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What are the kinematic formulas?

Here are the main equations you can use to analyze situations with constant acceleration.

What are the kinematic formulas?

The kinematic formulas are a set of formulas that relate the five kinematic variables listed below.
ΔxDisplacement
tTime interval 
v0  Initial velocity 
v   Final velocity 
a   Constant acceleration 
If we know three of these five kinematic variables—Δx,t,v0,v,a—for an object under constant acceleration, we can use a kinematic formula, see below, to solve for one of the unknown variables.
The kinematic formulas are often written as the following four equations.
1.v=v0+at
2.Δx=(v+v02)t
3.Δx=v0t+12at2
4.v2=v02+2aΔx
Since the kinematic formulas are only accurate if the acceleration is constant during the time interval considered, we have to be careful to not use them when the acceleration is changing. Also, the kinematic formulas assume all variables are referring to the same direction: horizontal x, vertical y, etc.

What is a freely flying object—i.e., a projectile?

It might seem like the fact that the kinematic formulas only work for time intervals of constant acceleration would severely limit the applicability of these formulas. However one of the most common forms of motion, free fall, just happens to be constant acceleration.
All freely flying objects—also called projectiles—on Earth, regardless of their mass, have a constant downward acceleration due to gravity of magnitude g=9.81ms2.
g=9.81ms2(Magnitude of acceleration due to gravity)
A freely flying object is defined as any object that is accelerating only due to the influence of gravity. We typically assume the effect of air resistance is small enough to ignore, which means any object that is dropped, thrown, or otherwise flying freely through the air is typically assumed to be a freely flying projectile with a constant downward acceleration of magnitude g=9.81ms2.
This is both strange and lucky if we think about it. It's strange since this means that a large boulder will accelerate downwards with the same acceleration as a small pebble, and if dropped from the same height, they would strike the ground at the same time.
It's lucky since we don't need to know the mass of the projectile when solving kinematic formulas since the freely flying object will have the same magnitude of acceleration, g=9.81ms2, no matter what mass it has—as long as air resistance is negligible.
Note that g=9.81ms2 is just the magnitude of the acceleration due to gravity. If upward is selected as positive, we must make the acceleration due to gravity negative ay=9.81ms2 for a projectile when we plug into the kinematic formulas.
Warning: Forgetting to include a negative sign is one of the most common sources of error when using kinematic formulas.

How do you select and use a kinematic formula?

We choose the kinematic formula that includes both the unknown variable we're looking for and three of the kinematic variables we already know. This way, we can solve for the unknown we want to find, which will be the only unknown in the formula.
For instance, say we knew a book on the ground was kicked forward with an initial velocity of v0=5 m/s, after which it took a time interval t=3 s for the book to slide a displacement of Δx=8 m. We could use the kinematic formula Δx=v0t+12at2 to algebraically solve for the unknown acceleration a of the book—assuming the acceleration was constant—since we know every other variable in the formula besides aΔx,v0,t.
Problem solving tip: Note that each kinematic formula is missing one of the five kinematic variables—Δx,t,v0,v,a.
1.v=v0+at(This formula is missing Δx.)
2.Δx=(v+v02)t(This formula is missing a.)
3.Δx=v0t+12at2(This formula is missing v.)
4.v2=v02+2aΔx(This formula is missing t.)
To choose the kinematic formula that's right for your problem, figure out which variable you are not given and not asked to find. For example, in the problem given above, the final velocity v of the book was neither given nor asked for, so we should choose a formula that does not include v at all. The kinematic formula Δx=v0t+12at2 is missing v, so it's the right choice in this case to solve for the acceleration a.

How do you derive the first kinematic formula, v=v0+at ?

This kinematic formula is probably the easiest to derive since it is really just a rearranged version of the definition of acceleration. We can start with the definition of acceleration,
a=ΔvΔt
Now we can replace Δv with the definition of change in velocity vv0.
a=vv0Δt
Finally if we just solve for v we get
v=v0+aΔt
And if we agree to just use t for Δt, this becomes the first kinematic formula.
v=v0+at

How do you derive the second kinematic formula, Δx=(v+v02)t?

