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## Physics library

### Course: Physics library > Unit 1

Lesson 4: Kinematic formulas and projectile motion- Average velocity for constant acceleration
- Acceleration of aircraft carrier take-off
- Airbus A380 take-off distance
- Deriving displacement as a function of time, acceleration, and initial velocity
- Plotting projectile displacement, acceleration, and velocity
- Projectile height given time
- Deriving max projectile displacement given time
- Impact velocity from given height
- Viewing g as the value of Earth's gravitational field near the surface
- What are the kinematic formulas?
- Choosing kinematic equations
- Setting up problems with constant acceleration
- Kinematic formulas in one-dimension

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# Average velocity for constant acceleration

Investigate the relationships between velocity, acceleration, and distance in a physics context. Learn how to use formulas to calculate final velocity and total distance traveled given an initial velocity, constant acceleration, and time.
Created by Sal Khan.

## Want to join the conversation?

- Why do you need the area of the square? Since we are measuring the distance covered by the vehicle between the time that it was at velocities of 5 m/s and 13 m/s, don't we just use the triangle?

So wouldn't it be s=(v(f)-v(i))times time which is s=(8)4=32?

Please help.(116 votes)- No. Displacement is area under the velocity-time graph.

If you didn't count the square, you'd go 1/2*4*8 = 16 m (Area of a triangle).

If you counted the square only, it would show you how much displament you had covered if you go at a constant velocity of 5 m/s. This is (as shown) 20 m.

But in the first case you're accelerating right? So your velocity is going to increase, i.e its going to be more than 5m/s. Thus it is impossible for you to go only 16m, which is less than 20m(and not 32 as you had calculated).

The total displacement is 20m + 16m = 36m.(128 votes)

- in my physics book i have another formula of distance:

it's initial distance + intial velocity x time +1/2 acceleration x time^2

is this formula other way of writting the formula of sal?(19 votes)- The formula you listed in the first question is a derivation of the sum of areas of the rectangle and triangle. Here S is displacement, u is initial velocity, v is final velocity, A is acceleration and t is time.

D = ut + (1/2)(v-u)t

(Multiplying and dividing t)

D = ut + (1/2)[(v-u)/t]t.t

[(v-u)/t is Acceleration]

D = ut + (1/2)A(t^2)

This can be also written as D = Initial Velocity x Time + 1/2 Acceleration x Time^2(2 votes)

- What is an increasing acceleration called (i.e. ∆a/∆t)?(16 votes)
- yes !, jerk is also known as change in its inertial position of an object by any force. In jerk ( impulse) FORCE(F) is crucial. so, we knew that F = mass * acceleration. Thus

by this force is proportional to acceleration. so the above statement is justified.!(4 votes)

- At1:35why did sal use a triangle before the t?(4 votes)
- The ∆ means "change" and is frequently used in physics. I think the "t" stands for time, so what he was trying to say is "change in time" meaning how much time has passed for a certain process to complete or for a certain event to take place.(17 votes)

- At12:40, I'm having trouble understanding why the equation can also be called as the average velocity. I remember from the previous videos that average velocity is the displacement over time. How can (1/2)(change in t)(v(i)+v(f)) be the average velocity?(7 votes)
- Sal does not include ∆t in the formula for average velocity. Velocity = Displacement/Time. Average Velocity only equals Displacement/Time when the Velocity is constant. When Velocity is not constant, Average Velocity = Initial Velocity + Final Velocity/2( the average of V(i) and V(f) )(5 votes)

- Am I correct in assuming the logic still holds by using the area of a trapezoid, instead of triangle plus square?

1/2 x (a+b) x h

So that'd be:

1/2 x (vi+vf) x t, which is what you got in the video.(4 votes) - When Sal used 8m/s in the beginning and in triangle formula....Where did he get it from?(2 votes)
- Sal writes 8m/s for the first time at4:39in the video. He started with the formula:

vi + (delta t)(a) = vf

5 + (4s)(2 m/s^2) = vf

5 m/s + 8 m/s = vf

13 m/s = vf

So 8 m/s = (delta t)(a) = vf - vi = delta v

This 8 m/s is the change in velocity (v increased by 2 m/s every second, for 4 seconds).

That's why 8 m/s also became the height of the triangle, since the bottom of the triangle on the graph was at vi = 5 m/s, and the top of the triangle was at vf = 13 m/s. The graph (velocity vs time) is a straight line that rises with a constant slope of 2, because the velocity is increasing at a steady rate of 2 m/s (graph goes up 2) per second (graph goes right 1).(6 votes)

- I'm new here, sorry if I ask a stupid question.

When Sal calculates the distance in the second half of the video. Wouldn't it be right to call it the displacement? My reasoning is that since he is calculating with velocity (a vector) and not just speed, the result wood be a vector again.

