Knowing my acceleration, what is my position? – Week 10 – Lecture 11 – Mooculus
[music] Maybe I don’t know my position and
my velocity, but I know my acceleration. Maybe my acceleration at sometime t is
just a constant 8 units per second squared, right?
So, I’m going to accelerate at a constant rate.
Is that enough information to determine my velocity?
We have to think back to how velocity and acceleration are related.
My definition and acceleration is the change to the derivative in velocity,
right. Acceleration is rate of change in
velocity. Or in other words, right, velocity is an
anti-derivative of acceleration. So, I know that a of t is 8.
Let’s now solve for v. Okay, so a of t, my acceleration, is just
constant function 8, and v of t is the anti-derivative of my acceleration, right.
If I anti-differentiate acceleration, I’m going to get velocity.
It means I’m anti-differentiating 8. Well what anti-differentiates to 8 or 8t,
plus some constant C? It’s that constant again.
Right? Knowing my acceleration doesn’t determine
my velocity, it only determines my velocity up to some constant.
I could be going really fast or really slow, but still accelerating at the same
rate. Well, in any case, I’ve at least got a
formula for my velocity with that constant.
Now, can I use that formula to determine my position?
Same kind of game. My velocity is the rate in change in my
position, right? My velocity is the derivative of position.
And that means position is an anti-derivative of velocity.
So, let’s anti-differentiate. So, p of t is an anti-derivative of my
velocity, and I figured out my velocity a minute ago.
My velocity is 8t plus c. So, I want to anti-differentiate 8t plus
c. Well, it’s an anti-derivative of a sum, so
it’s the sum of the anti-derivatives. And what’s an anti-derivative for 8 times
something? Well, that’s a constant multiple, so it’s
8 times the anti-derivative of t, plus an anti-derivative for C.
Now, 8 times, what’s an anti-derivative for t?
Well one of them is t squared over 2. And what’s an anti-derivative for C?
Well, C times t is an anti-derivative for C.
Then, I should add some constant here. I’m going to call that constant big D.
Right. So, here’s my position.
I guess I could write this a little bit more nicely, coz the 8 and the dividing by
2 simplify that I could write it as 4t squared, plus Ct, plus D.
Well, that’s kind of weird, right? Why are there two constants in my answer?
Just knowing that my acceleration is 8 units per second squared, doesn’t tell me
my initial velocity, right? This quantity C here, is really v of 0.
Right? It’s my initial velocity.
And I could be accelerating at a rate of 8 units per second squared.
But starting with any of a range of possible initial velocities, right?
I could be going really fast at first or really slow at first.
But still, always accelerating at 8 units per second squared.
So, knowing this doesn’t nail down C. Similarly, just knowing my velocity
doesn’t nail down my initial position. Right?
This D here is really p of 0. Right?
If I plug in t equals 0, I just get D. And that means that D is really providing
my initial position. So, if I know my initial position and my
initial velocity and my acceleration, then I can nail down an exact formula for my
position. I can get rid of these mystery constants.
The important lesson to take away here is that anti-differentiating can actually be
useful. Anti-differentiating let’s us take, say,
velocity information and produce the position data.
Or even better, in this case, it let us take acceleration information, figure out
the velocity, and then, take that velocity information to figure out my position.