A cool way to visually derive this kinematic formula is by considering the velocity graph for an object with constant acceleration—in other words, a constant slope—and starts with initial velocity v0 as seen in the graph below.
The area under any velocity graph gives the displacement Δx. So, the area under this velocity graph will be the displacement Δx of the object.
Δx= total area
We can conveniently break this area into a blue rectangle and a red triangle as seen in the graph above.
The height of the blue rectangle is v0 and the width is t, so the area of the blue rectangle is v0t.
The base of the red triangle is t and the height is vv0, so the area of the red triangle is 12t(vv0).
The total area will be the sum of the areas of the blue rectangle and the red triangle.
Δx=v0t+12t(vv0)
If we distribute the factor of 12t we get
Δx=v0t+12vt12v0t
We can simplify by combining the v0 terms to get
Δx=12vt+12v0t
And finally we can rewrite the right hand side to get the second kinematic formula.
Δx=(v+v02)t
This formula is interesting since if you divide both sides by t, you get Δxt=(v+v02). This shows that the average velocity Δxt equals the average of the final and initial velocities v+v02. However, this is only true assuming the acceleration is constant since we derived this formula from a velocity graph with constant slope/acceleration.

How do you derive the third kinematic formula, Δx=v0t+12at2?

There are a couple ways to derive the equation Δx=v0t+12at2. There's a cool geometric derivation and a less exciting plugging-and-chugging derivation. We'll do the cool geometric derivation first.
Consider an object that starts with a velocity v0 and maintains constant acceleration to a final velocity of v as seen in the graph below.
Since the area under a velocity graph gives the displacement Δx, each term on the right hand side of the formula Δx=v0t+12at2 represents an area in the graph above.
The term v0t represents the area of the blue rectangle since Arectangle=hw.
The term 12at2 represents the area of the red triangle since Atriangle=12bh.
That's it. The formula Δx=v0t+12at2 has to be true since the displacement must be given by the total area under the curve. We did assume the velocity graph was a nice diagonal line so that we could use the triangle formula, so this kinematic formula—like all the rest of the kinematic formulas—is only true under the assumption that the acceleration is constant.

Here's the alternative plugging-and-chugging derivation. The third kinematic formula can be derived by plugging in the first kinematic formula, v=v0+at, into the second kinematic formula, Δxt=v+v02.
If we start with second kinematic formula
Δxt=v+v02
and we use v=v0+at to plug in for v, we get
Δxt=(v0+at)+v02
We can expand the right hand side and get
Δxt=v02+at2+v02
Combining the v02 terms on the right hand side gives us
Δxt=v0+at2
And finally multiplying both sides by the time t gives us the third kinematic formula.
Δx=v0t+12at2
Again, we used other kinematic formulas, which have a requirement of constant acceleration, so this third kinematic formula is also only true under the assumption that the acceleration is constant.

How do you derive the fourth kinematic formula, v2=v02+2aΔx?

To derive the fourth kinematic formula, we'll start with the second kinematic formula:
Δx=(v+v02)t
We want to eliminate the time t from this formula. To do this, we'll solve the first kinematic formula, v=v0+at, for time to get t=vv0a. If we plug this expression for time t into the second kinematic formula we'll get
Δx=(v+v02)(vv0a)
Multiplying the fractions on the right hand side gives
Δx=(v2v022a)
And now solving for v2 we get the fourth kinematic formula.
v2=v02+2aΔx

What's confusing about the kinematic formulas?

People often forget that the kinematic formulas are only true assuming the acceleration is constant during the time interval considered.
Sometimes a known variable will not be explicitly given in a problem, but rather implied with codewords. For instance, "starts from rest" means v0=0, "dropped" often means v0=0, and "comes to a stop" means v=0. Also, the magnitude of the acceleration due to gravity on all freely flying projectiles is assumed to be g=9.81ms2, so this acceleration will usually not be given explicitly in a problem but will just be implied for a freely flying object.
People forget that all the kinematic variables—Δx,vo,v,a—except for t can be negative. A missing negative sign is a very common source of error. If upward is assumed to be positive, then the acceleration due to gravity for a freely flying object must be negative: ag=9.81ms2.
The third kinematic formula, Δx=v0t+12at2, might require the use of the quadratic formula, see solved example 3 below.
People forget that even though you can choose any time interval during the constant acceleration, the kinematic variables you plug into a kinematic formula must be consistent with that time interval. In other words, the initial velocity v0 has to be the velocity of the object at the initial position and start of the time interval t. Similarly, the final velocity v must be the velocity at the final position and end of the time interval t being analyzed.

