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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.
delta, x, start text, D, i, s, p, l, a, c, e, m, e, n, t, end text
t, start text, T, i, m, e, space, i, n, t, e, r, v, a, l, end text, space
v, start subscript, 0, end subscript, space, space, start text, I, n, i, t, i, a, l, space, v, e, l, o, c, i, t, y, end text, space
v, space, space, space, start text, F, i, n, a, l, space, v, e, l, o, c, i, t, y, end text, space
a, space, space, start text, space, C, o, n, s, t, a, n, t, space, a, c, c, e, l, e, r, a, t, i, o, n, end text, space
If we know three of these five kinematic variables—delta, x, comma, t, comma, v, start subscript, 0, end subscript, comma, v, comma, 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, point, v, equals, v, start subscript, 0, end subscript, plus, a, t
2, point, delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t
3, point, delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared
4, point, v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x
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, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.
g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, start text, left parenthesis, M, a, g, n, i, t, u, d, e, space, o, f, space, a, c, c, e, l, e, r, a, t, i, o, n, space, d, u, e, space, t, o, space, g, r, a, v, i, t, y, right parenthesis, end text
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, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.
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, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, no matter what mass it has—as long as air resistance is negligible.
Note that g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction 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, start subscript, y, end subscript, equals, minus, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction 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 v, start subscript, 0, end subscript, equals, 5, start text, space, m, slash, s, end text, after which it took a time interval t, equals, 3, start text, space, s, end text for the book to slide a displacement of delta, x, equals, 8, start text, space, m, end text. We could use the kinematic formula delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared 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 adelta, x, comma, v, start subscript, 0, end subscript, comma, t.
Problem solving tip: Note that each kinematic formula is missing one of the five kinematic variables—delta, x, comma, t, comma, v, start subscript, 0, end subscript, comma, v, comma, a.
1, point, v, equals, v, start subscript, 0, end subscript, plus, a, t, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, delta, x, point, right parenthesis, end text
2, point, delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, a, point, right parenthesis, end text
3, point, delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, v, point, right parenthesis, end text
4, point, v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x, start text, left parenthesis, T, h, i, s, space, f, o, r, m, u, l, a, space, i, s, space, m, i, s, s, i, n, g, space, t, point, right parenthesis, end text
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 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared 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, $v=v_0+at$v, equals, v, start subscript, 0, end subscript, plus, a, t ?

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, equals, start fraction, delta, v, divided by, delta, t, end fraction $\quad$
Now we can replace delta, v with the definition of change in velocity v, minus, v, start subscript, 0, end subscript.
a, equals, start fraction, v, minus, v, start subscript, 0, end subscript, divided by, delta, t, end fraction
Finally if we just solve for v we get
v, equals, v, start subscript, 0, end subscript, plus, a, delta, t
And if we agree to just use t for delta, t, this becomes the first kinematic formula.
v, equals, v, start subscript, 0, end subscript, plus, a, t

## How do you derive the second kinematic formula, ${\Delta x}=(\dfrac{v+v_0}{2})t$delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, 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 v, start subscript, 0, end subscript as seen in the graph below.
The area under any velocity graph gives the displacement delta, x. So, the area under this velocity graph will be the displacement delta, x of the object.
delta, x, equals, start text, space, t, o, t, a, l, space, a, r, e, a, end text
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 v, start subscript, 0, end subscript and the width is t, so the area of the blue rectangle is v, start subscript, 0, end subscript, t.
The base of the red triangle is t and the height is v, minus, v, start subscript, 0, end subscript, so the area of the red triangle is start fraction, 1, divided by, 2, end fraction, t, left parenthesis, v, minus, v, start subscript, 0, end subscript, right parenthesis.
The total area will be the sum of the areas of the blue rectangle and the red triangle.
delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, t, left parenthesis, v, minus, v, start subscript, 0, end subscript, right parenthesis
If we distribute the factor of start fraction, 1, divided by, 2, end fraction, t we get
delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, v, t, minus, start fraction, 1, divided by, 2, end fraction, v, start subscript, 0, end subscript, t
We can simplify by combining the v, start subscript, 0, end subscript terms to get
delta, x, equals, start fraction, 1, divided by, 2, end fraction, v, t, plus, start fraction, 1, divided by, 2, end fraction, v, start subscript, 0, end subscript, t
And finally we can rewrite the right hand side to get the second kinematic formula.
delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t
This formula is interesting since if you divide both sides by t, you get start fraction, delta, x, divided by, t, end fraction, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis. This shows that the average velocity start fraction, delta, x, divided by, t, end fraction equals the average of the final and initial velocities start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction. 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, $\Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared?

