleration. Since the average acceleration over a time interval is constant, the equation \( a=\frac{v^2-{v_o}^2}{2\Delta{x}} \) allows us to calculate the average acceleration from the velocity and distance.
We can verify that the derived equation is also reducible to the definition of average acceleration.
$$\begin{aligned}a&=\frac{v^2-{v_o}^2}{2\Delta{x}} \\a&=\frac{v^2-{v_o}^2}{2\Delta{t}(v_{\text{avg}})}\\ a&=\frac{(v+v_o)-(v-v_o)}{2\Delta{t}(\frac{v_o +v}{2})}\\a&=\frac{(v-v_o)}{\Delta{t}}\\a&=\frac{\Delta{v}}{\Delta{t}}.\\\end{aligned}$$
Note that \( v_{\text{avg}}=\frac{\Delta{x}}{\Delta{t}} \).
Now, in the above derivation, we found an expression for acceleration given the velocity and distance. We took the third kinematic equation as a starting point and isolated on the left-hand side the quantity we wanted. We could just as well have manipulated the same equation to solve for another quantity.
The example below illustrates this point. In it, you are given acceleration and distance and are asked to solve for the final velocity.
A ball, dropped from a building, travels \( 23\,\mathrm{m} \) to the ground under the force of gravity. What is the average velocity of the ball?
Dropping a ball to demonstrate average velocity and average acceleration.CC-Chegg
Based on the problem, we are given the following:
As a result, we can identify and use the equation, \( v^2={v_o}^2 +2g\Delta{x} \) to solve this problem. Therefore, our calculations are:
$$\begin{aligned}v^2&={v_o}^2+2g\Delta{x} \\v^2-{v_o}^2&=2g\Delta{x}\\ a\Delta{v}&=\sqrt{2g\Delta{x}}\\\Delta{v}&=\sqrt{2(9.81\,\mathrm{\frac{m}{s^2}})(23\,\mathrm{m})}\\\Delta{v}&= 21.24\,\mathrm{\frac{m}{s}}.\\\end{aligned}$$
The average velocity of the ball is \( 21.24\,\mathrm{\frac{m}{s}} \).
Zero Velocity and a Nonzero Average Acceleration
Is it possible to have zero velocity and a nonzero average acceleration? The answer to this question is yes. Imagine throwing a ball straight up into the air. Due to gravity, the ball will have a constant non-zero acceleration throughout its flight. However, when the ball reaches the highest vertical point of its path, its velocity will momentarily be zero. The figure below illustrates this.
A diagram demonstrating zero velocity and nonzero acceleration.CC-Mathsgee
Average Velocity and Acceleration - Key takeaways
- Average velocity is defined as an object's change in position with respect to time.
- Average velocity can be calculated in three ways: the formulas \(\ v_{\text{avg}}=\frac{\Delta{x}}{\Delta{t}} \) or \( v_{\text{avg}}= v_o + \frac{1}{2}at \) as well as the use of an acceleration-time graph in which the area under the acceleration curve is representative of the change in velocity.
- Average acceleration is defined as an object's change in velocity with respect to time.
- Average acceleration can be calculated in two ways: the formulas \( a_{\text{avg}}=\frac{\Delta{v}}{\Delta{t}} \) or \( a=\frac{v^2-{v_o}^2}{2\Delta{x}} \).
- Average velocity and average acceleration are not the same things as one describes an object's change in position with respect to time while the other describes an object's change in velocity with respect to time.
- It is possible for an object to have zero velocity and a nonzero average acceleration.
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