What do solved examples involving the kinematic formulas look like?

Example 1: First kinematic formula, v=v0+at

A water balloon filled with Kool-Aid is dropped from the top of a very tall building.
What is the velocity of the water balloon after falling for t=2.35 s?
Assuming upward is the positive direction, our known variables are
v0=0 (Since the water balloon was dropped, it started at rest.)
t=2.35 s (This is the time interval after which we want to find the velocity.)
ag=9.81ms2(This is implied since the water balloon is a freely falling object.)
The motion is vertical in this situation, so we'll use y as our position variable instead of x. The symbol we choose doesn't really matter as long as we're consistent, but people typically use y to indicate vertical motion.
Since we don't know the displacement Δy and we weren't asked for the displacement Δy, we'll use the first kinematic formula v=v0+at, which is missing Δy.
v=v0+at(Use the first kinematic formula since it’s missing Δy.)
v=0 m/s+(9.81ms2)(2.35 s)(Plug in known values.)
v=23.1 m/s(Calculate and celebrate!)
Note: The final velocity was negative since the water balloon was heading downward.

Example 2: Second kinematic formula, Δx=(v+v02)t

A leopard is running at 6.20 m/s and after seeing a mirage that's taken the form of an ice cream truck; the leopard then speeds up to 23.1 m/s in a time of 3.3 s.
How much ground did the leopard cover in going from 6.20 m/s to 23.1 m/s?
Assuming the initial direction of travel is the positive direction, our known variables are
v0=6.20 m/s (The initial speed of the leopard)
v=23.1 m/s (The final speed of the leopard)
t=3.30 s (The time it took for the leopard to speed up)
Since we do not know the acceleration a and were not asked for the acceleration, we'll use the second kinematic formula for the horizontal direction Δx=(v+v02)t, which is missing a.
Δx=(v+v02)t(Use the second kinematic formula since it’s missing a.)
Δx=(23.1 m/s+6.20 m/s2)(3.30 s)(Plug in known values.)
Δx=48.3 m(Calculate and celebrate!)

Example 3: Third kinematic formula, Δx=v0t+12at2

A student is fed up with doing her kinematic formula homework, so she throws her pencil straight upward at 18.3 m/s.
How long does it take the pencil to first reach a point 12.2 m higher than where it was thrown?
Assuming upward is the positive direction, our known variables are
v0=18.3 m/s (The initial upward velocity of the pencil)
Δy=12.2 m (We want to know the time when the pencil moves through this displacement.)
a=9.81 m s2 (The pencil is a freely flying projectile.)
Since we don't know the final velocity v and we weren't asked to find the final velocity, we will use the third kinematic formula for the vertical direction Δy=v0yt+12ayt2, which is missing v.
Δy=v0yt+12ayt2(Start with the third kinematic formula.)
Normally we would just solve our expression algebraically for the variable we want to find, but this kinematic formula can not be solved algebraically for time if none of the terms are zero. That's because when none of the terms are zero and t is the unknown variable, this equation becomes a quadratic equation. We can see this by plugging in known values.
12.2 m=(18.3 m/s)t+12(9.81 m s2)t2(Plug in known values.)
To put this into a more solvable form of the quadratic equation, we move everything onto one side of the equation. Subtracting 12.2 m from both sides we get
0=12(9.81 m s2)t2+(18.3 m/s)t12.2 m(Put it into the form of the quadratic equation.)
At this point, we solve the quadratic equation for time t. The solutions of a quadratic equation in the form of at2+bt+c=0 are found by using the quadratic formula t=b±b24ac2a. For our kinematic equation a=12(9.81 m s2), b=18.3 m/s, and c=12.2 m.
So, plugging into the quadratic formula, we get
t=18.3 m/s±(18.3 m/s)24[12(9.81 m s2)(12.2 m)]2[12(9.81 m s2)]
Since there is a plus or minus sign in the quadratic formula, we get two answers for the time t: one when using the + and one when using the . Solving the quadratic formula above gives these two times:
t=0.869 s and t=2.86 s
There are two positive solutions since there are two times when the pencil was 12.2 m high. The smaller time refers to the time required to go upward and first reach the displacement of 12.2 m high. The larger time refers to the time required to move upward, pass through 12.2 m high, reach a maximum height, and then fall back down to a point 12.2 m high.
So, to find the answer to our question of "How long does it take the pencil to first reach a point 12.2 m higher than where it was thrown?" we would choose the smaller time t=0.869 s.

