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\Delta x\quad\text{Displacement}
tTime interval t\qquad\text{Time interval}~
v0  Initial velocity v_0 ~~\quad\text{Initial velocity}~
v   Final velocity v\quad ~~~\text{Final velocity}~
a   Constant acceleration a \quad~~ \text{ Constant acceleration}~
The tt in these formulas really represents the interval of time it took the object to be displaced Δx\Delta x. It would probably be conceptually clearer to write the time interval as Δt\Delta t but the kinematic formulas are already so cluttered that people usually leave off the Δ\Delta sign in front of the time tt for the sake of cleanliness.
We just have to make sure we know that by writing tt, we really mean the time interval Δt\Delta t.
If we know three of these five kinematic variables—Δx,t,v0,v,a\Delta x, t, v_0, 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.
In this section, we're just stating the kinematic formulas. In the sections below, we'll derive each kinematic formula individually so we can see where they come from and why they're only true assuming constant acceleration.
In the solved examples section below, we will also solve an example problem using each individual formula in order.
1.v=v0+at\Large 1. \quad v=v_0+at
2.Δx=(v+v02)t\Large 2. \quad {\Delta x}=(\dfrac{v+v_0}{2})t
3.Δx=v0t+12at2\Large 3. \quad \Delta x=v_0 t+\dfrac{1}{2}at^2
4.v2=v02+2aΔx\Large 4. \quad v^2=v_0^2+2a\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 xx, vertical yy, etc.
If you plug in a horizontal initial velocity v0xv_{0x} into a kinematic formula, the rest of the variables you plug into that formula—Δx,vx,ax\Delta x, v_x, a_x—should also be for the horizontal direction.
If you plug in a vertical initial velocity v0yv_{0y} into a kinematic formula, the rest of the variables you plug into that formula—Δy,vy,ay\Delta y, v_y, a_y—should also be for the vertical direction.
In other words, to be really clear, the kinematic formulas for a given direction—for example, xx—should really be written with a subscript to denote that direction:
vx=v0x+axt v_x=v_{0x}+a_xt
Δx=v0xt+12axt2 \Delta x=v_{0x} t+\dfrac{1}{2}a_xt^2
vx2=v0x2+2axΔx v_x^2=v_{0x}^2+2a_x\Delta x
vx+v0x2=Δxt\dfrac{v_x+v_{0x}}{2}=\dfrac{\Delta 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 formulas 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 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=9.81ms2g=9.81\dfrac{\text{m}}{\text{s}^2}.
g=9.81ms2(Magnitude of acceleration due to gravity)\Large g=9.81\dfrac{\text{m}}{\text{s}^2}\quad \text{(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=9.81ms2g=9.81\dfrac{\text{m}}{\text{s}^2}.
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.
The force of gravity is larger on the boulder, which tends to increase the acceleration of the boulder. But the inertia—resistance to acceleration—is also larger for the boulder, which tends to decrease the acceleration of the boulder.
These confounding effects exactly cancel and the result is that the mass of an object does not affect its acceleration due to gravity.
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=9.81ms2g=9.81\dfrac{\text{m}}{\text{s}^2}, no matter what mass it has—as long as air resistance is negligible.
Note that g=9.81ms2g=9.81\dfrac{\text{m}}{\text{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 ay=9.81ms2a_y=-9.81\dfrac{\text{m}}{\text{s}^2} 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/sv_0=5\text{ m/s}, after which it took a time interval t=3 st=3\text{ s} for the book to slide a displacement of Δx=8 m\Delta x=8\text{ m}. We could use the kinematic formula Δx=v0t+12at2\Delta x=v_0 t+\dfrac{1}{2}at^2 to algebraically solve for the unknown acceleration aa of the book—assuming the acceleration was constant—since we know every other variable in the formula besides aaΔx,v0,t\Delta x, v_0, t.
Problem solving tip: Note that each kinematic formula is missing one of the five kinematic variables—Δx,t,v0,v,a\Delta x, t, v_0, v, a.
1.v=v0+at(This formula is missing .)Δx1. \quad v=v_0+at \quad \text{(This formula is missing $\Delta x$.)}
2.Δx=(v+v02)t(This formula is missing .)a2. \quad {\Delta x}=(\dfrac{v+v_0}{2})t\quad \text{(This formula is missing $a$.)}
3.Δx=v0t+12at2(This formula is missing .)v3. \quad \Delta x=v_0 t+\dfrac{1}{2}at^2\quad \text{(This formula is missing $v$.)}
4.v2=v02+2aΔx(This formula is missing .)t4. \quad v^2=v_0^2+2a\Delta x\quad \text{(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 vv of the book was neither given nor asked for, so we should choose a formula that does not include vv at all. The kinematic formula Δx=v0t+12at2\Delta x=v_0 t+\dfrac{1}{2}at^2 is missing vv, so it's the right choice in this case to solve for the acceleration aa.
Yes, there is.
5.Δx=vt12at2(This formula is missing .)v05.\quad \Delta x=vt-\dfrac{1}{2}at^2 \quad \text{(This formula is missing $v_0$.)}
The fifth kinematic formula looks just like the third kinematic formula Δx=v0t+12at2\Delta x=v_0 t+\dfrac{1}{2}at^2 except with the initial velocity v0v_0 replaced with final velocity vv and the plus sign replaced with a minus sign. It can be derived by plugging the first kinematic formula into the third kinematic formula.
This fifth kinematic formula often isn't as useful since the initial velocity is typically known/given in many situations.

