blog 3

Nonconstant acceleration lab 

In this activity, we were asked to solve a kinematics problem, namely, how far an elephant will roll before coming to rest with an initial velocity and a force pushing it in the opposite direction of this velocity. With constant acceleration, this is a simple problem which the basic kinematics equations relating acceleration, velocity and position can solve easily. However, in this situation, acceleration is not constant, and so different methods must be used to find the solution.

The problem is as follows:

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.

As Newtons second law states, F=ma, or in a form more useful to our problem, a=F/m. The m (total mass) of the system is changing with time, because the rocket on the elephants back is burning fuel at a rate of 20 kg/s. In order to solve this problem, then, calculus must be used. Asking how far the elephant rolls before coming to a rest is essentially the same as asking at what position x will v=0. We know that v= integral(a) so we can set up the equation for v as follows:

v = v initial + integral(a) 

We are given that v initial = 25 m/s, in what I will take to be the positive direction, we are also given that a constant force of 8000 n is pushing the elephant in what is now to be considered the negative direction. We can model mass over time as m= (5000+1500-20t)kg or m=6500-20t, plugging these values into our equation we get: 

v = 25 - integal (from 0 to t) ( 8000/(6500-20t) ) dt

We can also call the part of this equation shaded in yellow delta v, which is solved as follows: 

This expression can be rearranged in order to solve for t by plugging in -25 for delta v, as shown:

This value, t=19.69075s can now be plugged into an equation for position, x. To get this equation, we integrate our formula for v as shown:

The first integral is simple because 25-400ln(325) is a calculable number, the other integral must be solved by more complex means, namely integration by parts.

This can now be solved by adding and subtracting 325 from the numerator and the denominator, as shown.

We then plug in our value for t obtained in the velocity equation: 

Which when calculated, gives us the value, x = 248.7 m, which is how far the elephant rolls from the bottom of the ramp before coming to a rest. 

This value can also be reached numerically, by plugging in values into a computer system like Microsoft excel. I set up my document as shown:

With t=time, a=acceleration at given t, a_avg=average acceleration, delta v=change in velocity, v=velocity at given t, delta x=change in position, and x=position at given t.

I set up t as increments of 0.1 seconds, and used a=-400/325-t from the calculations carried out above. I then set up a_average as corresponding cell in the "a" column + the cell above it divided by 2 or:

Delta v then, would be:

With delta v sorted, v is then: 

And so we can calculate x(t) as follows:

After setting this up, we can use this document to find the time and position at which the elephant stops by creating enough rows for v to equal 0. In this case, v=0 at some value between 19.6 and 19.7 seconds as shown:

 Giving us a position of between 248.692 and 248.698 m, which is fairly close to the value we achieved with integrals, that is x=248.7 m