(I am aware that in the given example distance and displacemet are the same).(2 votes)- Since, we are working with "
**constant acceleration**" (magnitude and direction of acceleration will not change) motion will be in a straight line. So, we can use "displacement" and "distance" interchangeably.(2 votes)

- Starting from6:02Sal breaks the integral of the velocity-time curve into two parts. But wouldn't it be easier if he used the formula A(trapezoid) = 1/2(side 1 + side 2)*height, which in this case would be 1/2(v1 + v2)*deltaT? One step versus two steps.(2 votes)
- Yes, but it is easier and more clear in his way, since in physics math is not the biggest challenge(4 votes)

- Do you also cancel units when you are multiplying? In this video, you cancel second (s) when you multiplied 4 seconds times 2 meters per second. I thought we only cancel units when we use division.(3 votes)
- You "cancel" units the same way you cancel numbers.

It's not really "cancelling". It's recognizing that n/n = 1.(2 votes)

## Video transcript

The goal of this video is to
explore some of the concepts, or formulas you might see in
a traditional physics class. But even more importantly to see
that they're really just common sense ideas. So let's just start
with a simple example. Let's say that-- and for
the sake of this video just so I don't
have to keep saying this is the magnitude
of the velocity, this is the direction of the
velocity, et cetera, et cetera. Let's just assume that if I
have a positive number that it means, for example, if I
have a positive velocity, it means that I am
going to the right. And let's say if I have a
negative number, which we won't see in this video, let's assume
that I'm going to the left. And that way I can just
write a number down, are we're only operating
in one dimension. You know that it
is specifying what the magnitude and
the direction is. If I said the velocity is
five meters per second, that means five meters
per second to the right. If I say it's negative
five meters per second, that means it's five meters
per second to the left. Now let's just say
just for simplicity, let's just say we start
with an initial velocity, of five meters per second. And once again, I am
specifying both the magnitude and the direction. Because of this convention here,
we know it is to the right. And let's say that we have
a constant acceleration. We have a constant
acceleration at play of two meters per
second, per second, or two meters per
second squared. And once again, since this is
positive, it is to the right. And let's say that we
do this for a duration. So my change in
time is- let's say we do this for a duration
of four-- I'll just use s. So I don't use seconds
and then sec, and then s in different places. So s for this video is seconds. So I want to do is think
about how far do we travel. Well there's two things. How fast are we going
after four seconds, and how far have we traveled
over the course of those four seconds. So let's draw ourselves a
little bit of a diagram here. So this is my velocity axis and
this over here is my time axis. I can draw a straighter
line than that. So that is my time axis. This is velocity. That is my velocity
right over there. And I'm starting off at
five meters per second. So this is five meters per
second right over here. So v i is equal to
five meters per second. And then every second
that goes by, I'm going two meters
per second faster. Because it's two meters
per second, per second. So every second that goes
by-- so after one second, I'm going to go two
meters per second faster. So I'll be at seven. Or another way to
think about it, is the slope of
this velocity line is my constant acceleration. Or I have a constant slope here. So it might look something
look something like that. So what has happened
after four seconds? So one, two, three, four. So this is my delta t. So my final velocity is
going to be right over there. I'll write it down
here just because it's getting in the way of
this word velocity. And so this is v. This
is my final velocity. And what would it be? Well I'm starting at
five meters per second. So I'm going to do it
both, using the variables, and using the concrete numbers. So starting at some
initial velocity. The subscript i
says i for initial. And then each second
that goes by, I'm getting this much faster. So if I want to see how
much faster have I gone, I multiply the
number of seconds-- I'll just multiply the
number of seconds that go by, times my acceleration. And once again, this
right here, I just wrote the subscript
c saying that it's a constant acceleration. And so that will tell
me how fast I've gone. If I start at this
point, and I multiply the duration times my
slope, I will get this high. I will get to my final velocity. And just to make it
clear with the numbers, these numbers really
could be anything, I'm just picking these to
make it concrete in your mind. You have five meters per
second, plus four seconds, plus-- I want to
do that in yellow-- plus four seconds times
our acceleration of two meters per second squared. And what is this
going to be equal to? You have the seconds canceling
out with one of the seconds down here. So let me write it. So we have five
meters per second, plus four times two is eight. This second's gone. We're just left with
meters per second. Eight meters per second. Or this is the same thing as
13 meters per second, which is going to be our
final velocity. And I want to take a pause here. And you can pause and
think about yourself. This hopefully
should be intuitive. We were starting at going
five meters per second. Every second that
goes by we're going to go two meters
per second faster. So after one second we will
be at seven meters per second. After two seconds we will be
at nine meters per second. After three seconds we will
be 11 meters per second. And then after four
seconds we will be at 13 meters per second. So you multiply how much time
passed times acceleration. This is how much faster,
we're going to be going. If we're already going
five meters per second. Five plus how much faster? 13 meters per second. So this right up here
is 13 meters per second. So, I'll take a
little pause here. Hopefully that's intuitive. And the whole point
of that is to show you that this formula that you'll
often see in many physics books-- It's not just something
that randomly popped out of the air-- it just makes
complete common sense. Now the next thing
I want to talk about is what is the total distance
that we would have traveled. Now we know from the last
video that distance is just the area under this
curve right over here. So it's just the area
under this curve. You say wait, this is kind of
a strange shape right here. How do I calculate its area? And we can just use a
