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The kinematic equations

Learn what the kinematic equations are and how you can use them.

What are the kinematic equations?

The kinematic equations relate the five kinematic variables listed below.

Δ x Displacement

t Time interval

v_{0} Initial velocity

v Final velocity

a Constant acceleration

The

t

in these formulas really represents the interval of time it took the object to be displaced

Δ x

. It would probably be conceptually clearer to write the time interval as

Δ t

but the kinematic formulas are already so cluttered that people usually leave off the

Δ

t

for the sake of cleanliness.

We just have to make sure we know that by writing

t

, we really mean the time interval

Δ t

If we know three of these five kinematic variables for an object undergoing constant acceleration, we can use a kinematic equation to solve for one of the unknown variables.

The kinematic equations are listed below.

1. v = v_{0} + a t

2. Δ x = (\frac{v + v_{0}}{2}) t

3. Δ x = v_{0} t + \frac{1}{2} a t^{2}

4. v^{2} = v_{0}^{2} + 2 a Δ x

Since the kinematic equations 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 equations assume all variables are referring to the same dimension: horizontal

(x)

, vertical

(y)

, etc.

If you plug in a horizontal initial velocity

v_{0 x}

into a kinematic equation, the rest of the variables you plug into that equation should also be for the horizontal dimension.

If you plug in a vertical initial velocity

v_{0 y}

into a kinematic equation, the rest of the variables you plug into that formula should also be for the vertical dimension.

In other words, to be really clear, the kinematic formulas for a given dimension—for example,

x

—should really be written with a subscript to denote that dimension:

v_{x} = v_{0 x} + a_{x} t

Δ x = v_{0 x} t + \frac{1}{2} a_{x} t^{2}

v_{x}^{2} = v_{0 x}^{2} + 2 a_{x} Δ x

\frac{v_{x} + v_{0 x}}{2} = \frac{Δ x}{t}

However, including all these subscripts can start to get messy. So, we often leave off the subscripts and just remember that the kinematic equations will work for any dimension in which the acceleration is constant, as long as you include only variables in that dimension.

What is a free falling object (a projectile)?

It might seem like the fact that the kinematic equations only work for time intervals of constant acceleration would severely limit the applicability of these equations. However one of the most common forms of motion, free fall, just happens to be constant acceleration.

All free falling objects—also called projectiles—on Earth, regardless of their mass, have a constant downward acceleration due to gravity of magnitude

g = 9.81 \frac{m}{s^{2}}

g = 9.81 \frac{m}{s^{2}} (Magnitude of acceleration due to gravity)

A free falling object is defined as any object that is accelerating only due to the influence of gravity. We often assume the effect of air resistance is small enough to ignore, in which case an object that is dropped, thrown, or otherwise falling freely has a constant downward acceleration of magnitude

g = 9.81 \frac{m}{s^{2}}

Note that

g = 9.81 \frac{m}{s^{2}}

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

(a_{y} = - 9.81 \frac{m}{s^{2}})

for a projectile when we plug into the kinematic equations.

How do you select and use a kinematic equation?

We choose the kinematic equation 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 equation.

For instance, say we knew a book on the ground was kicked forward with an initial velocity of

v_{0} = 5 \frac{m}{s}

, after which it took a time interval

t = 3 s

for the book to slide a displacement of

Δ x = 8 m

. We could use the kinematic equation

Δ x = v_{0} t + \frac{1}{2} a t^{2}

to algebraically solve for the unknown acceleration

a

of the book—assuming the acceleration was constant—since we know every other variable in the equation besides

a .

Problem solving tip: Note that each kinematic equation is missing one of the five kinematic variables—

Δ x, t, v_{0}, v, a

1. v = v_{0} + a t (This equation is missing Δ x .)

2. Δ x = (\frac{v + v_{0}}{2}) t (This equation is missing a .)

3. Δ x = v_{0} t + \frac{1}{2} a t^{2} (This equation is missing v .)

4. v^{2} = v_{0}^{2} + 2 a Δ x (This equation is missing t .)

To choose the kinematic equation 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 equation that does not include

v

at all. The kinematic equation

Δ x = v_{0} t + \frac{1}{2} a t^{2}

is missing

v

, so it's the right choice in this case to solve for the acceleration

a

Try it!

Example 1

A water balloon is dropped from the top of a very tall building. Assume air resistance is negligible. 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

v_{0} = 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.)

a_{g} = - 9.81 \frac{m}{s^{2}}

(This is implied since the water balloon is a free falling object.)

If you wait long enough, yes, but we won't use that in our equation for two reasons.

First, we are only considering the time interval of

2.35 s

, which is presumably less than the time required to reach the ground.

