4-Sept-2014 Non-Constant Acceleration Lab #3

For this activity we were given this scenario:

A 5000-kg elephant on frictionless roller skates is going 25 m/s when it gets to the bottom of a
hill and arrives on level ground. At that point a rocket mounted on the elephant’s back generates
a constant 8000 N thrust opposite the elephant’s direction of motion.
The mass of the rocket changes with time (due to burning the fuel at a rate of 20 kg/s) so that the
m(t) = 1500 kg – 20 kg/s·t.

Find how far the elephant goes before coming to rest.

We were given some suggestions to help us solve this equation:

1) Newtons second law gives us the acceleration of the elephant plus the rocket system as a function of time:

2) You can integrate from 0 to 1 to find delta v and then derive an equation for v(t):

3) You can integrate the velocity from 0 to 1t to find delta x and then derive an equation for x(t):

4)You can solve v(t) to find the time at which v=0
5)Then you can use the time you derived above in 4 and plug that into your expression for x(t) to find how far the elephant goes.

Here are all of the calculations done to get the distance 248.7m

Then once that is determined we had try to determine it through excel (numerically).

1. We first created a row of time which incremented by 0.1 seconds for 258 rows. 
2. With the next column we input the formula to calculate acceleration at any time.
3. In the third column we calculate the average velocity.
4. The fourth row calculates the change in velocity
5. The fifth row calculate the instance Velocity.
6. The sixth row calculates change in distance.
7. The last row calculate the instance distance.
8. With all the columns going down to the 258 row
   

This table is continued on until the 258th row.

To figure out the distance numerically we have to find the velocity right before it turns negative in the table. Then once we have this we change the time interval in the tables to 1 sec and 0.05sec in order to compare the differences.

1 second interval

0.05 second interval

Conclusion:

1) The information gathered doing the problem analytically and numerically are relatively the same. It is probably off by .3 or less.
2) The way I knew that the results were small enough was that the distance was pretty close to what I calculated. I knew when to stop when the velocity was close to zero before turning negative.