What we are trying to do and accomplish in this lab is to first find the constant velocities of each buggy. Once that is found, we must use those velocities to calculate, from a certain distance apart from each other, where they would meet in centimeters. The distance would be measured centimeters.
Plan:
We decided we would first find the constant speed of the slow buggy and then the speed of them in comparison to the other buggy, the faster buggy. We would do this by starting each one on one side of a meter stick. Then we would see how long it would take in second for the slow buggy to reach the end of the meter stick also known as 100 centimeters. We would use our stopwatch on our phones to calculate the time taken. We would do the same with the faster buggy. Once we did that, we would find that the slower buggy had a constant speed of 18.58 cm/s We got that by taking the distance the buggy traveled, 100 cm, and divided it by the time it took the buggy to get to the end of the 100 cm. As seen on the white board at the end of the post, it took the slower buggy 5.38 seconds to go 100cm. Therefore we took the distance - 100cm and divided it by the time - 5.38 seconds. We did the same thing with the faster buggy which went 100 cm in 2.5 seconds. Therefore we did 100cm/2.5 seconds and found that that buggy would go 40 centimeters in one second. how did you get that? what was your time and how did you calculate the speed? and the constant speed of the faster buggy was calculated as 40 cm/s. We were given that the distance from each buggy would be 140 cm apart. After that we would graph both y=mx+b and find where on the graph the two lines would meet. The x value would then equal at what time in seconds the two would meet and the y value of the intercept would be where on the meter stick they would meet. good
can't see the picture for some reason?? try to fix these....
This would mean that the formula for the fast buggy would be moving at 40 cm per second and would be starting at a position of 0 cm. This would mean that to reach 100 cm would take this buggy 2.5 seconds. However, the end point would be 180 cm. This buggy would be starting at 0cm and moving in a positive direction, and since the opposite buggy is moving in a negative direction towards this buggy, they would meet at a certain point. confusing.....This does not show the intersection of the two points but gives the constant speed of the buggy. The formula being Position = (40 cm/s)time + 0
This would mean that the formula for the slow buggy would be moving at 18.58 cm per second and would be starting at a position of 140 cm. The starting position of the slower buggy would be 180 cm because it is starting on the opposite side of the meter sticks. Therefore, as opposed to the other buggy which starts at 0cm, this buggy would begin in the opposite direction, moving in a negative direction. is it 180 or 140????This would mean that to reach 100 cm, it would take a little less that 4 seconds. This graph does not show and intersection either however the formula being position = (18.57 cm/s)time + 180cm units...?
The graph above shows exactly where the two would intersect, at the point of (3.07,122.91). This would mean that if the buggies were started at the exact same time on either side of 140 cm, they would then intersect at 3.07 seconds and at 122.91 cm. This would mean that the fast buggy would have to travel a full 122.91 centimeters to intersect with the relatively slower constant speed of the other buggy that would only move 57.09 centimeters over the course of 3.07 seconds.good
After testing our theory of whether the buggies would meet at that exact point we found that the buggies did meet at the relative point within the same amount of time. As you can see from the photo below, they met at about 123 cm within the time of 3.71 seconds. This shows that our calculations we very close however our time was a bit off. To improve next time, we can test the constant speed of the buggies multiple times to have a more precise constant speed and therefore more accurate calculation.
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