PHY G7 Lab3 - Lab PDF

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Cart on a Ramp Lab Report

Part I 1.

Export, print, or sketch the three motion graphs. To view the acceleration vs. time graph,

change the y-axis of either graph to Acceleration. The graphs you have recorded are fairly complex and it is important to identify different regions of each graph. Record your answers directly on the printed or sketched graphs.

As we can observe from this velocity graph, the first segment is when the cart being pushed. And the second segment is when the cart rolling up freely and reached the top of it motion. Finally, the third segment is when the cart rolling down freely.

As we can observe from this acceleration graph, the first segment is when the cart being pushed. And the second segment is when the cart rolling up freely and reached the top of it motion. Finally, the third segment is when the cart rolling down freely.

As we can observe from this position graph, the first segment is when the cart rolling freely and moving up the incline. And the second segment is when rolling freely and moving down the incline.

On the velocity vs. time graph, just as the cart was released, the cart had its maximum velocity at 0.7s.

On the position vs. time graph, the highest point of the cart on the incline is 0.658m. And the velocity of the cart at the highest point is 0 m/s, and the acceleration is -0.57 m/s2

2. The motion of an object in constant acceleration is modeled by the equation x = ½ at2 + v0t + x0, where x is the position, a is the acceleration, t is time, and v0 is the initial velocity. This is a quadratic equation whose graph is a parabola. If the cart moved with constant acceleration while it was rolling, your graph of position vs. time will be parabolic

The above curve of best fit represents a range of time from 0.58 seconds to 3.54 seconds. This range is rather large, and to account for any imperfections, I will instead use a shorter range, which we see has nearly identical equations.

I will be using this equation instead, representative of a smaller range, but still fitting the curver almost perfectly. This is because the graph possesses the most quadratic properties towards the

vertex, and as it strays towards its limits, it loses its quadratic nature. In short, the smaller range should help represent the true quadratic portion as directed. The equation of best fit as calculated by Vernier Graphical Analysis (where y = position in meters and x = time in seconds) is: y = -0.2773x2 + 1.125x - 0.4835 The equation for position of an object with constant acceleration is: x = ½ at2 + v0t + x0 This suggests that the acceleration of our system is represented by the equation: ½ a = -0.2773 Finally, the calculated value for acceleration is -0.5546 m/s2.

The true representation of acceleration in the same interval of time (0.98s - 3.06s) is modelled by the linear equation a = 0.02556t - 0.6043. The slope of that equation is nearly zero, which suggests that the acceleration is nearly constant. Of course, there are imperfections, noise (as seen in the graph there are consistent raises and

drops in acceleration throughout the examined time), and factors such as friction that make it impossible to have a perfectly constant acceleration. Found in the middle of the interval (0.98 and 3.06) is 2.02 seconds. Using the equation a = 0.02556t - 0.6043, where time is 2.02s, we find acceleration to be approximately equal to -0.5527 m/s2. The calculated value for acceleration was -0.5546 m/s2. The difference in accelerations is around 0.3%. This difference is small enough to consider the acceleration constant throughout this interval, (where the graph for the position is quadratic). Again, there are factors that contribute to noise in the graph, but overall, the acceleration is relatively constant....