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## Physics library

### Course: Physics library > Unit 1

Lesson 2: Displacement, velocity, and time- Intro to vectors and scalars
- Introduction to reference frames
- What is displacement?
- Calculating average velocity or speed
- Solving for time
- Displacement from time and velocity example
- Instantaneous speed and velocity
- What is velocity?
- Position vs. time graphs
- What are position vs. time graphs?
- Average velocity and average speed from graphs
- Instantaneous velocity and instantaneous speed from graphs

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# Displacement from time and velocity example

AP.PHYS:

INT‑3.A (EU)

, INT‑3.A.1 (EK)

, INT‑3.A.1.1 (LO)

, INT‑3.A.1.2 (LO)

, INT‑3.A.1.3 (LO)

Worked example of calculating displacement from time and velocity. Created by Sal Khan.

## Want to join the conversation?

- i still dont understand the displacement

can you help(73 votes)- The simplest way to remember what is the meaning of displacement is to just remember this phrase- "Displacement is the shortest distance between two points in space."

To illustrate, imagine a body which travels 4m north and turns right at 90 degrees to the east, and travels 3m. So the shortest distance or DISPLACEMENT is actually the hypotenuse, or 5m. Best of Luck!(45 votes)

- Why is s used for displacement, and not d with a half-arrow over it? How did it come about that we used s instead?(14 votes)
- The answer to this question is calculus. In calculus, Godfrey Wilhelm Leibniz used the notation d/dx to indicate when he was taking a derivative of an equation, as opposed to Isaac Newton who simply used a hash mark. (As it turns out, Leibniz's notation is much simpler to use in calculus when dealing with derivatives and integrals and the like.)

So, the reason s is used is so that when you start doing calculus in physics, you don't make a confuse the displacement of an object with taking a derivative of an object and get a really bad answer.

I hope that helped.(39 votes)

- I was taught that since the answer is "south" it should have a negative and a y-hat symbol. Is that wrong?(8 votes)
- Maybe you mean j-hat? Sal has videos on that, it's called engineering notation, and it's not wrong. Though actually I don't see any reason why you can't use y-hat, as long as you're consistent throughout the problem. It's just a name, after all.(11 votes)

- What's the difference between displacement and distance?(5 votes)
- The first answer was correct, displacement does need a quantity and direction. Displacement can be calculated by measuring the final distance away from a point, and then subtracting the initial distance. Displacement is key when determining velocity (which is also a vector). Velocity = displacement/time whereas speed is distance/time. If I walked to school, then i realized that I forgot my homework and ran back home (all of which took me 20 min. and I live 500 meters away from school), then my average velocity would be 0meters/20min. My average speed on the other-hand would be 1,000meters/20min.(14 votes)

- Why is it not -300 m? Or is it just 300 m because Sal includes that the direction is to the south? If he didn't mention that the direction is south would it be -300 m?(8 votes)
- Yes, it is just 300m because he used 'south' when describing her displacement. If there is no direction then only will you put a sign (this is to keep it a vector). Plus, there is no way of knowing which direction '-' is. Is it south, or is it west? Maybe it is forward or backward. Unless it states it in the problem, the safe bet would be to simply put the direction.(1 vote)

- At4:05, Sal cancels out the seconds... How does that work? How come we can cancel out the units? What about the corresponding numbers?(3 votes)
- Anything divided by itself is 1.

How many seconds are there in 1 second? 1.

(1 second)/(1 second) = 1

(13.7 seconds)/(second) = 13.7(5 votes)

- my teacher talked about m/s/s or m/s2 what does that mean?(2 votes)
- This is the unit for acceleration.

You can think of it like this. Speed is the change in distance over time, so its units are meters per second. Acceleration is the change in speed over time, so its units are (meters per second) per second. Written another way it is (m/s)/s = m/s^2.(6 votes)

- Hello,

Can I get an example of when "time" is not actually "change in time"? In every problem I can think of, "time" pretty much means "change in time".