Example 4: Fourth kinematic formula, v2=v02+2aΔx

A European motorcyclist starts with a speed of 23.4 m/s and, seeing traffic up ahead, decides to slow down over a length of 50.2 m with a constant deceleration of magnitude 3.20 m s2. Assume the motorcycle is moving forward for the entire trip.
What is the new velocity of the motorcyclist after slowing down through the 50.2 m?
Assuming the initial direction of travel is the positive direction, our known variables are
v0=23.4 m/s (The initial forward velocity of the motorcycle)
a=3.20 m s2 (Acceleration is negative since the motorcycle is slowing down and we assumed forward is positive.)
Δx=50.2 m (We want to know the velocity after the motorcycle moves through this displacement.)
Since we don't know the time t and we weren't asked to find the time, we will use the fourth kinematic formula for the horizontal direction vx2=v0x2+2axΔx, which is missing t.
vx2=v0x2+2axΔx(Start with the fourth kinematic formula.)
vx=±v0x2+2axΔx(Algebraically solve for the final velocity.)
Note that in taking a square root, you get two possible answers: positive or negative. Since our motorcyclist will still be going in the direction of motion it started with and we assumed that direction was positive, we'll choose the positive answer vx=+v0x2+2axΔx.
Now we can plug in values to get
vx=(23.4 m/s)2+2(3.20 m s2)(50.2 m)(Plug in known values.)
vx=15.0 m/s(Calculate and celebrate!)

Want to join the conversation?

  • starky ultimate style avatar for user lolchessru
    I understand that these equations are only for acceleration being constant. I looked ahead and I noticed that acceleration being constant is a lot of the content ahead. Will there be any equations where we can find the other variables (time, distance, etc) where the acceleration is not constant? And if so, what are those equations and how can we get them?
    (22 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user THE GEEQ
    How to derive equations of motion by using calculus ?
    (9 votes)
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    • starky ultimate style avatar for user mr.swedishfish
      We usually start with acceleration to derive the kinematic equations.

      We know that acceleration is approximately -9.8 m/s^2 (we're just going to use -9.8 so the math is easier) and we know that acceleration is the derivative of velocity, which is the derivative of position. We can use this knowledge (and our knowledge of integrals) to derive the kinematics equations.

      First, we need to establish that acceleration is represented by the equation a(t) = -9.8.

      Because velocity is the antiderivative of acceleration, that means that v'(t) = a(t) and v(t) = int[a(t)]. Simplifying the integral results in the equation v(t) = -9.8t + C_1, where C_1 is the initial velocity (in physics, this the initial velocity is v_0). This means that for every second, the velocity decreases by -9.8 m/s.

      To find the position equation, simply repeat this step with velocity. Position is the antiderivative of velocity, so that means that x'(t) = v(t) and x(t) = int[v(t)]. Simplifying the integral results in the position equation x(t) = -4.9t^2 + (C_1)t + C_2, where C_1 is the initial velocity and C_2 is the initial position (in physics, C_2 is usually represented by x_0).

      In order to make this equation more universal, the position equation can be generalized as x(t) = 1/2(at^2) + v_0 + x_0
      (14 votes)
  • leaf orange style avatar for user RAQ2016
    the gravity magnitude for the free fall is always -9.81 ?
    (7 votes)
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    • leaf green style avatar for user Mark Zwald
      Near the surface of the Earth, yes. Not other places. Pretty much all high school physics problems will assume the Earth's gravity will be constant near the surface of the Earth. In reality, the acceleration will get weaker the further from the surface you get, but accounting for this change makes the problems considerably more difficult. But the approximation of g as a constant 9.81 m/s² is a very good one as long as your distance from the Earth's surface is very small compared to the radius of the Earth.
      (23 votes)
  • spunky sam blue style avatar for user Arunabh Bhattacharya
    In example 3, where the pencil is being thrown upward, should g = +9.8 m/s^2 or -9.8 m/s^2 ? I thought it should be positive (upward), but here it is negative. Could someone explain this to me.
    (6 votes)
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    • blobby green style avatar for user Teacher Mackenzie (UK)
      For this situation (any most situations), any vector that point UP or to the RIGHT is taken as positive.

