Instead of the acceleration being 9.81 m/s^2 as I hypothesized, it was actually 6.73 m/s^2 when calculated due to calculation and experimental errors. My hypothesis was only partially correct. I concluded that the velocity-time graph did go in a straight line. A random error that occurred was that one group member released the object, not in synchronization with the ticker timer which may have hindered the process of dots being recorded on the tape. The anticipated error was that the numbers would have been rounded and not completely accurate.Īnother systematic error is that the ruler is not 100% accurate when measuring it may be off by. used in teaching laboratories, like for example spark timer-based systems 20. One error might be in the calculations in Table 1. The acceleration of gravity was determined with different experimental. ErrorsĪfter performing the experiment, several errors were noted. We know that velocity changes with t = time and displacement changes with v = velocity. Given the formula vf = vi +at and d = (vf +vi/2). Due to a change in velocity, there was a change in displacement. This was because as time increases, velocity changed at a constant rate. The change in velocity of an object in free fall was directly proportional to the displacement. The velocity position graph (refer to graph 2) was curved because changing velocity means change in slope.ĥ. The slope of the velocity time graph was accelerating 6.74 m/s^2. The relationship between the change in velocity and elapsed time was that the velocity increases constantly as the time increases therefore they were proportionate.ģ. It was heading in a positive direction (refer to graph 1).Ģ. The velocity-time graph is in a straight line, meaning it was going at a constant acceleration. Therefore the acceleration throughout was 6.74 m/s^2 this may be caused by experimental errors and calculation errors.ġ. = 674 cm/s^2 (*note: To convert into meters, I divided (674 cm/s^2) by 100) From the reference spark, carefully measure the positions of all the. The time interval between sparks is then t 0:0333s. The spark timer is timed to spark 30 times per second. Label the posi-tion there as y 0 0:000m and the time associated with it as t 0 0:000s. Therefore at 0.2s, the average velocity was 92cm/s.Īcceleration was found by finding the slope of the velocity time graph. Select the rst clear and well dened spark as your reference spark. To find average velocity I did the following: One example calculation would be at 0.2 seconds, when displacement is 9.2cm. Position from the start (cm)Īverage velocity was found by dividing change in time by displacement. I measured and recorded the position from the start of the tape corresponding to each half time interval and then plotted a velocity-position graph using the velocities I recorded. I calculated the slope of the velocity-time graph, which gave me acceleration in cm/s^2.
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