Thanks a lot!(4 votes) - if we denote west as (-) and east as (+)

then what do we denote north and south with .

the above infor mation was from the video before this one

solving of time.

please tell!(2 votes)- Normally in a problem, you will be given only the forward and backward directions which are opposite to each other. This is
**One dimensional motion**(motion in a straight line). Usually, you will be given only opposite directions.(like east and west, north and south, up and down, etc.) When you are given east and west, you won't have north and south in your problem and vice-versa.

North, east are normally denoted as positive.

South, west are normally denoted as negative.

You usually won't get any problems having north, east or south, west mixed up. Ignore signs in such cases.(5 votes)

- At1:40in the video what does he mean by the delta?(3 votes)
- Delta is a symbol used in math that means "change in". Δt would mean the change in time. This means the difference between an initial time and a final time. However, since it's almost always easier to just think of the initial time as 0, we can replace Δt with just t, which represents the final time.

Δt = t_f - t_i = t_f - 0 = t(3 votes)

## Video transcript

Let's do one more
example dealing with displacement,
velocity, and time. So we have if Marcia travels for
1 minute at 5 meters per second to the south, how much
will she be displaced? Let me do it this way. We know that velocity
is equal to displacement divided by time. And it's really, once
again, it is change in time. But we'll just say time. That's implicitly
change in time. And if you manipulate
this a little bit, you really just multiply
both sides by time. You just multiply both
sides by the variable t. You get displacement. Because this cancels out. You get displacement. And I'll flip this around. What's on the right-hand
side, I'll write on the left. So you get displacement
is equal to time times velocity or
velocity times time. Is equal to velocity times
time or velocity times change in time. So over here, they're
asking us for displacement. They're asking us how much
did Marcia get displaced? And they're saying that
she travels for 1 minute. So this 1 minute right over
here, this is her time. Sometimes you could view
that as her change in time. Or it really is
her change in time. If it said 0 minutes on her
stopwatch when she started, at the end it'll say 1 minute. Or if it said 5
minutes, if maybe it was 3:05 when she
started, it would be 3:06 when she finished. So it was really
the change in time. Once again, I won't
write the delta there just because this is the way
you most frequently see it. But I want to tell
you that these are the same thing for the
purpose of this problem because sometimes you'll
see the delta there. So the 1 minute, so the t
right over here is 1 minute. At 5 meters per
second to the south. This right over here
is the velocity. They give us the magnitude,
which is 5 meters per second. Or you could say
that's the speed. And they also give us the
direction, to the south. So this right over here is 5
meters per second to the south. So we might just say, look, if
we want displacement, that's just going to be equal to 5
meters per second to the south times 1 minute. The problem here
is that when we're talking about
displacement, we're going to think about a magnitude
of how much it's moved. So it'll be a
distance of some kind. And some direction. We have our direction
here, but we don't want any
other units there. And if we just multiply
this over here, we have 1 minute over here. But we have seconds
in the denominator. You can't just cancel out
a minute and a second. So you can't just
say that you're going to get 5 and have
some weird thing here. So in order for it
to all work out, you have to either convert
the 5 meters per second to 5 meters per minute. Or let me phrase
that another way. You have to convert
the 5 meters per second to some amount of
meters per minute, not 5 meters per minute. It's going to be different. Or you convert the
1 minute to seconds. So at least in my
mind, it's easier to convert 1 minute to seconds. So let's do that. So this is the same thing. 1 minute times. And we want to get
rid of the minute. And the minute is essentially
in the numerator right now. We could put this over 1. But it's essentially
in the numerator. So we want to divide by minutes. And we want to
multiply by seconds. We want seconds
in the numerator. And so how many seconds
are there per minute? You have 60 seconds
for every 1 minute. And so you have a minute
cancelling out with the minutes. And so now you have
5 meters per second to the south times 60 seconds. This is now cool because you
have seconds and seconds. I wrote "sec" there,
but this is also sec. So now you have
seconds over seconds. Those cancel out. And so your
displacement is going to be equal to 5 times 60. And then your units
left are meters. All the time units have
cancelled out and then it's meters to the south. So meters to the south. And this is equal to 5 times
60 is 300 meters to the south. And we are done. That's how much she
has been displaced. If they just wanted
the distance, you could say that she
traveled 300 meters. Just that part. The magnitude of
the displacement, that is the distance
that she traveled.