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## 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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# Solving for time

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)

Rate of change in position, or speed, is equal to distance traveled divided by time. To solve for time, divide the distance traveled by the rate. For example, if Cole drives his car 45 km per hour and travels a total of 225 km, then he traveled for 225/45 = 5 hours. Created by Sal Khan.

## Video transcript

Let's work through another
few scenarios involving displacement, velocity, and
time, or distance, rate, and time. So over here we have, Ben is
running at a constant velocity of three 3 meters per
second to the east. And just as a review,
this is a vector quantity. They're giving us the
magnitude and the direction. If they just said 3
meters per second, then that would just be speed. So this is the magnitude,
is 3 meters per second. And it is to the east. So they are giving
us the direction. So this is a vector quantity. And that's why it's
velocity instead of speed. How long will it take
him to travel 720 meters. So let's just remind
ourselves a few things. And I'll do it both with
the vector version of it. And maybe they should
say, how long will it take them to travel
720 meters to the east, to make sure, to make it clear,
that it is a vector quantity. So that it's displacement,
as opposed to just distance, but we'll do it both ways. So one way to think about it, if
we think about just the scalar version of it, we said
already that rate or speed is equal to the distance that
you travel over some time. I might write t there. But it's really
a change in time. So sometimes some people
would write a little triangle, a delta there, which
means change in time. But that's implicitly
meant when you just write over time like that. So rate or speed is equal
to distance divided by time. Now, if you know-- they're
giving us in this problem, they're giving us the rate. If we think about the
scalar part of it, they're telling us that
that is 3 meters per second. And they're also
telling us the time. Or, sorry, they're not
telling us the time. They are telling
us the distance, and they want us to
figure out the time. So they tell us the
distance is 720 meters. And so we just have to
figure out the time. we So if we just do the
scalar version of it, we're not dealing with
velocity and displacement. We're dealing with the
rate or speed and distance. So we have 3 meters
per second is equal to 720 meters over
some change in time. And so we can algebraically
manipulate this. We can multiply both
sides times time. Multiply time right over there. And then we could,
if we all-- well, let's just take it
one step at a time. So 3 meters per second times
time is equal to 720 meters because the times on the
right will cancel out right over there. And that makes sense,
at least units-wise, because time is going
to be in seconds, seconds cancel out the
seconds in the denominator, so you'll just get meters. So that just makes sense there. So if you want to
solve for time, you can divide both sides
by 3 meters per second. And then the left
side, they cancel out. On the right hand
side, this is going to be equal to 720
divided by 3 times meters. That's meters in the numerator. And you had meters per
second in the denominator. If you bring it out
to the numerator, you take the inverse of this. So that's meters-- let me do
the meters that was on top, let me do that in green. Let me color code it. So 720 meters. And now you're dividing
by meters per second. That's the same
thing as multiplying by the inverse, times
seconds per meters. And so what you're
going to get here, the meters are
going to cancel out, and you'll get 720
divided by 3 seconds. So what is that? 720 divided by 3. 72 divided by 3 is 24. So this is going to be 240. This part right over
here is going to be 240. And it's going to
be 240 seconds. That's the only unit
we're left with, and on the left hand side,
we just had the time. So the time is 240 seconds. Sometimes you'll see it. And just to show you,
in some physics classes, they'll show you
all these formulas. But one thing I really
want you to understand as we go through this
journey together, is that all of those
formulas are really just algebraic
manipulations of each other. So you really shouldn't
memorize any of them. You should always
say, hey, that's just manipulating one of those other
formulas that I got before. And even these formulas are,
hopefully, reasonably common sense. And so you can start from
very common sense things-- rate is distance
divided by time-- and then just
manipulate it to get other hopefully
common sense things. So we could have done it here. So we could have multiplied
both sides by time before we even put
in the variables, and you would have gotten--
So if you multiplied both sides by time here,
you would have got, on the right hand side, distance
is equal to time times rate, or rate times time. And this is one
of-- you'll often see this as kind of
the formula for rate, or the formula for motion. So if we flip it around, you get
distance is equal to rate times time. So these are all
saying the same things. And then if you wanted
to solve for time, you could divide
both sides by rate, and you get distance divided
by rate is equal to time. And that's exactly what we got. Distance divided by
rate was equal to time. So if your distance
is 720 meters, your rate is 3
meters per second, 720 meters divided by
3 meters per second will also give you a
time of 240 seconds. If we wanted to do
the exact same thing, but the vector version
of it, just the notation will look a little
bit different. And we want to keep track
of the actual direction. So we could say we
know that velocity-- and it is a vector quantity,
so I put a little arrow on top. Velocity is the same
thing as displacement. Let me pick a nice color
for displacement-- blue. As displacement-- Now, remember,
we use s for displacement. We don't want to
use d because when you start doing calculus,
especially vector calculus-- well, any
type of calculus-- you use d for the
derivative operator. If you don't know what that is,
don't worry about it right now. But this right here,
s is displacement. At least this is convention. You could kind of use
anything, but this is what most people use. So if you don't want
to get confused, or if you don't want to be
confused when they use s, it's good to practice with it. So it's the
displacement per time. So it's displacement
divided by time. Sometimes, once again,
you'll have displacement per change in time,
which is really a little bit more correct. But I'll just go with
the time right here because this is the convention
that you see, at least in most beginning physics books. So once again, if we
want to solve for time, you can multiply
both sides by time. And you get-- this cancels
out-- and I'll flip this around. Well, actually, I'll
leave it like this. So you get displacement
is equal to-- I can flip these around--
velocity times change in time, I should say. Or we could just say time
just to keep things simple. And if you want
to solve for time, you divide both
sides by velocity. And then that gives you time
is equal to displacement divided by velocity. And so we can apply that
to this right over here. Our displacement is
720 meters to the east. So in this case, our time
is equal to 720 meters to the east. 720 meters east is
our displacement, and we want to divide that
by the given velocity. Well, they give us the
velocity of 3 meters per second per the east. And once again, 720 divided
by 3 will give you 240. And then when you take
meters in the numerator, and you divide by
meters per second in the denominator, that's
the same thing as multiplying by seconds per meter,
those cancel out. And you are just left
with seconds here. One note I want to give you. In the last few problems, I've
been making vector quantities by saying to the
east, or going north. And what you're
going to see as we go into more complex problems--
and this is what you might see in typical physics
classes, or typical books, is that you define a convention. That maybe you'll say, the
positive direction, especially when we're just dealing
with one dimension, whether you can either
go forward or backwards, or left or right. We'll talk about other
vector quantities when we can move in two
or three dimensions. But they might take
some convention, like positive means maybe
you're moving to the east, and maybe negative means
you're moving to the west. And so that way-- well,
in the future, we'll see, the math will
produce the results that we see a little bit better. So this would just be
a positive 720 meters. This would be a positive
3 meters per second. And that implicitly tells
us that that's the east. If it was negative, it
would then be to the west. Something to think about. We're going to start exploring
that a little bit more in future videos. And maybe we might say positive
is up, negative is down, or who knows. There's different
ways to define it when you're dealing
in one dimension.