Obtaining position from a curved velocity vs time graph

Question

I have just entered AP Physics and Im struggling with the following: I need to obtain position from a curved position vs time graph, i.e. the acceleration slope is not constant.

I first attempted to use the displacement formula $$x = v_0t + \frac{1}{2}at^2$$ where the initial velocity at time $0$ is $0\ \mathrm{m/s}$. First I knew I had to get instantaneous acceleration - never done that before but I drew tangent lines to the points at the different intervals and did: $$a = \frac{v_f-v_i}{t}$$ So far, this is what I've done. Acceleration is on the left side:

However this is not correct because on the 2nd interval at $20\ \mathrm{s}$ where velocity has just started going backwards, I got $-80\ \mathrm{m}$. It can't jump $450\ \mathrm{m}$ to $80\ \mathrm{m}$ in 10 seconds based on the graph.

My thinking is I'm not plugging in the right value for $v_0$, meaning original velocity. With $20\ \mathrm{m}$ should I be using the velocity from $10\ \mathrm{s}$?

EDIT: this is how I solved it and the graph:

I used formula

x = (vi+vf)/2 * t

, although Im sure thats wrong.

Answer 1

What you are doing wrong is using equations that apply only when the acceleration is constant to a situation where the acceleration is variable.

If you had a function that gave the velocity vs time, you could integrate that from $t_0$ to $t_{final}$.

EDIT: For example. suppose the velocity, $v$, as a function of time is given by:$$v=18-12t+0.1t^2$$Then the displacement, $d$, at a time $T$, is given by:$$d=\int_{0}^{T}{v}dt=\int_{0}^{T}{18-12t+0.1t^2}dt=18T-6T^2+\frac{0.1}{3}T^3$$

Given a graph, one solution is to plot the curve very carefully on some graph paper, and then count the squares between the velocity curve and the x-axis (the time axis). Remember that squares below the x-axis are negative...

Answer 2

The area between the x axis and the curve on a velocity-time graph represents displacement. When this area is above the x axis the displacement is +ve; when the area is below the x axis the displacement is -ve.

The last formula you quoted is correct for calculating this area :
$\Delta x \approx \frac12(v_i+v_f)\Delta t$.
This formula should be applied for each interval. Ideally you should aim to choose intervals over which the curve is approx. a straight line; then the formula is exact.

For each interval you have $\Delta t=5s$. For the 1st interval $(0-5s)$ you have $v_i=0m/s$ and $v_f=3m/s$ so then $\Delta x \approx \frac12(0+3)*5=7.5m$. For the 2nd interval $(5-10s)$ you have $v_i=3m/s$ and $v_f=9m/s$ so $\Delta x \approx \frac12(3+9)*5=30m$ giving a cumulative displacement of $37.5m$ at the end of $t=10s$.

Alternatively you can count the number of rectangles under the curve, estimating fractions. This is what I would do. You need to exercise care when the area becomes -ve; when velocity becomes -ve the area is above the curve.

For your graph I would use intervals $\Delta t$ of $5s$. The unit of area (au=displacement) is $5s \times 2m/s=10m$. The estimates which I make are :

In the 3rd column I have converted 1au to 10m and accumulated the distance - ie added the interval amounts to a running total.