8-Mar-2017: Non-Constant acceleration

During the time we just learned motion, we know that acceleration is the change of velocity, and therefore, we can find out the velocity by the function ∆v = a • ∆t. It's easy because we can get the velocity by multiple the acceleration and time, but what if acceleration is changing from time to time? It is no longer fit this function. So how can we get the velocity by acceleration? The following lab tells a simple example to calculate the velocity and gives a way to solve them.

Purpose:

Solve a problem that its acceleration is changing all the time, and think a way to solve other acceleration changing problems.

The problem:

A 5000-kg elephant on frictionless roller skates is going 25m/s when it gets to the bottom of a hill and arrives on level ground. At that point a 1500-kg rocket mounted on 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

m(t) - 1500 kg - 20 kg/s • t

Find how far the elephant goes before coming to rest. 

Set up: 

Because during a very small time we could think acceleration's change is very small that we can ignore. Therefore the function ∆v = a • ∆t still works. Then in the same time, we can think the change velocity is also ignorable, which gives us the distance traveled. As a result, (By F = ma, ∆X = v∆t)

t              ∆a              a            ∆v              v            ∆X              X

0              0               a1             0              25             0                0

∆t           ∆a1          a1 +∆a1     a1• ∆t     25 - ∆v1     v1•∆t           ∆X1

2∆t         ∆a2          a2 +∆a2     a2• ∆t     25 - ∆v2     v2•∆t        X +∆X2

…            …             …          …              …           …               …

After we plug them into Excel, we get

Then, we found it's between 19s and 20s. But we can tell that the change within 1s can not be ignore, because it is pretty large. So we changed the ∆t to get some accurate value. 

In this two ∆t, four beginning digit number remain the same. Then we can say the ∆t is small enough, and the answer is 248.7 m.After this, we did a similar problem that is changed the value of mass (7000 kg in total), speed of burn (40 kg/s), and force (13000 N). We can still use the same way to solve it. 

 

Then, we found out the answer is 164.0 m.

Analyze:

This process is actually the same as integral, when we change ∆t into infinitely approaching to 0, we can get the function:

In first problem, because F = ma, m(t) = 1500 - 20t. a(t) = 8000/(6500 - 20t)
After we solved the integral, we could find out the time that makes v(t) = 0.
Then plug the number into X(t), we got Xf = 248.7 m, which is the same as what we got.
Then in similar way, we can get the answer of second problem.

Conclusion:
Today, we learned a way to solve the problems of changing acceleration, which is by dividing the change into many small part, and then find the sum of them. This idea is the same as integral. It means that the relationships among X, v, a can be written into  
                   
Moreover, the idea of integral can be used in many other value changing problems.