There are a couple ways to derive the equation delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared. 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 v, start subscript, 0, end subscript 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 delta, x, each term on the right hand side of the formula delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared represents an area in the graph above.
The term v, start subscript, 0, end subscript, t represents the area of the blue rectangle since A, start subscript, r, e, c, t, a, n, g, l, e, end subscript, equals, h, w.
The term start fraction, 1, divided by, 2, end fraction, a, t, squared represents the area of the red triangle since A, start subscript, t, r, i, a, n, g, l, e, end subscript, equals, start fraction, 1, divided by, 2, end fraction, b, h.
That's it. The formula delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared 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, equals, v, start subscript, 0, end subscript, plus, a, t, into the second kinematic formula, start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction.
start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction
and we use v, equals, v, start subscript, 0, end subscript, plus, a, t to plug in for v, we get
start fraction, delta, x, divided by, t, end fraction, equals, start fraction, left parenthesis, v, start subscript, 0, end subscript, plus, a, t, right parenthesis, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction
We can expand the right hand side and get
start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction, plus, start fraction, a, t, divided by, 2, end fraction, plus, start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction
Combining the start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction terms on the right hand side gives us
start fraction, delta, x, divided by, t, end fraction, equals, v, start subscript, 0, end subscript, plus, start fraction, a, t, divided by, 2, end fraction
And finally multiplying both sides by the time t gives us the third kinematic formula.
delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared
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, $v^2=v_0^2+2a\Delta x$v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x?

To derive the fourth kinematic formula, we'll start with the second kinematic formula:
delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t
We want to eliminate the time t from this formula. To do this, we'll solve the first kinematic formula, v, equals, v, start subscript, 0, end subscript, plus, a, t, for time to get t, equals, start fraction, v, minus, v, start subscript, 0, end subscript, divided by, a, end fraction. If we plug this expression for time t into the second kinematic formula we'll get
delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, left parenthesis, start fraction, v, minus, v, start subscript, 0, end subscript, divided by, a, end fraction, right parenthesis
Multiplying the fractions on the right hand side gives
delta, x, equals, left parenthesis, start fraction, v, squared, minus, v, start subscript, 0, end subscript, squared, divided by, 2, a, end fraction, right parenthesis
And now solving for v, squared we get the fourth kinematic formula.
v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x

## 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 v, start subscript, 0, end subscript, equals, 0, "dropped" often means v, start subscript, 0, end subscript, equals, 0, and "comes to a stop" means v, equals, 0. Also, the magnitude of the acceleration due to gravity on all freely flying projectiles is assumed to be g, equals, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, 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—delta, x, comma, v, start subscript, o, end subscript, comma, v, comma, 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: a, start subscript, g, end subscript, equals, minus, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.
The third kinematic formula, delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared, 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 v, start subscript, 0, end subscript 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, $v=v_0+at$v, equals, v, start subscript, 0, end subscript, plus, a, t