How do you derive the first kinematic formula, v=v0+atv=v_0+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=ΔvΔta=\dfrac{\Delta v}{\Delta t} \quad
Yes. If the time interval Δt\Delta t is large, the formula a=ΔvΔta=\dfrac{\Delta v}{\Delta t} is the average acceleration over the time interval.
So, the derivation we are going through below really assumes that the aa in the formula is the average acceleration aavga_{avg}. However, if the acceleration is constant, the instantaneous acceleration ainsta_{inst} will equal the average acceleration aavga_{avg}, in which case we don't have to worry about being specific and can just call it aa.
So, for the kinematic formula we will derive to be accurate, the acceleration should be constant. Or else the aa in the formula is referring to the average acceleration and not the instantaneous acceleration.
Now we can replace Δv\Delta v with the definition of change in velocity vv0v-v_0.
a=vv0Δta=\dfrac{v_-v_0}{\Delta t}
Finally if we just solve for vv we get
v=v0+aΔtv=v_0+a\Delta t
And if we agree to just use tt for Δt\Delta t, this becomes the first kinematic formula.
v=v0+at\Large v=v_0+at

How do you derive the second kinematic formula, Δx=(v+v02)t{\Delta x}=(\dfrac{v+v_0}{2})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 v0v_0 as seen in the graph below.
The area under any velocity graph gives the displacement Δx\Delta x. So, the area under this velocity graph will be the displacement Δx\Delta x of the object.
Δx= total area\Delta x=\text{ 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 v0v_0 and the width is tt, so the area of the blue rectangle is v0tv_0t.
The base of the red triangle is tt and the height is vv0v-v_0, so the area of the red triangle is 12t(vv0)\dfrac{1}{2}t(v-v_0).
The total area will be the sum of the areas of the blue rectangle and the red triangle.
Δx=v0t+12t(vv0)\Delta x=v_0t+\dfrac{1}{2}t(v-v_0)
If we distribute the factor of 12t\dfrac{1}{2}t we get
Δx=v0t+12vt12v0t\Delta x=v_0t+\dfrac{1}{2}vt-\dfrac{1}{2}v_0t
We can simplify by combining the v0v_0 terms to get
Δx=12vt+12v0t\Delta x=\dfrac{1}{2}vt+\dfrac{1}{2}v_0t
And finally we can rewrite the right hand side to get the second kinematic formula.
Δx=(v+v02)t\Large \Delta x=(\dfrac{v+v_0}{2})t
This formula is interesting since if you divide both sides by tt, you get Δxt=(v+v02)\dfrac{\Delta x}{t}=(\dfrac{v+v_0}{2}). This shows that the average velocity Δxt\dfrac{\Delta x}{t} equals the average of the final and initial velocities v+v02\dfrac{v+v_0}{2}. 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\Delta x=v_0 t+\dfrac{1}{2}at^2?

There are a couple ways to derive the equation Δx=v0t+12at2\Delta x=v_0 t+\dfrac{1}{2}at^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 v0v_0 and maintains constant acceleration to a final velocity of vv as seen in the graph below.
Since the area under a velocity graph gives the displacement Δx\Delta x, each term on the right hand side of the formula Δx=v0t+12at2\Delta x=v_0 t+\dfrac{1}{2}at^2 represents an area in the graph above.
The term v0tv_0t represents the area of the blue rectangle since Arectangle=hwA_{rectangle}=hw.
The term 12at2\dfrac{1}{2}at^2 represents the area of the red triangle since Atriangle=12bhA_{triangle}=\dfrac{1}{2}bh.
The base of the red triangle is given by tt. The height of the red triangle is given by vv0v-v_0.
But this height can be rewritten as atat since vv0=atv-v_0=at, from the definition of acceleration.
So, Atriangle=12bh=12t(vv0)=12t(at)=12at2A_{triangle}=\dfrac{1}{2}bh=\dfrac{1}{2}t(v-v_0)=\dfrac{1}{2}t(at)=\dfrac{1}{2}at^2
That's it. The formula Δx=v0t+12at2\Delta x=v_0 t+\dfrac{1}{2}at^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 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+atv=v_0+at, into the second kinematic formula, Δxt=v+v02\dfrac{\Delta x}{t}=\dfrac{v+v_0}{2}.
If we start with second kinematic formula
Δxt=v+v02\dfrac{\Delta x}{t}=\dfrac{v+v_0}{2}
and we use v=v0+atv=v_0+at to plug in for vv, we get
Δxt=(v0+at)+v02\dfrac{\Delta x}{t}=\dfrac{(v_0+at)+v_0}{2}
We can expand the right hand side and get
Δxt=v02+at2+v02\dfrac{\Delta x}{t}=\dfrac{v_0}{2}+\dfrac{at}{2}+\dfrac{v_0}{2}
Combining the v02\dfrac{v_0}{2} terms on the right hand side gives us
Δxt=v0+at2\dfrac{\Delta x}{t}=v_0+\dfrac{at}{2}
And finally multiplying both sides by the time tt gives us the third kinematic formula.
Δx=v0t+12at2\Large \Delta x=v_0 t+\dfrac{1}{2}at^2
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+2aΔxv^2=v_0^2+2a\Delta x?

