# What would the position vs. time graph look like if the velocity vs. time graph was a curve?

In my physics class we are learning about velocity vs. time graphs and position vs. time graphs for an object. We have learned that if the object is moving with constant velocity (this would be a diagonal line on the position vs. time graph) that the velocity vs. time graph would be a horizontal line. We have also learned that if the object is accelerating and is not moving with constant velocity (a curved position vs. time graph) that the velocity vs. time graph would be a diagonal line. However, what would the graph of the Position vs. Time graph look like if the Velocity vs. Time graph was a curve?

• @Craig That seems like an answer, rather than a comment. – rob Jan 14 at 23:28
• @Craig I know how to find the integral but not any details like that – Jodast Jan 14 at 23:41

The displacement $$s$$ between time $$t= t_1$$ and $$t=t_2$$ is the area under a velocity against time graph $$s = \int^{t_2}_{t_1}v \: dt$$ as shown below.
The reverse process is that the gradient of a displacement against time graph $$\frac{ds}{dt}$$ is equal to the velocity $$v$$.
Let’s say the speed was described by a curve that is itself described by some function of time $$v = f(t)$$. If you know what the initial position $$x_0$$ is, then the curve of position vs. time would be described by
$$x(t) =x_0 +\int_0^t f(t’)dt’$$
From a conceptual standpoint, you are calculating the total area below the curve $$f(t)$$ between the point where time is zero and some point $$t$$, adding the value of the initial position, and jotting down that value as the position of the system at time $$t$$. Doing this for all points in time give you the position curve. (Remember to count the areas below the speed equals zero axis as negative!)