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Meaning of the Kinematic Equations
The kinematic equations are a set of three equations that describe the motion of objects without considering the forces that cause them to move. We can derive them from Newton's second law by considering the motion of projectiles in the presence of a constant gravitational field. This set of equations can answer all of the above questions and more. The three kinematic equations are the linear kinematic equation
\[v_x = v_{x0} + a_xt,\]
the quadratic kinematic equation
\[x = x_0 + v_{x0}t + \frac{1}{2}a_xt^2,\]
and the time-independent kinematic equation
\[v_x^2 = v_{x0}^2 + 2a_x(x-x_0).\]
In these equations, \(t\) is time, \(x\) is position, \(v_x\) is velocity, and \(a_x\) is acceleration. We use the subscript \(_0\) to denote the initial value of the quantity. Note that by convention, the kinematic equations represent motion in the \(x\)-direction. However, they are applicable to any spatial direction.
Kinematic Equations in Constant Acceleration
These equations only apply to situations where we have constant acceleration. This requirement means that the rate of change of velocity is constant. Thus, velocity looks like a straight line when we look at the velocity vs. time graph. Note this does not necessarily mean that the slope is 0. Indeed, the diagram below depicts a situation of constant non-zero acceleration:
Velocity vs. Time graph showing constant acceleration, StudySmarter Originals
Can you guess what is physically happening based on the graph above? Our velocity starts off positive, slows to zero, and goes negative.
One of the most common examples of constant acceleration is the acceleration due to the Earth's gravitational field where \(a = g = -9.8 \;\text{m}/\text{s}^2\). The minus sign comes from choosing a reference frame in which the acceleration points downwards. Another assumption we make is that air resistance is negligible.
Applications of the Kinematic Equations
The applications can be varied and complex, utilizing one, two, or three dimensions, describing some form of projectile motion. Examples include playing catch, shooting an arrow, or dropping a ball off the Empire State Building. Here we will go through the most simple one-dimensional case and a more complicated two-dimensional case.
Projectile Motion in One Dimension
Consider our question earlier in which we dropped our keys. Ignore air resistance and assume a height of \(135\;\text{cm}\). Your keys fall from rest in your hand. How long does it take for the keys to hit the ground?
The first step in solving any physics problem is drawing a diagram including all relevant information.. In this case, we have a fairly simple diagram. We have a height \(135\;\text{cm}\), an initial velocity of \(0 \;\text{m}/\text{s}\), and we assume that the acceleration is due to Earth's gravity.
Problem diagram for dropped keys, StudySmarter Originals
Now we can apply our formula. We already have an initial velocity, a displacement, and an acceleration. We'd like to solve for time. Thus, we can use the quadratic kinematic equation:
\[x = x_0 + v_{x0}t + \frac{1}{2}a_xt^2.\]
We first apply it to our motion in the \(y\)-direction, then rearrange it to get an equation of time in terms of our given variables.
\[\begin{align} y &= v_{y0}t + \frac{1}{2}a_yt^2 \\ 0 &= \frac{1}{2}a_yt^2 + v_{y0}t - y.\end{align}\]
We can recognize the last line as a case of the quadratic equation. Apply the quadratic formula to solve for \(t\). Recall that the roots of the quadratic formula are
\[t_{1,2} = \frac{-b \pm \sqrt{b^2 - 4Ac}}{2A}\]
where \(A\), \(b\), and \(c\)are our coefficients. Note that here we wrote a capital \(A\) to avoid confusing the quadratic coefficient with the acceleration. Next, we can plug in our coefficients:
\[\begin{align}t_{1,2} &= \frac{-(v_{y0}) \pm \sqrt{(v_{y0})^2 - 4\left(\frac{1}{2}a_y\right)(-y)}}{2\left(\frac{1}{2}a_y\right)}\\ &= \frac{-v_{y0} \pm \sqrt{v_{y0}^2 + 2a_yy}}{a_y}\end{align}\]
Now we can plug in our numerical values:
\[t_{1,2} = \frac{-(0 \;\text{m}/\text{s}) \pm \sqrt{(0 \;\text{m}/\text{s})^2 + 2\cdot (9.8 \;\text{m}/\text{s}^2)(1.35 \;\text{m})}}{9.8 \;\text{m}/\text{s}^2}.\]
Simplifying the above gives us
\[\begin{align} t_{1,2} &= \pm \frac{\sqrt{26.46 \;\text{m}^2/\text{s}^2}}{9.8 \;\text{m}/\text{s}^2}\\ &= \pm \frac{5.14 \;\text{m}/\text{s}}{9.8 \;\text{m}/\text{s}^2}\\ &= \pm 0.52 \;\text{s}.\end{align}\]
Note that we have two solutions: \(t_1 = -0.52 \;\text{s}\) and \(t_2 = 0.52 \;\text{s}\). However, only one of these makes physical sense, so we discard the negative time and are left with our answer.
