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 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

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??

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!

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

Why isn't the fifth kinematic formula known a lot(or like used) and also how do you derive it?

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.

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?

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).

Main content

Course: Mechanics (Essentials) - Class 11th > Unit 4

Lesson 9: Building equations to analyze motion

What are the kinematic equations?

Google Classroom

Learn where the kinematic equations come from, and how you can use them to analyze scenarios involving constant acceleration.

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 direction: 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 direction.

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 direction.

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

x

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

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 textbooks, courses, and professors often leave off the subscripts and just remember that the kinematic equations will work for any direction—even diagonal—in which the acceleration is constant, as long as you include only variables in that particular direction.

What is a free falling object—i.e., 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.

Warning: Forgetting to include a negative sign is one of the most common sources of error when using 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

Yes, there is.

5. Δ x = v t - \frac{1}{2} a t^{2} (This formula is missing v_{0} .)

The fifth kinematic equation looks just like the third kinematic equation

Δ x = v_{0} t + \frac{1}{2} a t^{2}

except with the initial velocity

v_{0}

replaced with final velocity

v

and the plus sign replaced with a minus sign. It can be derived by plugging the first kinematic equation into the third kinematic equation.

This fifth kinematic equation often isn't as useful since the initial velocity is typically known/given in many situations.

How do you derive the first kinematic equation, $v = v_{0} + a t$ ‍ ?

This kinematic equation is 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 = \frac{Δ v}{Δ t}

Yes. If the time interval

Δ t

is large, the formula

a = \frac{Δ v}{Δ t}

is the average acceleration over the time interval.

So, the derivation we are going through below really assumes that the

a

in the equation is the average acceleration

a_{a v g}

. However, if the acceleration is constant, the instantaneous acceleration

a_{i n s t}

will equal the average acceleration

a_{a v g}

, in which case we don't have to worry about being specific and can just call it

a

So, for the kinematic equation we will derive to be accurate, the acceleration should be constant. Or else the

a

in the equation is referring to the average acceleration and not the instantaneous acceleration.

Now we can replace

Δ v

with the definition of change in velocity

v - v_{0}

a = \frac{v - v_{0}}{Δ t}

Finally if we just solve for

v

we get

v = v_{0} + a Δ t

And if we agree to just use

t

for

Δ t

, this becomes the first kinematic equation.

v = v_{0} + a t

How do you derive the second kinematic equation, $Δ x = (\frac{v + v_{0}}{2}) t$ ‍?

A cool way to visually derive this kinematic equation is by considering the velocity graph for an object with constant acceleration—in other words, a constant slope—and starts with initial velocity

v_{0}

as seen in the graph below.

The area under any velocity graph gives the displacement $Δ x$ ‍. So, the area under this velocity graph will be the displacement

Δ x

of the object.

Δ x = 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

v_{0}

and the width is

t

, so the area of the blue rectangle is

v_{0} t

.
The base of the red triangle is

t

and the height is

v - v_{0}

, so the area of the red triangle is

\frac{1}{2} t (v - v_{0})

The total area will be the sum of the areas of the blue rectangle and the red triangle.

Δ x = v_{0} t + \frac{1}{2} t (v - v_{0})

If we distribute the factor of

\frac{1}{2} t

we get

Δ x = v_{0} t + \frac{1}{2} v t - \frac{1}{2} v_{0} t

We can simplify by combining the

v_{0}

terms to get

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

And finally we can rewrite the right hand side to get the second kinematic equation.

Δ x = (\frac{v + v_{0}}{2}) t

This equation is interesting since if you divide both sides by

t

, you get

\frac{Δ x}{t} = \frac{v + v_{0}}{2}

. This shows that the average velocity

\frac{Δ x}{t}

equals the average of the final and initial velocities

\frac{v + v_{0}}{2}

. However, this is only true assuming the acceleration is constant since we derived this equation from a velocity graph with constant slope/acceleration.

How do you derive the third kinematic equation, $Δ x = v_{0} t + \frac{1}{2} a t^{2}$ ‍?

There are a couple ways to derive the equation

Δ x = v_{0} t + \frac{1}{2} a t^{2}

. 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_{0}

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

Δ x

, each term on the right hand side of the equation

Δ x = v_{0} t + \frac{1}{2} a t^{2}

represents an area in the graph above.

