Ramp 1 1.65 degree
This is the first ramp (not drawn to scale). The length of the long leg is 228.2 cm. To find the height (short leg), we took the height from the ramp on either side and then subtracted them. From there we went to find the angle of elevation. We took arc tan(6.2/228.2) and this came out to theta = 1.56 degrees.
Ramp 2 3.6 degree
This is the second ramp (not drawn to scale). This one had larger angle of elevation. We did the same thing to both sides of the ramp to find the height and it came out to be 14.3 cm. Then we took arc tan(14.3/228.2). Theta = 3.6 degrees.
Graph 1
This is the position verses time graph. It is a parabola because the car went up the ramp and then back down. The car went up the ramp where the velocity was decreasing, and then the car went back down the ramp where the velocity increased.
Graph 2
This is the velocity verses time graph. For this line, there are two different accelerations. This is because there are two different events happening, the car increasing in velocity and the car decreasing in velocity. It is linear because it shows the slope of the parabola from the position verses time graph.
Table 1.56 degree
Trial for 1.56 degree incline | a1 (motion up) m/s2 | a2 (motion down) m/s2 | Gexp m/s2 | % difference |
1 | 0.33 | 0.18 | 9.4 | 4.10% |
2 | 0.36 | 0.18 | 9.9 | 1.00% |
3 | 0.34 | 0.19 | 9.7 | 1.00% |
Table 3.6 degree
Trial for 3.6 degree incline | a1 (motion up) m/s2 | a2 (motion down) m/s2 | Gexp m/s2 | % difference |
1 | 0.69 | 0.5 | 9.5 | 3.30% |
2 | 0.69 | 0.54 | 9.8 | 0.00% |
3 | 0.72 | 0.52 | 9.9 | 1.00% |
Raychel,
ReplyDeleteNice start. Please include more labels on figures, etc. You need sample calculations. How did you get Gexp?
Also comment on errors in your conclusions. Why do you think your results are better with a bigger angle?
grade for now == s-
Let me know when you've fixed the above comments and I'll regrade.