# Obtaining position from a curved velocity vs time graph

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.

• Please show the whole question, including the graph which you are asking about. Sep 3, 2016 at 21:58
– blue
Sep 4, 2016 at 14:42

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...

• I really think my teacher wants us to use an equation - what happens if I plug the last position in as the x original? Would that work? Id really like to not resort to counting squares on a graph
– blue
Sep 3, 2016 at 12:59
• Any way you'd recommend involving an equation?
– blue
Sep 3, 2016 at 13:00
• Because velocity and acceleration vary however, I don't know how to find an equation please look at the edit above
– blue
Sep 4, 2016 at 14:42

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.

• Thank you - I am going by 10 sec intervals. So vi is the last velocity i.e. for 20 s it would be velocity at 10-velocity at 20s?
– blue
Sep 4, 2016 at 17:01
• $v_i$ is initial velocity, ie at $t=10s$. $v_f$ is final velocity, ie at $t=20s$. Sep 4, 2016 at 17:12
• Ok thank you, just checking so for 50s for example I have (-10+-11/2)*50 = -525m as position correct? As I am going at 10s, 20s, 30s, etc
– blue
Sep 4, 2016 at 17:24
• Not quite right. $t$ in your formula should be $\Delta t$ which is the time interval over which $v_i$ and $v_f$ are measured, not the absolute time at the end of that interval. So each calculation should use $\Delta t=10s$. For the interval $45-50s$ the displacement for this interval is $\Delta x=\frac12(-10-11)*10=-105m$. This needs to be added to the position at the end of the last interval. This gives you the cumulative displacement at $t=50s$. Your answers should be very roughly the same as in my table. Sep 4, 2016 at 17:36