Physics Help Forum An unorthodox Speed vs Time to Speed vs Distance question
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 Dec 3rd 2017, 03:07 AM #1 Junior Member   Join Date: Dec 2017 Posts: 1 An unorthodox Speed vs Time to Speed vs Distance question Hi There, I'm doing some analysis work on a lab I ran recently for some of my undergrad thesis work.. and I think I've had a brain fart. I've got some recorded data which shows the speed of an object (in this case a towing carriage) over time. I need to get this speed-time plot to a speed-distance plot (speed y, distance x). It's a little trickier because I'm looking at varying linear acceleration cases, and what I've worked out so far doesn't seem to be right.. or maybe it is.. I've included some snippets of what I'm talking about. Am I wrong in assuming that because the acceleration over time is linear.. the acceleration over distance will be quadratic? Would anyone be able to point me in the right direction please? Keep in mind this is done using filtered raw data in Excel. Any help would be greatly appreciated!! Cheers, Attached Thumbnails     Last edited by kegz91; Dec 3rd 2017 at 03:09 AM.
 Dec 3rd 2017, 03:56 AM #2 Senior Member   Join Date: Aug 2010 Posts: 336 Acceleration is, of course, the derivative of speed with respect to time: $\displaystyle a= \frac{dv}{dt}$. If you have a as a function of time, say, a= f(t), then you can get to v by integrating: $\displaystyle v= \int a dt= \int f(t)dt$. Given a graph or table of values of a, you can use numerical methods, such as Riemann sums, the trapezoid method, or Simpson's rule.
 Dec 6th 2017, 08:03 AM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,287 From the first plot is appears that velocity is essentially a linear function of time, (i.e. has constant slope with respect to time), and hence acceleration is constant. So this is pretty straight forward: using standard velocity versus distance formula for constant acceleration: $\displaystyle v^2(x)-v_i^2 = 2 a x$ where v_i is initial velocity and a = acceleration. For the initial time period of constant acceleration v_i = 0 and a is equal to the slope of the plot of velocity versus time. Total distance covered in that first phase can be determined by solving for x using the final velocity for v(x), which you can read off the graph. Then you can plot velocity versus distance by rearranging the above equation into: v(x) = sqrt {2 a x} for the range of velocity from v=0 to v= v_f.
 Dec 7th 2017, 02:29 AM #4 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 194 Since $\displaystyle v = \frac{dx}{dt}$ then $\displaystyle \int dx = \int v(t) dt$ $\displaystyle x = \int v(t) dt$ So take the speed curve you have and integrate it with respect to time to get the distance curve. If you have data instead of a function, you can approximate the velocity curve as a set of piece-wise trapezoids and then work out the area of each trapezoid under the curve, giving you the distance travelled under each trapezoid. Then, you add up the values in a cumulative manner to get the total distance travelled as a curve with as many points as you have trapezoids. Be sure to position the x-points correctly... For example, consider a velocity curve: v(t) = [0.2, 0.4, 0.6, 0.8] t = [0, 1, 2, 3] The distances travelled over each little time duration can be calculated as: $\displaystyle \Delta x = \frac{1}{2}\Delta t (v_1 + v_0)$ where $\displaystyle \Delta t = 1$ second Therefore $\displaystyle \Delta x(1/2) = \frac{1}{2} \times 1 \times (0.2 + 0.4) = 0.3$ $\displaystyle \Delta x(3/2) = \frac{1}{2} \times 1 \times (0.4 + 0.6) = 0.5$ $\displaystyle \Delta x(5/2) = \frac{1}{2} \times 1 \times (0.6 + 0.8) = 0.7$ Which finally gives $\displaystyle x(1/2) = 0 + 0.3 = 0.3$ $\displaystyle x(3/2) = 0.3 + 0.5 = 0.8$ $\displaystyle x(5/2) = 0.5 + 0.7 = 1.5$ EDIT: For the speed-distance plot you can work out the above and then plot velocity versus distance travelled. If you want to get the x-points over the same mesh as the v-points, you can use linear interpolation of the velocity or distance points. Last edited by benit13; Dec 7th 2017 at 08:01 AM.
Dec 13th 2017, 07:08 PM   #5
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 Originally Posted by kegz91 Hi There, I'm doing some analysis work on a lab I ran recently for some of my undergrad thesis work.. and I think I've had a brain fart. I've got some recorded data which shows the speed of an object (in this case a towing carriage) over time. I need to get this speed-time plot to a speed-distance plot (speed y, distance x). It's a little trickier because I'm looking at varying linear acceleration cases, and what I've worked out so far doesn't seem to be right.. or maybe it is.. I've included some snippets of what I'm talking about. Am I wrong in assuming that because the acceleration over time is linear.. the acceleration over distance will be quadratic? Would anyone be able to point me in the right direction please? Keep in mind this is done using filtered raw data in Excel. Any help would be greatly appreciated!! Cheers,
If the acceleration is not constant then there is no unique plot of speed verses distance. To see this note that for a non-constant acceleration, a(t) = dv/dt = f(t) = a function of time. To get x(t) you need to integrate that expression twice. Doing that will demonstrate that the plot that you're looking for will be dependent on the exact nature of f(t).

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