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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:

table showing position and time values

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: enter image description here

I used formula

x = (vi+vf)/2 * t

, although Im sure thats wrong.

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  • $\begingroup$ Please show the whole question, including the graph which you are asking about. $\endgroup$ – sammy gerbil Sep 3 '16 at 21:58
  • $\begingroup$ Please see the graph $\endgroup$ – skyguy Sep 4 '16 at 14:42
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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 :

enter image description here

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

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  • $\begingroup$ 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? $\endgroup$ – skyguy Sep 4 '16 at 17:01
  • $\begingroup$ $v_i$ is initial velocity, ie at $t=10s$. $v_f$ is final velocity, ie at $t=20s$. $\endgroup$ – sammy gerbil Sep 4 '16 at 17:12
  • $\begingroup$ 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 $\endgroup$ – skyguy Sep 4 '16 at 17:24
  • $\begingroup$ 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. $\endgroup$ – sammy gerbil Sep 4 '16 at 17:36
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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...

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  • $\begingroup$ 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 $\endgroup$ – skyguy Sep 3 '16 at 12:59
  • $\begingroup$ Any way you'd recommend involving an equation? $\endgroup$ – skyguy Sep 3 '16 at 13:00
  • $\begingroup$ Because velocity and acceleration vary however, I don't know how to find an equation please look at the edit above $\endgroup$ – skyguy Sep 4 '16 at 14:42

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