\[\boxed{t = 0.52 \; \text{s}.}\]
It takes our keys \(0.52\) seconds to hit the ground. As a last check, always make sure the answer makes physical sense. If we had gotten a solution of an hour, we would know something was wrong. Half a second makes physical sense, though. So, right away, we know we didn't do anything drastically wrong.
Consider carefully which kinematic equations to apply based on the values the problem gives you.
Projectile Motion in Two Dimensions
Next, we will consider a more difficult problem. Suppose that, instead of dropping your keys, you are tossing them to a friend. They are standing \(4\;\text{m}\) away and you throw the keys at a \(30^{\circ}\) angle. How fast do you have to throw the keys for them to reach your friend?
As always, the first step is drawing a diagram with all relevant information.
Diagram for two-dimensional projectile motion problem, StudySmarter Originals
This is a more difficult problem, so let's break it down into simpler problems. The first thing is to realize we have both an \(x\) and a \(y\)-direction that we are dealing with, and these two are independent of one another so that we can deal with them separately.
Let's first consider the \(y\)-direction. Note that we will have to break up our initial velocity in its components. For the \(y\)-component, we use trigonometry to get
\[v_{y0} = v_y\sin(\theta).\]
Now let's consider the keys rising. At their peak, they change direction, so we have a final velocity of
\[v_{y} = 0 \;\text{m}/\text{s}.\]
Now we apply this information to the kinematic equations.
\[\begin{align}v_y &= v_{y0} + a_yt_r \\0 &= v_0\sin(\theta) + a_yt_r \\v_0 &= \frac{-a_yt_r}{\sin(\theta)}. \end{align}\]
We've subscripted \(t\) with an \(r\) so we remember that this is only the time it takes to get to the top while it is rising. To find our total time in the air, we have to double this, so
\[\begin{align}t &= 2t_r \\ t_r &= \frac{1}{2}t. \end{align}\]
By substituting this in, we have an equation for our initial velocity in terms of \(t\):
\[v_0 = \frac{-a_yt}{2\sin(\theta)}.\]
Let's see if we can find \(t\) using the \(x\)-direction. In the \(x\)-direction, we have no forces acting on our object. Gravity only operates in the \(y\)-direction. This means our acceleration is \(0 \;\text{m}/\text{s}^2\). Thus
\[\begin{aligned}v_x &= v_{x0} + a_xt \\ &= v_{x0}.\end{aligned}\]
Because our velocity in the \(x\)-direction does not change, we can drop the subscript for the initial velocity. We plug this information in and see if we can solve for \(t\).
\[\begin{aligned}x &= x_0 + v_{x0}t + \frac{1}{2}a_xt^2 \\ &= 0 + v_xt + 0 \\ &= v_xt.\end{aligned}\]
Using trigonometry again, we get that the \(x\)-component of our velocity is
\[v_x = v_0\cos(\theta),\]
which allows us to solve for\(t\) in terms of \(v_0\):
\[\begin{aligned}x &= v_xt \\ &= v_0\cos(\theta)t \\ t &= \frac{x}{v_0\cos(\theta)}.\end{aligned}\]
Next, we go back and plug this into our velocity equation:
\[\begin{aligned} v_0 &= \frac{-a_yt}{2\sin(\theta)}\\ &= \frac{-a_y}{2\sin(\theta)}\cdot \frac{x}{v_0\cos(\theta)}.\end{aligned}\]
Note that we have \(v_0\) on both sides, so we can multiply it to the left side and take the square root:
\[\begin{aligned} v_0^2 &= \frac{-a_yx}{2\sin(\theta)\cos(\theta)}\\ v_0 &= \pm \sqrt{\frac{-a_yx}{2\sin(\theta)\cos(\theta)}}.\end{aligned}\]
This is our equation, in terms of known variables, for the initial velocity with which we need to throw our keys so that they reach our friend. Note that the acceleration due to gravity is negative based on our choice of coordinates, so the value under the square root won't cause any trouble. Finally, we plug in our numbers.