To derive the fourth kinematic formula, we'll start with the second kinematic formula:
Δx=(v+v02)t{\Delta x}=(\dfrac{v+v_0}{2})t
We want to eliminate the time tt from this formula. To do this, we'll solve the first kinematic formula, v=v0+atv=v_0+at, for time to get t=vv0at=\dfrac{v-v_0}{a}. If we plug this expression for time tt into the second kinematic formula we'll get
Δx=(v+v02)(vv0a){\Delta x}=(\dfrac{v+v_0}{2})(\dfrac{v-v_0}{a})
Multiplying the fractions on the right hand side gives
Δx=(v2v022a){\Delta x}=(\dfrac{v^2-v_0^2}{2a})
And now solving for v2v^2 we get the fourth kinematic formula.
v2=v02+2aΔx\Large v^2=v_0^2+2a\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 v0=0v_0=0, "dropped" often means v0=0v_0=0, and "comes to a stop" means v=0v=0. Also, the magnitude of the acceleration due to gravity on all freely flying projectiles is assumed to be g=9.81ms2g=9.81\dfrac{\text{m}}{\text{s}^2}, 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\Delta x, v_o, v, a—except for tt 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=9.81ms2a_g=-9.81\dfrac{\text{m}}{\text{s}^2}.
The third kinematic formula, Δx=v0t+12at2\Delta x=v_0 t+\dfrac{1}{2}at^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 formula must be consistent with that time interval. In other words, the initial velocity v0v_0 has to be the velocity of the object at the initial position and start of the time interval tt. Similarly, the final velocity vv must be the velocity at the final position and end of the time interval tt being analyzed.

What do solved examples involving the kinematic formulas look like?

Example 1: First kinematic formula, v=v0+atv=v_0+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 st=2.35 \text{ s}?
Assuming upward is the positive direction, our known variables are
v0=0v_0=0 \quad (Since the water balloon was dropped, it started at rest.)
t=2.35 st=2.35\text{ s} \quad (This is the time interval after which we want to find the velocity.)
ag=9.81ms2a_g=-9.81\dfrac{\text{m}}{\text{s}^2} \quad(This is implied since the water balloon is a freely falling object.)
If you wait long enough, yes, but we won't use that in our formula for two reasons.
First, we are only considering the time interval of 2.35 s2.35 \text{ s}, which is presumably less than the time required to reach the ground.
Second, we can only apply the kinematic formulas 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 formula is ag=9.81ms2a_g=-9.81\dfrac{\text{m}}{\text{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 formulas 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 yy as our position variable instead of xx. The symbol we choose doesn't really matter as long as we're consistent, but people typically use yy to indicate vertical motion.
Since we don't know the displacement Δy\Delta y and we weren't asked for the displacement Δy\Delta y, we'll use the first kinematic formula v=v0+atv=v_{0}+at, which is missing Δy\Delta y.
v=v0+at(Use the first kinematic formula since it’s missing .)Δyv=v_{0}+at \quad \text{(Use the first kinematic formula since it's missing $\Delta y$.)}
v=0 m/s+(9.81ms2)(2.35 s)(Plug in known values.)v=0\text{ m/s} +(-9.81\dfrac{\text{m}}{\text{s}^2})(2.35\text{ s}) \quad \text{(Plug in known values.)}
v=23.1 m/s(Calculate and celebrate!)v=-23.1 \text { m/s}\quad \text{(Calculate and celebrate!)}
Note: The final velocity was negative since the water balloon was heading downward.
Yes, we're free to call downward the positive direction for any problem, but we have to be consistent with that decision. If we call downward positive, then every vector pointing downward has to be considered positive.
So the acceleration would have to be +9.8ms2+9.8\dfrac{\text{m}}{\text{s}^2}, and we would end up getting a positive final velocity instead of a negative final velocity.

