12-Sept, 2016 . Lab #3: Non-Constant acceleration problem

Lab #3: Non-Constant acceleration problem

Dahlia - Ariel - Carlos

12-Sept-2016

The purpose of the lab#3 is to help students know how to use excel to calculate the distance of  an object can go (in a problem in which the acceleration of the object moving is non-constant and decreased over time.) or in other words, this lab will help students know how to calculate integration for kinematics by using excel.

The Problem:
A 5000-kg elephant on friction-less roller skates is going 25 m/s when its 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. 

Analysis:

From the question, I knew that F(net) = 8000 N

                                                  total mass = 5000 + 1500 = 6500 kg

So, from the Newton's law, F(net) = m*a(t)

                                                   a(t)=F(net)/m

                                                         = -8000/(6500-20t)

                                                         = -400/(325-t)

 Now, to find velocity, I need to integrate a(t) with the bounded from 0 to t.

after I had velocity, to find the distance, I need to integrate v(t) with the bounded from 0 to t.

And the result of the integration was:

x(t) =[25-400ln(325)]t + 400[(t-325)ln(325-t)-t+325ln(325)]

To find how far the elephant goes before coming to rest, I needed to find the time at which the velocity of the elephant was zero.

From ∆v = 400ln(325-t)-400ln(325)=-25

=>t=19.69s

Plug this time to the expression for x(t), I had x = 248.7m

Thus, I open a sheet of excel and use this to find the time when the velocity become zero.

From this table, 

1. Column A: Time begins at 0, so start by setting time increased by 1 seconds (= A8+$B$5)

2. Column B: Since acceleration is not constant, so B8 (= -400/(325-A8))

3. Column C: a_avg = (a1+a2)/2 ----------------------C9 (=(B8+B9)/2)

4. Column D: ∆v = a_avg*t --------------------------- D9 (= C9*$B$5)

5. Column E: v = vf + vo. ------------------------------E9 (=E8+D9)

6. Column F: v_avg = (vf + vo)/2 ---------------------F9 (=(E8+E9)/2)

7. Column G: ∆x = v_avg*t ----------------------------G9  (=F9*$B$5)

8. Column H: xf = xo + ∆x -----------------------------H9 (=H8+G9)

Then, I filled down all of these columns until I saw the value of velocity was zero.

and I got velocity became zero when t =19s - 20s, and ∆x = 248 m

To make the result became more accuracy, I changed the change of time become 0.1s

Now, I got the better value. Velocity became zero when t =19.6s, and ∆x = 248.69 m

Conclusion:

1. the result I got from doing the problem analytically and doing it numerically was the same. Velocity became zero when t =19.69 s, and ∆x = 248.7 m

2. From the problem, when I use t=1s, the result would not accuracy than 0.1s. From the above table, I could see the distance did not change so much in 19.6, 19.7, and 19.8s, that mean the time 0.1s was small enough.