\[\begin{aligned} v_0 &= \pm \sqrt{\frac{-a_yx}{2\sin(\theta)\cos(\theta)}}\\ &= \pm \sqrt{\frac{-(-9.8 \;\text{m}/\text{s}^2)\cdot 4 \;\text{m}}{2\sin(30^\circ)\cos(30^\circ)}} \\ &= \pm \sqrt{\frac{39.2\;\text{m}^2/\text{s}^2}{2\cdot 0.5 \cdot 0.87}} \\ &= \pm \sqrt{45.5 \;\text{m}^2/\text{s}^2} \\ &= \pm 6.7 \;\text{m}/\text{s}.\end{aligned}\]
Here we can discard the negative value as we know that doesn't make physical sense in this situation. Therefore our final value is
\[\boxed{v_0 = 6.7\;\text{m}/\text{s}}.\]
We need to throw our keys at \(6.7\;\text{m}/\text{s}\) in order for them to reach our friend.
Time-Independent Example
As a final example. Let's consider a problem that requires the time-independent kinematic equation. Suppose that you are skiing down a hill with an angle of inclination of \(30^\circ\). You start from rest at the top, and accelerate at a rate of \(0.5\;\text{m}/\text{s}^2\). If your final velocity is \(9\;\text{m}/\text{s}\), how tall is the hill?As usual, we draw a picture first.
Skier on a slope problem setup for time-independent kinematic equation, StudySmarter Originals
Next, we have all the information required to apply our formula, so lets plug in the numbers.
$$\begin{aligned} v_x^2 &= v_{x0}^2+2a_x(x-x_0) \\ (9\;\text{m}/\text{s})^2 &= (0\;\text{m}/\text{s})^2 + 2\cdot 0.5\;\text{m}/\text{s}^2(x-x_0) \\ x-x_0 &= \frac{(9\;\text{m}/\text{s})^2}{2\cdot 0.5\;\text{m}/\text{s}^2} \\ x-x_0 &= \frac{81\;\text{m}^2/\text{s}^2}{2\cdot 0.5\;\text{m}/\text{s}^2} \\ x-x_0 &= 81\;\text{m}\end{aligned}$$
Note that this distance is how far we went along the hill, but we are looking for the height of the hill, so we need to use a bit of trigonometry to get our final answer.
$$\begin{aligned} \sin(30^\circ) &= \frac{h}{81\;\text{m}} \\ h &= 81\;\text{m}\cdot\sin(30^\circ) \\ h &= 40.5\;\text{m}\end{aligned}$$
Thus, our final height of the hill is
$$\boxed{h=40.5\;\text{m}}$$
Derivation of the Kinematic Equations
Below we briefly outline how to derive each of the three kinematic equations.
Deriving the Linear Kinematic Equation
We can derive the kinematic equations from Newton's second law. However, doing so requires solving differential equations. Another method is to derive them algebraically using our knowledge of the world around us to develop a starting point. To do that, we note that the acceleration is the change in velocity over the change in time, so
\[a = \frac{\Delta v}{\Delta t}\]
where \(\Delta v = v - v_0\), and we denote our time frame by \(t\). Thus, we have:
\[a = \frac{v - v_0}{t}.\]
Now to finish deriving our first kinematic equation, we consider the \(x\)-direction, and we rearrange our terms:
\[\begin{align}a_x &= \frac{v_x - v_{x0}}{t} \\ a_xt &= v_x - v_{x0}. \end{align}\]
Hence, we arrive at the linear kinematic equation.
\[\boxed{v_x = v_{x0} + a_xt.}\]
Deriving the Quadratic Kinematic Equation
Note that similarly, we have that the average velocity is the change in position over the change in time.
\[v = \frac{\Delta x}{\Delta t}\]
where \(\Delta x = x - x_0\). We can also express the average velocity as the mean of the initial and final velocity. We plug this into our equation and solve for \(v_x\):
\[\begin{align} v_{\text{avg}} &= \frac{x - x_0}{t} \\ \frac{v_x + v_{x0}}{2} &= \frac{x - x_0}{t} \\ v_x &= \frac{2(x - x_0)}{t} - v_{x0}. \end{align}\]
We can plug this into the linear kinematic equation:
\[\begin{align} v_x &= v_{x0} + a_xt \\ \frac{2(x-x_0)}{t} - v_{x0} &= v_{x0} + a_xt \\ \frac{2(x-x_0)}{t} &= 2v_{x0} + a_xt \\ x - x_0 &= v_{x0}t + \frac{1}{2}a_xt^2.\end{align}\]
Finally, by rearranging, we get the quadratic kinematic equation.
\[\boxed{x = x_0 + v_{x0}t + \frac{1}{2}a_xt^2.}\]
Deriving the Time-Independent Kinematic Equation
Finally, we can derive our third equation by solving the first equation for \(t\) and plugging it into the second equation.