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, equals, 2, point, 35, start text, space, s, end text?
Assuming upward is the positive direction, our known variables are
v, start subscript, 0, end subscript, equals, 0 (Since the water balloon was dropped, it started at rest.)
t, equals, 2, point, 35, start text, space, s, end text (This is the time interval after which we want to find the velocity.)
a, start subscript, g, end subscript, equals, minus, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction(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 delta, y and we weren't asked for the displacement delta, y, we'll use the first kinematic formula v, equals, v, start subscript, 0, end subscript, plus, a, t, which is missing delta, y.
v, equals, v, start subscript, 0, end subscript, plus, a, t, start text, left parenthesis, U, s, e, space, t, h, e, space, f, i, r, s, t, space, k, i, n, e, m, a, t, i, c, space, f, o, r, m, u, l, a, space, s, i, n, c, e, space, i, t, apostrophe, s, space, m, i, s, s, i, n, g, space, delta, y, point, right parenthesis, end text
v, equals, 0, start text, space, m, slash, s, end text, plus, left parenthesis, minus, 9, point, 81, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction, right parenthesis, left parenthesis, 2, point, 35, start text, space, s, end text, right parenthesis, start text, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis, end text
v, equals, minus, 23, point, 1, start text, space, m, slash, s, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, !, right parenthesis, end text
Note: The final velocity was negative since the water balloon was heading downward.

### Example 2: Second kinematic formula, ${\Delta x}=(\dfrac{v+v_0}{2})t$delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, 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
v, start subscript, 0, end subscript, equals, 6, point, 20, start text, space, m, slash, s, end text (The initial speed of the leopard)
v, equals, 23, point, 1, start text, space, m, slash, s, end text (The final speed of the leopard)
t, equals, 3, point, 30, start text, space, s, end text (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 delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t, which is missing a.
delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t, start text, left parenthesis, U, s, e, space, t, h, e, space, s, e, c, o, n, d, space, k, i, n, e, m, a, t, i, c, space, f, o, r, m, u, l, a, space, s, i, n, c, e, space, i, t, apostrophe, s, space, m, i, s, s, i, n, g, space, a, point, right parenthesis, end text
delta, x, equals, left parenthesis, start fraction, 23, point, 1, start text, space, m, slash, s, end text, plus, 6, point, 20, start text, space, m, slash, s, end text, divided by, 2, end fraction, right parenthesis, left parenthesis, 3, point, 30, start text, space, s, end text, right parenthesis, start text, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis, end text
delta, x, equals, 48, point, 3, start text, space, m, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, !, right parenthesis, end text

### Example 3: Third kinematic formula, $\Delta x=v_0 t+\dfrac{1}{2}at^2$delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared

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
v, start subscript, 0, end subscript, equals, 18, point, 3, start text, space, m, slash, s, end text (The initial upward velocity of the pencil)
delta, y, equals, 12, point, 2, start text, space, m, end text (We want to know the time when the pencil moves through this displacement.)
a, equals, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction (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 delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared, which is missing v.
delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared, start text, left parenthesis, S, t, a, r, t, space, w, i, t, h, space, t, h, e, space, t, h, i, r, d, space, k, i, n, e, m, a, t, i, c, space, f, o, r, m, u, l, a, point, right parenthesis, end text
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, point, 2, start text, space, m, end text, equals, left parenthesis, 18, point, 3, start text, space, m, slash, s, end text, right parenthesis, t, plus, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, t, squared, start text, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis, end text
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, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, t, squared, plus, left parenthesis, 18, point, 3, start text, space, m, slash, s, end text, right parenthesis, t, minus, 12, point, 2, start text, space, m, end text, start text, left parenthesis, P, u, t, space, i, t, space, i, n, t, o, space, t, h, e, space, f, o, r, m, space, o, f, space, t, h, e, space, q, u, a, d, r, a, t, i, c, space, e, q, u, a, t, i, o, n, point, right parenthesis, end text
At this point, we solve the quadratic equation for time t. The solutions of a quadratic equation in the form of a, t, squared, plus, b, t, plus, c, equals, 0 are found by using the quadratic formula t, equals, start fraction, minus, b, plus minus, square root of, b, squared, minus, 4, a, c, end square root, divided by, 2, a, end fraction. For our kinematic equation a, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, b, equals, 18, point, 3, start text, space, m, slash, s, end text, and c, equals, minus, 12, point, 2, start text, space, m, end text.
So, plugging into the quadratic formula, we get
t, equals, start fraction, minus, 18, point, 3, start text, space, m, slash, s, end text, plus minus, square root of, left parenthesis, 18, point, 3, start text, space, m, slash, s, end text, right parenthesis, squared, minus, 4, open bracket, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, left parenthesis, minus, 12, point, 2, start text, space, m, end text, right parenthesis, close bracket, end square root, divided by, 2, open bracket, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, close bracket, end fraction
Since there is a plus or minus sign in the quadratic formula, we get two answers for the time t: one when using the plus and one when using the minus. Solving the quadratic formula above gives these two times:
t, equals, 0, point, 869, start text, space, s, end text and t, equals, 2, point, 86, start text, space, s, end text
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, equals, 0, point, 869, start text, space, s, end text.