The term

v_{0} t

represents the area of the blue rectangle since

A_{r e c t a n g l e} = h w

The term

\frac{1}{2} a t^{2}

represents the area of the red triangle since

A_{t r i a n g l e} = \frac{1}{2} b h

That's it. The equation

Δ x = v_{0} t + \frac{1}{2} a t^{2}

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 equation—like all the rest of the kinematic equations—is only true under the assumption that the acceleration is constant.

Here's the alternative plugging-and-chugging derivation. The third kinematic equation can be derived by plugging in the first kinematic equation,

v = v_{0} + a t

, into the second kinematic equation,

\frac{Δ x}{t} = \frac{v + v_{0}}{2}

If we start with second kinematic equation

\frac{Δ x}{t} = \frac{v + v_{0}}{2}

and we use

v = v_{0} + a t

to plug in for

v

, we get

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

We can expand the right hand side and get

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

Combining the

\frac{v_{0}}{2}

terms on the right hand side gives us

\frac{Δ x}{t} = v_{0} + \frac{a t}{2}

And finally multiplying both sides by the time

t

gives us the third kinematic equation.

Δ x = v_{0} t + \frac{1}{2} a t^{2}

Again, we used other kinematic equations, which have a requirement of constant acceleration, so this third kinematic equation is also only true under the assumption that the acceleration is constant.

How do you derive the fourth kinematic equation, $v^{2} = v_{0}^{2} + 2 a Δ x$ ‍?

To derive the fourth kinematic equation, we'll start with the second kinematic equation:

Δ x = (\frac{v + v_{0}}{2}) t

We want to eliminate the time

t

from this equation. To do this, we'll solve the first kinematic equation,

v = v_{0} + a t

, for time to get

t = \frac{v - v_{0}}{a}

. If we plug this expression for time

t

into the second kinematic equation we'll get

Δ x = (\frac{v + v_{0}}{2}) (\frac{v - v_{0}}{a})

Multiplying the fractions on the right hand side gives

Δ x = \frac{v^{2} - v_{0}^{2}}{2 a}

And now solving for

v^{2}

we get the fourth kinematic equation.

v^{2} = v_{0}^{2} + 2 a Δ x

What's confusing about the kinematic equations?

People often forget that the kinematic equations 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_{0} = 0

, "dropped" often means

v_{0} = 0

, and "comes to a stop" means

v = 0

. Also, the magnitude of the acceleration due to gravity on all free falling projectiles is assumed to be

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

, so this acceleration will usually not be given explicitly in a problem but will just be implied for a free falling object.

People forget that all the kinematic variables—

Δ x, v_{o}, 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 free falling object must be negative:

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

The third kinematic equation,

Δ x = v_{0} t + \frac{1}{2} a t^{2}

, 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 equation must be consistent with that time interval. In other words, the initial velocity

v_{0}

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 equations look like?

Example 1: First kinematic equation, $v = v_{0} + a t$ ‍

A water balloon 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

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) and we would be contradicting ourselves.

If we knew the acceleration upon impact, and if that acceleration was constant we could apply the kinematic equations during the time interval of impact. But that's not the question we're considering here, and in most free fall problems we'll only consider the time after release and the time before impact.

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 was heading downward.

Example 2: Second kinematic equation, $Δ x = (\frac{v + v_{0}}{2}) t$ ‍

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 $6.20 \frac{m}{s}$ ‍ to $23.1 \frac{m}{s} ?$ ‍

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: Third kinematic equation, $Δ x = v_{0} t + \frac{1}{2} a t^{2}$ ‍

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

If the quadratic formula gives a negative time and a positive time, then the object in question only reaches the specified displacement once after being thrown. In that case, we can simply choose the positive time and neglect the negative time.

The negative time is referring to a time before the object was thrown when it would have been at the specified displacement. In other words, assume the object had been on its trajectory before being thrown and just happened to reach the point of being thrown with the initial velocity specified.

