Question

At t=0 an object is stationary and experiencing an acceleration of -9.8m/s^2 due to gravity. At t=2 the object has an acceleration of -4m/s^2.

Find the function x(t).

Answer 1

First we have to find an function for the acceleration .
Then we use that to find the function of the position, x
Let's start with the acceleration.
How much did the acceleration change?

Answer 2

It changed by 5.8 m/s^2 in the positive direction.

Answer 3

\[\dfrac{\Delta a}{\Delta t}=\dfrac{5}{2}\dfrac{m}{s^3}\] I guess that would be good. I'm pretty sure we know that the object likely moved downwards.

Answer 4

x'(0) = 0
x''(0) = -9.8
x''(2) = -4

looks like free falling body with air resistance

Answer 5

Well really the problem is more general than I've led on to believe, the particle can be at any place. The only information we can know about the particle is what the acceleration is at different points in time and how much time has passed between points to predict the path of a particle. So for example, I might have something unpredictable like:

a(0)=9.8
a(1.2)=5
a(2)=8
a(3.3)=-4
etc...

Answer 6

And the initial location is also known.

Answer 7

You said "the initial location is also known". Can you type in what the initial location is?

Answer 8

Here's sort of a visual representation in 2 dimensions of what I imagine this might look like but I'm not sure. Each corner is a point where we know the acceleration.

@Miracrown unfortunately it won't really help much. I think the answer is supposed to be fairly general, like an iterative process.
\[\Large x(0)=x_0\]

Answer 9

Well, the problem is worded a little poorly I believe. The acceleration to gravity is always in the y direction, but it is asking us for the function x(t).

However, I think they want us to use x for the vertical direction in this problem.
Either way, let's focus on this actual problem and what it states.

Answer 10

As you said before, the acceleration increases by 5.8 m/s^2 over a time of 2 seconds.
That means it is going up by 2.9 m/s^2 every second, correct?

Answer 11

Yes, that's right. However at the next point the acceleration could be something completely different. This is partly supposed to be a programming problem which is why it might seem a little weird.

Answer 12

yeah...unfortunately, we can only assume things that are actually in the problem.
it does not tell us that the object's position. It also doesn't tell us that the acceleration is random, or changing. It only tells us the initial speed, and the acceleration at two times.

Answer 13

No, I am telling you that is what the actual problem says. I originally said what I said as a simplification since I thought it would be easier to answer for just one period of time.

Answer 14

Do you agree that the acceleration went up by 5.8 m/s^2?