Solve for \(t\),
\[\begin{aligned}v_x &= v_{x0} + a_xt \\ t &= \frac{v_x - v_{x0}}{a_x},\end{aligned}\]
and plug it into the quadratic kinematic equation:
\[\begin{aligned} x &= x_0 + v_{x0}t + \frac{1}{2}a_xt^2 \\ &= x_0 + v_{x0}\left(\frac{v_x - v_{x0}}{a_x}\right) + \frac{1}{2}a_x\left(\frac{v_x - v_{x0}}{a_x}\right)^2.\end{aligned}\]
The next step is to pull out the common factor from the expression and simplify the fractions:
\[\begin{aligned} x - x_0 &= \left(\frac{v_x - v_{x0}}{a_x}\right)\left(v_{x0} + \frac{1}{2}a_x\left(\frac{v_x - v_{x0}}{a_x}\right)\right) \\ &= \left(\frac{v_x - v_{x0}}{a_x}\right)\left(\frac{2v_{x0}}{2} + \frac{v_x - v_{x0}}{2}\right) \\ &= \left(\frac{v_x - v_{x0}}{a_x}\right)\left(\frac{2v_{x0} + v_x - v_{x0}}{2}\right) \\ &= \left(\frac{v_x - v_{x0}}{a_x}\right)\left(\frac{v_{x0} + v_x}{2}\right).\end{aligned}\]
Multiply the factors on the right-hand side and notice that the non-quadratic velocity terms cancel out:
\[\begin{aligned} x - x_0 &= \left(\frac{v_x^2 + v_xv_{x0} - v_{x0}v_x - v_{x0}^2}{2a_x}\right) \\ &= \frac{v_x^2 - v_{x0}^2}{2a_x} \end{aligned}\]
so
\[2a_x(x-x_0) = v_x^2 - v_{x0}^2.\]
Rearranging this last line results in the time-independent kinematic equation:
\[\boxed{v_x^2 = v_{x0}^2 + 2a_x(x-x_0).}\]
This completes the derivation of the kinematic equations.
Angular Kinematic Equations
In rotational kinematics, there is an analog to the kinematic equations we have discussed. These equations look identical in form, except that the displacement, \(x\), is replaced with the angle, \(\theta\), the velocity, \(v\), is replaced with the angular velocity, \(\omega\), and the acceleration, \(a\), is replaced with the angular acceleration, \(\alpha\). Thus the linear angular kinematic equation is
\[\omega = \omega_0+\alpha t.\]
The quadratic angular kinematic equation is
\[\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2,\]
and finally, the time-independent angular kinematic equation is
\[\omega^2 = \omega_0^2 + 2\alpha (\theta - \theta_0).\]
Kinematic Equations - Key takeaways
- There are three kinematic equations.
- The kinematic equations can be applied to motion in one-dimensional and two-dimensional or three-dimensional problems.
- The kinematic equations are used to describe the motion of an object under constant acceleration.
- The kinematic equations are derived from Newton's second law of motion, however, they can also be derived experimentally.
- There is a set of rotational kinematic equations that is analogous to the linear kinematic equations.
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Frequently Asked Questions about Kinematic Equations
How do you derive the kinematic equations?
The kinematic equations can be derived mathematically from Newton's second law using some more advanced math; however, they can be derived experimentally by taking detailed data and developing relationships between the values, or they can be derived algebraically using some things we know about the world around us.
What are kinematic equations?
A set of equations that describe the motion of projectiles.
What are the three kinematics equations?
- The linear kinematic equation.
- The quadratic kinematic equation.
- The time-independent kinematic equation.
What are the kinematic equations in constant acceleration?
One of the assumptions for the kinematic equations is that the acceleration remains constant throughout the timeframe we are interested in, so the kinematic equations require constant acceleration.
How do you manipulate the kinematic equations?
Typically, problems involve algebraic manipulations. Frequently, this consists in applying one of the kinematic equations, then using that value to solve a second kinematic equation for the value we are interested in, though problems can be quite complex involving two or three spacial dimensions.
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