little simple geometry to break it down into
two different areas, that are very easy to
calculate their areas. Or two simple shapes. You can break it down into
this blue part, this rectangle right over here. Easy to figure out the
area of a rectangle. And we could break it down
into this purple part. This triangle right here. Easy to figure out the
area of a triangle. And that will be the
total distance we travel. And even this will hopefully
make some intuition. Because this blue
area is how far would we have traveled if we
were not accelerating. If we just went five meters
per second, for four seconds. So if you go five meters
per second times-- so this is one second, two
second, three second, four second. So you're going from
times zero to time four, your change in time
is four seconds. So if you go five meters per
second, for four seconds, you're going to go 20 meters. This right here is 20 meters. That is the area
of this right here. Five times four. This purple or
magenta area tells you how further than
this, are you going, because you are
accelerating, because you kept going faster and
faster and faster. And it's pretty easy
to calculate this area. The base here is still
five, because that's five seconds have gone by. And what's the height here? The height here is
my final velocity minus my initial velocity. Or it's the change in velocity
due to the acceleration. And the change in velocity
due to the acceleration. 13 minus five is eight. Or it's this 8 right over here. It is eight meters per second. So this height right over here
is eight meters per second. The base right over
here is four seconds. That's the time that past. So what's the area
of this triangle? Area of a triangle is
one half, times the base, which is four seconds,
times the height, which is eight meters per second. Seconds cancel out. One half times four
is two, times eight is equal to 16 meters. So the total distance we
traveled is 20 plus 16, is 36 meters. That is the total, or I could
say the total displacement, and once again, it's to the
right, since it is positive. So this is our displacement. Now what I want to do is do
this exact same calculation, but keep it in variable
form and that'll give another formula that many
people, many often memorize. But I want you to
understand it it's a completely intuitive
formula, and it just comes out of the logical
flow of reasoning that we went through
in this video. What is the area? Once again, if we just
think about the variables. Well, the area of
this rectangle right here is our initial velocity,
times our change in time. So this is the blue
rectangle right over here. And then plus-- what
do we have to do-- we have the change
in time, once again, we have the change in
time, times this height. Which is our final velocity. Which is times
our final velocity minus our initial velocity. These are all vectors. They're just positive, telling
us we're going to the right. And we don't just multiply
the base times the height. That would give us the area
of this entire rectangle. We have to take it by half
because a triangle is only half of that rectangle. So times one half. And so this is the area. This is the purple area. That's not purple. This is the purple
area, right over here. This is the area of this. This is the area of that. And let's simplify
this a little bit. Let's factor out a delta t. So if you factor
out a delta t, you get delta t times, a
bunch of stuff, v sub i. So your initial velocity. We factored this out. Plus this stuff. Plus this thing right over here. And we can distribute
the one half. We factored the delta t's out. And let's multiply the one half
times each of these things. So it's going to be
plus, one half times v f, times our final velocity. That's not the right
color, let me actually do it in the right color so
you understand what I'm doing. So this is the one half. So plus one half times
our final velocity. Minus one half times
our initial velocity. I want to do that in blue. Sorry, I'm having trouble
changing colors today. Minus one half times
our initial velocity. And now what does
this simplify to? We have something plus one
half times something else, minus one half times
that original something. So what is v i minus
one half times v i. So anything minus
half of it, you're just going to have a
positive half left. So these two terms,
this term and this term will simplify to one half v i. One half the initial
velocity plus one half times the final velocity. And all of that is
being multiplied by our change in time. Or the time that has gone by. And this tells us the distance. The distance that we travelled. Or another way to
think about it-- let's factor out this one half. You get distance is equal
to change in time, times-- factoring out the one
half-- v i plus v f. No, that's not the right color. v i plus v f. So this is interesting. The distance we traveled
is equal to one half of the initial velocity
plus the final velocity. So this is really--
if you just took, this quantity right
over here is just the arithmetic-- I always
have trouble saying that word. It's the arithmetic mean
of these two numbers. And so I'm going to define
this as something new. I'm going to call this
the average velocity. But we have to be very
careful with this. This right here is
the average velocity. But the only reason
why I can just take the starting velocity and
the ending velocity and, adding them together, and
then divide by two. Essentially taking the
average of these two things which would be
someplace over here. And I take that as
the average velocity, is because my
acceleration is constant. Which is usually an assumption
in most introductory physics classes. But it's not always
the assumption. But if you do have a constant
acceleration like this, you can assume that
the average velocity is going to be the average
of the initial velocity and the final velocity. If this was a curve or if the
acceleration was changing, you could not do that. But what's useful
about this-- is if you want to figure out the
distance that was traveled you just need to know
the initial velocity and the final velocity. Average the two, and
then multiply that times the time that goes by. So in this situation,
our final velocity is 13 meters per second. Our initial velocity was
five meters per second. So you have 13 plus
5 is equal to 18. You divide that by 2. Your average velocity
is 9 meters per second, if you take the
average of 13 and 5. And then 9 meters per
second times 4 seconds gives you 36 meters. So hopefully that
doesn't confuse you. I just want to show you
where some of these things that you'll see in
your physics class, some of these formulas, why
you shouldn't memorize them, and they can all be deduced.