2. The purpose of this lab was to find a numerical approach that allowed us to find the time and distance it takes for an object to stop. The exact question was,"Find how far the elephant goes before coming to rest."
3. In order to find the distance that it takes for the object in this problem to stop we were given some information about the problem. The information that was given to us was: V_initial= 25 m/s, Force= 8000 N, Mass = 6500 kg, and that B_burning rate = 20 kg/s. With this given information we were able to find the distance and time it took for the elephant to stop. We were able to do this two ways analytically, and numerically. The analytical way was by finding an equation for the acceleration and then integrating it too find the v(t) equation and the x(t) equation which allowed us to find the two thing we were looking for. The second way we were able to do it was by plugging in the information that was given to us into excel and using those numbers to find the acceleration, average acceleration, change in velocity, velocity, average velocity, change in displacement, and distance all in a certain time. Once we had this done we were able to check in our excel sheet for the time at which the velocity was closest to zero and find the distance that the elephant had traveled at that time.
4. First, we started off by creating some set variables whose values weren't going to be changing. These were going to be those initial values that were given to us and change in time. After that, we set a row that showed us what variable we were going to be calculating in each column. Our first column was going to calculate time, followed by acceleration, average acceleration, change in velocity, velocity, average velocity, change in displacement, and distance. We then calculated all of these by using a = F/(m-bt), a_avg = (a_f - a_i)/2, delta v = a_avg* delta t, v = v_i - delta v, v_avg = (v_f - v_i)/2, delta x = v_avg* delta t, and x = x_i + delta x. This allowed us to find the time and distance it took for the elephant to come to rest by looking at the the point where the velocity was closest to zero.
5. Using delta t = 1 we get that the elephant stops at 19 s and travels 248.4 m
Using delta t = 0.1 we get that the elephant stops at 19.6 s and travels 248.7 m
Using delta t = 0.05 we get that the elephant stops at 19.65 s and travels 248.7 m
With the analytical approach we had gotten t = 19.69 s and x = 248.7 m
In conclusion our excel sheet did a really good job in helping us find the time and distance it took for the elephant to come to rest. Our closest result gave us a time of 19.65 and a distance of 248.7. This result agreed with the answer we got with an analytical approach. The way one would be able to tell if our time interval was small enough is by looking at how much the distance changes by with the different time interval. In our case we knew that using a delta t of 0.1 was good enough because there were more than ten numbers times where the velocity was close to zero that gave us 248.7 m when the distance was rounded. If the elephant were to change and have an initial mass of 55000 kg, the burning rate was to change to 40 kg/s, and the force was to go up to 13000 N, the elephant would travel a distance of 164.0 m before coming to rest.
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