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An vs

Physics

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An vs. time graph plots acceleration values on the y-axis, and time values on the x-axis.

As an example, check out the following graph which shows a constant acceleration:

In this graph, the acceleration remains constant at 2.0 m/s/s for 10 seconds.

Now, let's consider how such an acceleration would affect an object's velocity and position. Acceleration is defined as a change in an object's velocity, so if an object is accelerating at 2 m/s/s, then every 1 second, the object will gain 2 m/s. After 1 second, the object will be moving at 2 m/s; after 2 seconds, it'll be moving at 4 m/s, and so on.

If we look at what the same object's velocity vs. time graph would be, it would look something like this:

Note that this graph shows a line which has a slope of ##2##. In fact, if you choose any point in time on a velocity vs. time graph, the slope at that point (derivative) will always be equal to the acceleration at that time.

So, if at some moment your acceleration is positive, then at that moment your velocity will be increasing. If your acceleration is negative, then your velocity will be decreasing. Finally, if your acceleration is zero, your velocity will not be changing, and it will remain constant.

If you want to know what this object's position vs. time graph will look like, you'll need to use the first equation of motion, which is:

##s(t) = 1/2 at^2 + v_i t + s_i##

##s(t)## gives the object's position at time ##t##, and ##v_i## is the object's starting velocity. Likewise, ##s_i## is the object's starting position.

Our graph above indicates a starting velocity of ##0##. We're really only interested in relative position in this example, so we'll use ##0## for that too. Plugging in ##2## for ##a##, we'll get:

##s(t) = 1/2 (2)t^2##

Simplified:

##s(t) = t^2##

Below is the graph of this function: