Going Further: 10.1 Accelerated Motion

We have seen that the distance traveled by an object moving with a constant velocity is given by:

\begin{equation} d=vt \end{equation}

We cannot generally use this expression if the velocity is changing. However, we can use a similar relation if we know the average speed a particle moves. For example, if you traveled 30 km/hr for half an hour, and then you sped up to 90 km/hr for an additional half hour, you would have traveled a total of 60 km in one hour:

\begin{equation} (30\,\rm{km/hr})(0.5\,\rm{hr})+(90\,\rm{km/hr})(0.5\,\rm{hr})= 15km + 45km = 60km \end{equation}

Since you completed the trip in one hour, and you traveled a total distance of 60 km, your average speed must have been 60 km/hr, though you did not spend any of the trip actually traveling 60 km/hr. In general, if we know the average speed of an object, then we can use the following relation to find how far it moves in a time t:

\begin{equation} d=\bar{v}t \end{equation}

The bar over the v means that we are using the average speed for a trip. In effect, this is one definition of what we mean by average speed.

We can rewrite the previous numerical example to show explicitly how the distance for that trip is related to the average speed for the trip. We start as we did before, then rearrange terms slightly by writing 0.5 hours as (1 hr/2):

\begin{equation} (30\,\rm{km/hr})(0.5\,\rm{hr})+(90\,\rm{km/hr})(0.5\,\rm{hr})=\left (\frac{30\,\rm{km/hr}+ 90\,\rm{km/hr}}{2}\right ) (1\,\rm{hr}) = 60\,\rm{km} \end{equation}

The term in the big parentheses is by definition the average speed for the trip in our example. Multiplying by the total time for the trip gives us the distance traveled during the trip.

This example was simple because there were only two speeds traveled during the trip. However, even if the speed constantly changes we can use this notion of average speed to deduce how far an object moves in a given time. If the object accelerates at a constant rate, then we can find a general expression for the distance it travels, as we now show.

First, imagine an object that starts from rest and accelerates at a constant rate. We do not have to start from rest, but it makes the argument a little simpler to follow: adding an initial speed would shift the distance up or down, but it would not change the outline of the argument we present.

For the sake of argument, imagine that the object accelerates at 1 m/s/s, for 10 seconds. How fast will this object be moving at the end of the 10 seconds? Well, after one second it is moving 1 m/s, after 2 seconds it is moving 2 m/s, and so on. After 10 seconds it will be moving 10 m/s. If we plot its speed against time, it will look like this:

Figure B.8.1. Graph shows velocity (vertical) vs. time (horizontal) for an object with constant acceleration.

The average speed for the trip is right in the middle: 5 m/s. Notice how the object spends as much time above its average speed as below it because the speed of the particle is symmetric about its average. Under these circumstances the average can be computed simply from the mean of the initial and final speeds. Check that this formula works for the example above.

\begin{equation} \bar{v}=\frac{v_0+v_f}{2} \end{equation}

But the final speed (v_f) is just the initial speed (v₀), plus whatever change in speed is due to the object’s acceleration, v_f = v₀ + at. So we can rewrite the last expression as follows:

\begin{equation} \bar{v}=\frac{v_0+(v_0+at)}{2} \end{equation}

\begin{equation} \bar{v}=\frac{v_o+v_o+at}{2} \end{equation}

And if we simplify this expression, we get

\begin{equation} \bar{v}=v_0+\frac{1}{2}at \end{equation}

Now we can substitute this into our earlier expression for the distance in terms of the average speed.

\begin{equation} d=\left(v_0+\frac{1}{2}at\right)t \end{equation}

This leads to the following expression for the distance traveled by an object that accelerates at a constant rate:

\begin{equation} d=v_0t+\frac{1}{2}at^2 \end{equation}

We can use this expression to find how far an object moves when it starts from some initial speed and accelerates at a constant rate. An example would be an object falling under the influence of gravity near Earth’s surface, where the gravitational acceleration is essentially constant.