      Whilst the initial vertical VELOCITY vector is upwards and therefore positive, the force of graviy is always downwards, and therefore (F=ma) the acceleration is always downward and negative.


      (Note: when the velocity vector and acceleration vectors are in opposite direct this means that the object is slowing down. When the pencil starts to fall, the velocity and acceleration are in the same direction and so it is speeding up)

      ok??
      (10 votes)
  • old spice man green style avatar for user mazzapants140894
    Could someone please explain in a step by step fashion; how to solve for Vf in the first kinematic equation:
    a=vf-v0/delta t.
    I cannot seem to be able to wrap my mind around the manipulation.

    Thank you.
    (2 votes)
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  • winston default style avatar for user Shravani
    How can two objects with different masses experience the same acceleration (-9.8m/s^2)? I thought heavier things fall to the ground quicker.
    (3 votes)
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    • piceratops ultimate style avatar for user Hecretary Bird
      You're close, but not quite there. While it's true that there is more gravitational force acting on a heavier object, this doesn't correspond to an increase in acceleration. In fact, the opposite is true. A heavier object has more inertia, which is a resistance to a change in motion.
      If you look at Newton's law of universal gravitation, you see that the force of attraction between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. If we take this equation and frame it in terms of somebody standing on the earth, we get a force due to gravitational attraction that is the product of their masses divided by the square of the earth's radius.
      From here, we can take Newton's second law of motion, f = ma. Here, we are looking for the force on the person from the earth. So, the mass on the left side of the following equation would refer to the person, which we can arbitrarily call m2. We can plug this in for the attraction force, giving us this:
      m2 * a = G (m1 * m2) / (r^2)
      Canceling out m2, we get a definition for gravitational acceleration:
      a = G (M1) / R^2
      From here, we have to use measurements to help us. Using measurements of the earth's mass and radius, as well as Newton's constant of gravitation, we can determine that the average value of a on the Earth's surface is about 9.81 m/s^2

      Hope this helped!
      (8 votes)
  • aqualine ultimate style avatar for user Leo Michaels
    In example 4 when plugging in the formulas there were no step by steps on how they got the answer.What are they someone please...
    (6 votes)
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  • blobby green style avatar for user Studyhere
    why did they assume a positive value of vx=15.0m/s and not vx=-15.0m/s when taking the root in the final example?
    (2 votes)
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  • piceratops ultimate style avatar for user Saif Jeelani
    Why isn't the fifth kinematic formula known a lot(or like used) and also how do you derive it?
    (3 votes)
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    • piceratops ultimate style avatar for user Hecretary Bird
      The fifth kinematic formula is the one without initial velocity. It's not used a lot because initial velocity is something you can usually control in an experiment, and is more often a known value than final velocity, for instance. It would then be useful to have an equation that has initial velocity in it, because this minimizes the values that you have to find in order to solve what you're solving for.
      You can derive the fifth formula by solving the first kinematic formula for initial velocity, and then plugging in that definition into the third kinematic formula to eliminate initial velocity.
      (3 votes)
  • blobby green style avatar for user Kyle Delaney
    I've seen a different list of five kinetic variables with position as one of them. Is there one standard list of "the" five kinetic variables that I should know? Or is it more subjective and situational?
    (1 vote)
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    • female robot ada style avatar for user Rodrigo Campos
      I don't think you should look for such a list, try and understand the most important concepts like position, velocity (how position changes over time) and acceleration (how velocity changes over time). We can visually understand these three simply, you probably already have an accurate intuition of how each should look like. Position is merely an object's location in space; its also easy to notice when something has some sort of velocity; About acceleration, its noticeable whenever the object's velocity is changing somehow (becoming faster or changing direction).
      (7 votes)