In the youtube comments, Steven F. Le Brun left a comment asking about how you derived the acceleration without knowing the initial velocity.

This got me thinking, and the problem is that when you are calculating the acceleration, you are measuring the time between b and d, and you would get a correct calculation if you knew the instantaneous velocity at point b and d. But the final velocity is not what you have measured; instead you have measured the average velocity over the period of time between a and b (velocity_1) and the average velocity between c and d (velocity_2).

I searched for some examples of calculations using a slotted card and a light gate, and although I found an example BBC Bitesize Measuring acceleration using a double mask, the description of the calculations donât discuss how the calculations are really done; it is assumed you will be using a data logger that does the calculations for you.

I also found Measuring Acceleration by XTronical, but it has the same calculation you are using, so I donât think it is correct either.

Because of acceleration, the velocity at point a and the velocity at b will not be identical. For others that want a quick review, this Khan Academy video is good Average velocity for constant acceleration, in fact the whole series is good.

If you measure the time difference between a and c, you will get a different value than the time between b and d, even though the widths of the cards are identical in your case. If there is a positive acceleration, the time between b and d will be less than the time between a and c.

To get the correct acceleration value (and assuming that acceleration is constant, as it would ideally be on an inclined plane), you need to save the time at a, b, c, and d. Then for acceleration time, use ((td-tb)+(tc-ta))/2. Then you will get the correct value for acceleration.

Edit: Although the previous equation will work, it is far from obvious why.

Here is a better explanation: Instead of using the timestamps at âbâ and âdâ, use the average of the timestamps at âaâ and âbâ (ta+tb)/2 and then subtract the average of the timestamps at âcâ and âdâ (tc+td)/2. Then you can also compute the acceleration between a and c, the acceleration between b and d and the acceleration between a and d, which should all be very close to each other (as long as acceleration is constant).

The following are the âequationsâ in the updated google spreadsheet.

acc_ac=(av_bc-av_ab)/((tb+tc)/2-(ta+tb)/2)

acc_bd=(av_cd-av_bc)/((tc+td)/2-(tb+tc)/2)

acc_ad=(av_cd-av_ab)/((tc+td)/2-(ta+tb)/2)

Note well: This is not the time at the midpoint of displacement between âaâ and âbâ (unless acceleration is zero). Finding the point where the instantaneous velocity is equal to the average velocity requires solving a quadratic equation, and it depends on the initial velocity and acceleration.

For some unknown reason, google sheets is adding an extra set of parenthesis (I think you call them brackets) around the denominator, but only on the last equation, so it gets changed to =(av_cd-av_ab)/**(**((tc+td)/2-(ta+tb)/2)**)**

I have a google sheet that allows you to enter values for acceleration, widths of first card, slot and second card, and then simulated the ideal values you would measure. Then using the âtimesâ and the widths, it then calculates the average velocities and acceleration. I sent a copy to the moderator. You can play with it to get an intuitive feel.

I didnât add charts or much fancy stuff to the sheet, just the caculations (which do use named cells which at least for me make the formulas mean more). In Google sheets you can also click on view â show formulas (or ctrl+` the back tick - the key to left of the 1 key), but if you do, you will probably need to expand the columns.

A graphic my son made to show what is happening as card moves from right to left and passes the lightgate at a, b, c, and d. In the spreadsheet time 0 is ta, âŚ time 3 is td.

Google spreadsheet screenshot with view formulas turned on

Here is the same sheet with view formulas turned off.

Note I entered the acceleration .9876543 so you can see it is recomputing the correct value. Also note that the card widths donât have to be the same size, but you will need to have the program know about the widths of at least the two cards. If you also specify the width of the slot, you can calculate accelerations over three different (but overlapping) periods, a-c, b-d, a-d (which should be very close in value). This is simulated, so there are no measurement jitter effects due to ISR latency, polling waits, IR response (although I would expect this to be very low especially if no bandpass filtering is being done by the IR receiver). In other words this is âperfectâ conditions, which canât be duplicated in the real world because of friction, measurement errors, etc.