Second, we can only apply the kinematic equations for a time interval of constant acceleration. When the water balloon hits the ground the acceleration changes dramatically from the collision. So if we are going to claim that the acceleration in our kinematic equation is

a_{g} = - 9.81 \frac{m}{s^{2}}

since the object is in free fall, then we can't claim that the final velocity is zero since that is a portion of the trip when the object was colliding with the ground (i.e. not in free fall).

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

Δ y

and we weren't asked for the displacement

Δ y

, we'll use the first kinematic equation

v = v_{0} + a t

, which is missing

Δ y

v = v_{0} + a t (Use the first kinematic equation since it’s missing Δ y .)

v = 0 \frac{m}{s} + (- 9.81 \frac{m}{s^{2}}) (2.35 s) (Plug in known values.)

v = - 23.1 \frac{m}{s} (Calculate and celebrate!)

Note: The final velocity was negative since the water balloon is moving downward.

Example 2

A leopard is running at

6.20 \frac{m}{s} .

The leopard then speeds up to

23.1 \frac{m}{s}

in a time of

3.3 s

. How much ground did the leopard cover in going from

6.20 \frac{m}{s}

23.1 \frac{m}{s} ?

Assuming the initial direction of travel is the positive direction, our known variables are

v_{0} = 6.20 \frac{m}{s}

(The initial speed of the leopard)

v = 23.1 \frac{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 equation for the horizontal direction

Δ x = (\frac{v + v_{0}}{2}) t

, which is missing

a

Δ x = (\frac{v + v_{0}}{2}) t (Use the second kinematic equaiton since it’s missing a .)

Δ x = (\frac{23.1 \frac{m}{s} + 6.20 \frac{m}{s}}{2}) (3.30 s) (Plug in known values.)

Δ x = 48.3 m (Calculate and celebrate!)

Example 3

A student throws her pencil straight upward at

18.3 \frac{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

v_{0} = 18.3 \frac{m}{s}

(The initial upward velocity of the pencil)

Δ y = 12.2 m

(We want to know the time when the pencil moves through this displacement.)

a = - 9.81 \frac{m}{s^{2}}

(The pencil is a free falling 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 equation for the vertical direction

Δ y = v_{0 y} t + \frac{1}{2} a_{y} t^{2}

, which is missing

v

Δ y = v_{0 y} t + \frac{1}{2} a_{y} t^{2} (Start with the third kinematic equation.)

Normally we would just solve our expression algebraically for the variable we want to find, but this kinematic equation cannot 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 \frac{m}{s}) t + \frac{1}{2} (- 9.81 \frac{m}{s^{2}}) t^{2} (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 = \frac{1}{2} (- 9.81 \frac{m}{s^{2}}) t^{2} + (18.3 \frac{m}{s}) t - 12.2 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

a t^{2} + b t + c = 0

are found by using the quadratic formula

t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

. For our kinematic equation

a = \frac{1}{2} (- 9.81 \frac{m}{s^{2}})

b = 18.3 \frac{m}{s}

, and

c = - 12.2 m

So, plugging into the quadratic formula, we get

t = \frac{- 18.3 \frac{m}{s} \pm \sqrt{(18.3 \frac{m}{s})^{2} - 4 \frac{1}{2} (- 9.81 \frac{m}{s^{2}}) (- 12.2 m)}}{2 (\frac{1}{2} (- 9.81 \frac{m}{s^{2}}))}

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 answer 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

A motorcyclist starts with a speed of

23.4 \frac{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 \frac{m}{s^{2}}

. What is the new velocity of the motorcyclist after slowing down over

50.2 m ?

Assuming the initial direction of travel is the positive direction, our known variables are

v_{0} = 23.4 \frac{m}{s}

(The initial forward velocity of the motorcycle)

a = - 3.20 \frac{m}{s^{2}}

(Acceleration is negative since the motorcycle is slowing down and we assumed forward is positive.)

Δ x = 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 equation for the horizontal direction

v_{x}^{2} = v_{0 x}^{2} + 2 a_{x} Δ x

, which is missing

t

v_{x}^{2} = v_{0 x}^{2} + 2 a_{x} Δ x (Start with the fourth kinematic equation.)

v_{x} = \pm \sqrt{v_{0 x}^{2} + 2 a_{x} Δ x} (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

v_{x} = + \sqrt{v_{0 x}^{2} + 2 a_{x} Δ x}

Now we can plug in values to get

v_{x} = \sqrt{(23.4 \frac{m}{s})^{2} + 2 (- 3.20 \frac{m}{s^{2}}) (50.2 m)} (Plug in known values.)

v_{x} = 15.0 \frac{m}{s} (Calculate and celebrate!)

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Course: Physics library > Unit 1

The kinematic equations

What are the kinematic equations?

What is a free falling object (a projectile)?

How do you select and use a kinematic equation?

More problem solving tips

Try it!

Example 1

Example 2

Example 3

Example 4

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