Example 4: Fourth kinematic equation, $v^{2} = v_{0}^{2} + 2 a Δ x$ ‍

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}}

. 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_{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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lolchessru
Posted 8 years ago. Direct link to lolchessru's post “I understand that these e...”
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?
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(22 votes)
Answer
- Arjun Kavungal
  Posted 8 years ago. Direct link to Arjun Kavungal's post “You will work with variab...”
  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.
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  (28 votes)
THE GEEQ
Posted 8 years ago. Direct link to THE GEEQ's post “How to derive equations o...”
How to derive equations of motion by using calculus ?
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(9 votes)
Answer
- mr.swedishfish
  Posted 4 years ago. Direct link to mr.swedishfish's post “We usually start with acc...”
  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
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  (14 votes)
RAQ2016
Posted 8 years ago. Direct link to RAQ2016's post “the gravity magnitude for...”
the gravity magnitude for the free fall is always -9.81 ?
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(7 votes)
Answer
- Mark Zwald
  Posted 8 years ago. Direct link to Mark Zwald's post “Near the surface of the E...”
  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.
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  (23 votes)
Arunabh Bhattacharya
Posted 7 years ago. Direct link to Arunabh Bhattacharya's post “In example 3, where the p...”
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.
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(6 votes)
Answer
- Teacher Mackenzie (UK)
  Posted 7 years ago. Direct link to Teacher Mackenzie (UK)'s post “For this situation (any m...”
  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??
  Comment on Teacher Mackenzie (UK)'s post “For this situation (any m...”
  (10 votes)
mazzapants140894
Posted 7 years ago. Direct link to mazzapants140894's post “Could someone please expl...”
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.
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(3 votes)
Answer
- Unstable Cesium
  Posted 3 years ago. Direct link to Unstable Cesium's post “We want to get Vf = somet...”
  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
  Comment on Unstable Cesium's post “We want to get Vf = somet...”
  (3 votes)
Shravani
Posted 4 years ago. Direct link to Shravani's post “How can two objects with ...”
How can two objects with different masses experience the same acceleration (-9.8m/s^2)? I thought heavier things fall to the ground quicker.
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(4 votes)
Answer
- Hecretary Bird
  Posted 4 years ago. Direct link to Hecretary Bird's post “You're close, but not qui...”
  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!
  Comment on Hecretary Bird's post “You're close, but not qui...”
  (8 votes)
Leo Michaels
Posted 7 years ago. Direct link to Leo Michaels's post “In example 4 when pluggin...”
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...
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(6 votes)
Answer
- peterbpesch
  Posted a year ago. Direct link to peterbpesch's post “(final velocity) squared ...”
  (final velocity) squared = some formula with known values
  
  therefore
  
  final velocity = positve or negative root of (that same formula with known values)
  Comment on peterbpesch's post “(final velocity) squared ...”
  (1 vote)
Studyhere
Posted a year ago. Direct link to Studyhere's post “why did they assume a pos...”
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?
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(3 votes)
Answer
- Charles LaCour
  Posted a year ago. Direct link to Charles LaCour's post “In the note in that examp...”
  In the note in that example it says: 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
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  (6 votes)
Saif Jeelani
Posted 4 years ago. Direct link to Saif Jeelani's post “Why isn't the fifth kinem...”
Why isn't the fifth kinematic formula known a lot(or like used) and also how do you derive it?
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(3 votes)
Answer
- Hecretary Bird
  Posted 4 years ago. Direct link to Hecretary Bird's post “The fifth kinematic formu...”
  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.
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  (3 votes)
Kyle Delaney
Posted 8 years ago. Direct link to Kyle Delaney's post “I've seen a different lis...”
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?
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(1 vote)
Answer
- Rodrigo Campos
  Posted 8 years ago. Direct link to Rodrigo Campos's post “I don't think you should ...”
  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).
  Comment on Rodrigo Campos's post “I don't think you should ...”
  (7 votes)

Mechanics (Essentials) - Class 11th

Course: Mechanics (Essentials) - Class 11th > Unit 4

What are the kinematic equations?

What is a free falling object—i.e., a projectile?

How do you select and use a kinematic equation?

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

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

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

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

What's confusing about the kinematic equations?

What do solved examples involving the kinematic equations look like?

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

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

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

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

Want to join the conversation?

How do you derive the first kinematic equation, $v = v_{0} + a t$ ‍ ?

How do you derive the second kinematic equation, $Δ x = (\frac{v + v_{0}}{2}) t$ ‍?

How do you derive the third kinematic equation, $Δ x = v_{0} t + \frac{1}{2} a t^{2}$ ‍?

How do you derive the fourth kinematic equation, $v^{2} = v_{0}^{2} + 2 a Δ x$ ‍?

Example 1: First kinematic equation, $v = v_{0} + a t$ ‍

Example 2: Second kinematic equation, $Δ x = (\frac{v + v_{0}}{2}) t$ ‍

Example 3: Third kinematic equation, $Δ x = v_{0} t + \frac{1}{2} a t^{2}$ ‍

Example 4: Fourth kinematic equation, $v^{2} = v_{0}^{2} + 2 a Δ x$ ‍