### Example 4: Fourth kinematic formula, $v^2=v_0^2+2a\Delta x$v, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x

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, point, 20, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction. 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
v, start subscript, 0, end subscript, equals, 23, point, 4, start text, space, m, slash, s, end text (The initial forward velocity of the motorcycle)
a, equals, minus, 3, point, 20, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction (Acceleration is negative since the motorcycle is slowing down and we assumed forward is positive.)
delta, x, equals, 50, point, 2, start text, space, m, end text (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 v, start subscript, x, end subscript, squared, equals, v, start subscript, 0, x, end subscript, squared, plus, 2, a, start subscript, x, end subscript, delta, x, which is missing t.
v, start subscript, x, end subscript, squared, equals, v, start subscript, 0, x, end subscript, squared, plus, 2, a, start subscript, x, end subscript, delta, x, start text, left parenthesis, S, t, a, r, t, space, w, i, t, h, space, t, h, e, space, f, o, u, r, t, h, space, k, i, n, e, m, a, t, i, c, space, f, o, r, m, u, l, a, point, right parenthesis, end text
v, start subscript, x, end subscript, equals, plus minus, square root of, v, start subscript, 0, x, end subscript, squared, plus, 2, a, start subscript, x, end subscript, delta, x, end square root, start text, left parenthesis, A, l, g, e, b, r, a, i, c, a, l, l, y, space, s, o, l, v, e, space, f, o, r, space, t, h, e, space, f, i, n, a, l, space, v, e, l, o, c, i, t, y, point, right parenthesis, end text
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, start subscript, x, end subscript, equals, plus, square root of, v, start subscript, 0, x, end subscript, squared, plus, 2, a, start subscript, x, end subscript, delta, x, end square root.
Now we can plug in values to get
v, start subscript, x, end subscript, equals, square root of, left parenthesis, 23, point, 4, start text, space, m, slash, s, end text, right parenthesis, squared, plus, 2, left parenthesis, minus, 3, point, 20, start fraction, start text, space, m, end text, divided by, start text, space, s, end text, squared, end fraction, right parenthesis, left parenthesis, 50, point, 2, start text, space, m, end text, right parenthesis, end square root, start text, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis, end text
v, start subscript, x, end subscript, equals, 15, point, 0, start text, space, m, slash, s, end text, start text, left parenthesis, C, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, !, right parenthesis, end text

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• 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?
• You will work with variable acceleration in calculus. You will learn how to do this when you do differential calculus. You will learn this when you apply derivatives.
• How to derive equations of motion by using calculus ?

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
• the gravity magnitude for the free fall is always -9.81 ?
• 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.
• 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.
• 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??
• 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.
• We want to get Vf = something, so we start by multiplying both sides of the equation:
A(delta t) = vf - v0.
The add v0 to both sides, and voila!
A(delta t) + v0 = vf
• Finally, I understood where these formulas came from. At school, I just memorized them without acknowledging their origins.
• 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...
• (final velocity) squared = some formula with known values

therefore

final velocity = positve or negative root of (that same formula with known values)
(1 vote)
• How can two objects with different masses experience the same acceleration (-9.8m/s^2)? I thought heavier things fall to the ground quicker.
• 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!