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# Worked example: Motion problems with derivatives

Finding (and interpreting) the velocity and acceleration given position as a function of time.

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• hmmm so if Speed is always the magnitude of the Velocity.....can it be said that Speed is always the absolute value of whatever the Velocity is?
• That is correct. However, a more rigorous way of saying it is the "modulus" instead of the "absolute value".
Velocity is a vector, which means it has both a magnitude and a direction, while speed is a scaler.
We call this modulus. The modulus of a vector is a positive number which is the measure of the length of the line segment representing that vector.
• At , can you please explain in more detail how can we get the particle's direction based on the velocity?
• Hi Jakub!

Velocity is a vector, which means it takes into account not only magnitude but direction. We can see this represented in velocity as it is defined as a change in position with regards to the origin, over time.

When the slope of a position over time graph is negative (the derivative is negative), we see that it is moving to the left (we usually define the right to be positive) in relation to the origin.

Hope this helps ;)

Please feel free to ask if anything is still unclear to you.
• How does distance play into all this? Like how would I find the distance travelled by the particle, using these same equations?
• If you want to find the displacement, you can subtract the final x from the starting x.

If you want to find the full length of the path, that's more challenging, and probably what you're asking for, so I'm going to show it. Let's do it from x = 0 to 3.

To do that, just like normal, we have to split the path up into when x is decreasing and when it's increasing. We can do that by finding each time the velocity dips above or below zero. Let's do just that:

v(t) = 3t^2 - 8t + 3 set equal to 0
t^2 - (8/3)t + 1 = 0

I'm gonna complete the square.

t^2 - (8/3)t + 16/9 - 7/9 = 0
(t - 4/3)^2 = 7/9
t - 4/3 = ±√(7/9)
t - 4/3 = (±√7)/3
t = (4 ± √7)/3

Now we know the t values where the velocity goes from increasing to decreasing or vice versa. if you put both t values in a calculator, you'll get 0.451 and 2.215, which are both in our range of 0 to 3. So, we have 3 areas to keep track of.

``<----------------------------------------> 0 to 0.451 | 0.451 to 2.215 | 2.215 to 3``

In each of these areas, we're guaranteed to be going in the same direction, so we don't have to worry anymore. Now we can just get the displacement in each of those and arrive at our answer.

0 to 0.451: x(0.451) - x(0) = -1.369 + 2 = 0.631
0.451 to 2.215: x(2.215) - x(0.451) = -4.113 + 1.369 = -2.744
2.215 to 3: x(3) - x(2.215) = -2 + 4.113 = 0.744

Since we just want to know the distance and not the direction, we can get rid of the negatives and add these distances up.

Distance traveled = 0.631 + 2.744 + 0.744 = 4.119

Ugh, why does everything I write end up being so long? Hope you stayed with me.
• That does not make any sense.
If speed is increasing or decreasing isn't that just acceleration? 1. Right?
So if the second derivative of position (aka acceleration) is positive doesn't that mean speed is increasing? 2. Right?
And cant speed increase in a positive or negative direction (aka positive/right or negative/left velocity)? 3. Right?
THUS, if velocity (1nd derivative) is negative and acceleration (2nd derivative) is positive. Doesn't that mean we are increase speed (aka accelerating) in a negative/left direction?

Just the different vs same signs comment between acceleration and velocity just completely through me off.
(1 vote)
• Your first three points are correct, but your conclusion is not. If velocity is negative, that means the object is moving in the negative direction (say, left).

If acceleration is also positive, that means the velocity is increasing. As a negative number increases, it gets closer to 0. So if we apply a constant, positive acceleration to an object moving in the negative direction, we would see it slow down, stop for an instant, then begin moving at ever-increasing speed in the positive direction.

This is what happens when you toss an object into the air. Gravity pulls constantly downward on the object, so we see it rise for a while, come to a brief stop, then begin moving downward again.
• I'm surprised no one has asked: why is x moving down "left" and moving up "right"? Like, in relation to what? I guess if I tilt my head to the left x is moving in those directions. Am I missing something?
• You are right that from a bystander's point of view the 𝑥-axis can be aligned in any direction, not necessarily left to right.

Therefore, if I were given this question on a test I would not answer that the particle is moving to the left, but rather that it is moving in the negative direction of the 𝑥-axis.
(1 vote)
• Wait a minute, I just realized something. please just hear me out.
So we can calculate the distance traveled by a particle by finding the area between velocity time graph because distance is velocity times time right? So if we were to know the equation of the velocity function with time as an input and somehow make a function from the velocity function such that our new function's derivative is the velocity function. So from definition, the derivative of the distance function is the velocity so our new function got to be the distance function of the velocity function right? So that means the area of the velocity time graph up to a time is equal to the distance function value at that point??
And if this true then it means we will be able find the area under EVERY DIFFERENTIABLE FUNCTION up to a point by just creating a new function whose derivative is our first function and calculating the value at that point?
• You're correct. Your observation is (half of) the fundamental theorem of calculus, that the area under a curve is described by the antiderivative of that function.

More exactly, if f(x) is differentiable, then for any constant a, ∫_a^x f'(t)dt=f(x).
• What if the velocity is 0 and the acceleration is a positive number both at t=2? Would the particle be speeding up, slowing down, or neither?
• If the velocity is 0 and the acceleration is positive, the magnitude of the particle's speed would be increasing so it is speeding up.
(1 vote)
• when we trying to find out whether an object is speeding up or slowing down, can we just find the derivative of absolute value of velocity function? if the derivative is positive, then the object is speeding up, if the derivative is negative, then the object is slowing down. is my assumption correct?
• Yes that is right. If derivative of the position function is > 0, velocity is increasing, and vice versa. Furthermore, to find if acceleration is increasing, you take the second derivative
• So what does the derivative of acceleration mean?
• derivative is just rate of change or in other words gradient.

So in this case derivative of acceleration does not mean anything as it is not clear what derivative is being taken with respect to i.e. what is the independent variable.