Example 2: Second kinematic formula, Δx=(v+v02)t{\Delta x}=(\dfrac{v+v_0}{2})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/sv_0= 6.20\text{ m/s} \quad (The initial speed of the leopard)
v=23.1 m/sv= 23.1\text{ m/s} \quad (The final speed of the leopard)
t=3.30 st=3.30\text{ s} \quad (The time it took for the leopard to speed up)
Since we do not know the acceleration aa and were not asked for the acceleration, we'll use the second kinematic formula for the horizontal direction Δx=(v+v02)t{\Delta x}=(\dfrac{v+v_{0}}{2})t, which is missing aa.
Δx=(v+v02)t(Use the second kinematic formula since it’s missing .)a{\Delta x}=(\dfrac{v+v_{0}}{2})t \quad \text{(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.){\Delta x}=(\dfrac{23.1\text{ m/s}+6.20\text{ m/s}}{2})(3.30\text{ s} ) \quad \text{(Plug in known values.)}
Δx=48.3 m(Calculate and celebrate!)\Delta x=48.3 \text{ m} \quad \text{(Calculate and celebrate!)}

Example 3: Third kinematic formula, Δx=v0t+12at2\Delta x=v_0 t+\dfrac{1}{2}at^2

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/sv_0=18.3 \text { m/s} \quad (The initial upward velocity of the pencil)
Δy=12.2 m\Delta y=12.2\text{ m} \quad (We want to know the time when the pencil moves through this displacement.)
a=9.81 m s2a=-9.81\dfrac{\text{ m}}{\text{ s}^2} \quad (The pencil is a freely flying projectile.)
Since we don't know the final velocity vv and we weren't asked to find the final velocity, we will use the third kinematic formula for the vertical direction Δy=v0yt+12ayt2\Delta y=v_{0y} t+\dfrac{1}{2}a_yt^2, which is missing vv.
Δy=v0yt+12ayt2(Start with the third kinematic formula.)\Delta y=v_{0y} t+\dfrac{1}{2}a_yt^2 \quad \text{(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 tt 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.)12.2\text{ m}=(18.3\text{ m/s})t+\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})t^2 \quad \text{(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.)0=\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})t^2+(18.3\text{ m/s})t -12.2\text{ m} \quad \text{(Put it into the form of the quadratic equation.)}
At this point, we solve the quadratic equation for time tt. The solutions of a quadratic equation in the form of at2+bt+c=0at^2+bt+c=0 are found by using the quadratic formula t=b±b24ac2at=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}. For our kinematic equation a=12(9.81 m s2)a=\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2}), b=18.3 m/sb=18.3\text{ m/s}, and c=12.2 mc=-12.2\text{ 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)]t=\dfrac{-18.3\text{ m/s}\pm\sqrt{(18.3\text{ m/s})^2-4[\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})(-12.2\text{ m})]}}{2[\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})]}
Since there is a plus or minus sign in the quadratic formula, we get two answers for the time tt: one when using the ++ and one when using the -. Solving the quadratic formula above gives these two times:
t=0.869 st=0.869\text{ s} and t=2.86 st=2.86\text{ 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 st=0.869\text{ 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 formula, v2=v02+2aΔxv^2=v_0^2+2a\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 3.20 m s23.20\dfrac{\text{ m}}{\text{ 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
v0=23.4 m/sv_0=23.4 \text { m/s} \quad (The initial forward velocity of the motorcycle)
a=3.20 m s2a=-3.20\dfrac{\text{ m}}{\text{ s}^2} \quad (Acceleration is negative since the motorcycle is slowing down and we assumed forward is positive.)
Δx=50.2 m\Delta x=50.2\text{ m} \quad (We want to know the velocity after the motorcycle moves through this displacement.)
Since we don't know the time tt and we weren't asked to find the time, we will use the fourth kinematic formula for the horizontal direction vx2=v0x2+2axΔxv_x^2=v_{0x}^2+2a_x\Delta x, which is missing tt.
vx2=v0x2+2axΔx(Start with the fourth kinematic formula.)v_x^2=v_{0x}^2+2a_x\Delta x \quad \text{(Start with the fourth kinematic formula.)}
vx=±v0x2+2axΔx(Algebraically solve for the final velocity.)v_x=\pm\sqrt{v_{0x}^2+2a_x\Delta x} \quad \text{(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+2axΔxv_x=+\sqrt{v_{0x}^2+2a_x\Delta x}.
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.)v_x=\sqrt{(23.4\text{ m/s})^2+2(-3.20\dfrac{\text{ m}}{\text{ s}^2})(50.2\text{ m})} \quad \text{(Plug in known values.)}
vx=15.0 m/s(Calculate and celebrate!)v_x=15.0\text{ m/s} \quad \text{(Calculate and celebrate!)}