Kinematics is the study of objects in motion, without asking anything about why the object is moving. We will focus on translational motion with constant acceleration.
12.1 One-dimensional kinematics
Say you are driving in a car along a long, straight road. If you know your current velocity, and you know how your velocity is changing (that is, you know your acceleration), you can determine your velocity at some later time. Assuming the acceleration is constant, we can multiply it by the elapsed time to find out how the velocity has changed in that amount of time. Then add that to whatever the velocity was to begin with. In symbols, we have
\[
v_{fx} = v_{ix} + a_xt
\]
We have essentially just used some unit analysis to figure this out:
\[ \left(\textrm{m/s}^2\right)\left(\textrm{s}\right) = \textrm{m/s} \]
We can use similar reasoning to find your position after some time, by multiplying velocity by time (since \((\textrm{m/s})(\textrm{s}) = \textrm{m}\)). Unfortunately, unit analysis leaves out any dimensionless (or unitless) factors. It turns out that we need to include a factor of \(\frac{1}{2}\) to accurately find position:
\[
x_f = x_i + v_{ix} t + \frac{1}{2}a_x t^2
\]
For this text, we don’t need to worry about where the \(\frac{1}{2}\) comes from; if you continue on in physics you’ll find out soon enough.
There is one more relationship that is often helpful:
\[
v_{fx}^2 = v_{ix}^2 + 2a_x(x_f – x_i)
\]
These relationships are known as the kinematic equations for constant acceleration. They tell you relationships between the different physical quantities that are used to describe an object’s motion (position, velocity, and acceleration). You can sit down and memorize these equations, but you will find that this happens on its own when you work through enough practice problems.
Note the use of subscripts: there is \(i\) for initial, \(f\) for final, and \(x\) to represent motion along the \(x\) axis. If you are dealing with motion in the \(y\) direction, simply replace all \(x\)s with \(y\)s. It seems tedious now, but when we are working with two dimensions it will be extremely important to keep the \(x\) and \(y\) components of motion distinguished from each other.
12.1.1 Problem solving steps
As with any physics problem, you should always start by drawing a picture. For kinematics problems, we tend to have lots of variables to keep track of— it is very helpful to get organized. After you have drawn your picture, make a table of what you know and what you don’t know. Then, you can look at the relationships between these variables to see what kinematics equations you need to use to solve the problem.
To summarize:
- Draw a picture.
- Organize what you know and what you don’t know.
- Use the kinematic equations to identify connections between knowns and unknowns. These equations are $$\begin{align*} x_f &= x_i + v_{ix}t + \frac{1}{2}a_xt^2 \tag{12.1} \\ v_{fx} &= v_{ix} + a_xt \tag{12.2} \\ v_{fx}^2 &= v_{ix}^2 + 2a_x(x_f – x_i) \tag{12.3} \end{align*}$$
- Solve with appropriate algebraic steps.
Example 12.1
A car driving down the road accelerates uniformly in a straight line from 25 m/s to 35 m/s in 3.5 s. What is the car’s acceleration? How far does the car travel while it accelerates?
| Known | Unknown |
|---|---|
| \(x_i = 0\) | \(a_x\) |
| \(v_{ix} = 25\ \textrm{m/s}\) | \(x_f\) |
| \(v_{fx} = 35\ \textrm{m/s}\) | |
| \(t = 3.5\ \textrm{s}\) |
Using equation (12.2), we can find the acceleration. Then use either other equation to find the distance traveled. I’ll use equation (12.1)
\[
\begin{align*}
v_{fx} &= v_{ix} + a_x t & x_f &= x_i + v_{ix} t + \frac{1}{2}a_xt^2 \\
\hookrightarrow a_x &= \frac{v_{fx} – v_{ix}}{t} & &= 0 + (25\ \textrm{m/s})(3.5\ \textrm{s}) + \frac{1}{2}\left(2.9\ \textrm{m/s}^2\right)(3.5\ \textrm{s})^2 \\
&= \frac{(35\ \textrm{m/s}) – (25\ \textrm{m/s})}{3.5\ \textrm{s}} & &= 105\ \textrm{m} \\
&= 2.9\ \textrm{m/s}^2
\end{align*}
\]
Example 12.2
A speed skater moving across frictionless ice at 8 m/s hits a 5 m long patch of rough ice. She slows steadily, then continues on at 6 m/s. What is her acceleration on the rough ice? How much time did it take her to travel across the rough spot?
| Known | Unknown |
|---|---|
| \(x_i = 0\) | \(a_x\) |
| \(x_f = 5\ \textrm{m}\) | \(t\) |
| \(v_{ix} = 8\ \textrm{m/s}\) | |
| \(v_{fx} = 6\ \textrm{m/s}\) |
Use equation (12.3) to determine the acceleration. Use either other equation to find the time. I’ll use equation (12.2).
\[
\begin{align*}
v_{f,x}^2 &= v_{i,x}^2 + 2a_x(x_f – x_i) & v_{f,x} &= v_{i,x} + a_x t \\
\hookrightarrow a_x &= \frac{v_{f,x}^2 – v_{i,x}^2}{2(x_f – 0)} & \hookrightarrow t &= \frac{v_{f,x} – v_{i,x}}{a_x} \\
&= \frac{(6\ \textrm{m/s})^2 – (8\ \textrm{m/s})^2}{2(5\ \textrm{m})} & &= \frac{(6\ \textrm{m/s}) – (8\ \textrm{m/s})}{-2.8\ \textrm{m/s}^2} \\
&=-2.8\ \textrm{m/s}^2 & &= 0.71\ \textrm{s}
\end{align*}
\]
Example 12.3
A driver slams on the brakes when they see a tree lying across the road. The car slows uniformly with an acceleration of 5.6 m/s2 for 4.2 s, making straight skid marks 62.4 m long, all the way to the tree. With what speed does the car strike the tree?
| Known | Unknown |
|---|---|
| \(x_i = 0\) | \(v_{ix}\) |
| \(x_f = 62.4\ \textrm{m}\) | \(v_{fx}\) |
| \(a_x = -5.6\ \textrm{m/s}^2\) | |
| \(t = 4.2\ \textrm{s}\) |
By defining \(x_i = 0\) and \(x_f = +62.4\ \textrm{m}\) we have defined the \(+x\) direction as the direction the car is moving. Since the car is slowing down its, acceleration must be in the opposite direction as its motion; this is why the acceleration is in the \(-x\) direction.
Use equation (12.1) to find the initial speed, then either other equation to find the final speed. I’ll use equation (12.2). Note that “final” here means “just before impact.”
\[
\begin{align*}
x_f &=x_i + v_{ix}t + \frac{1}{2}a_xt^2 & v_{fx} &= v_{ix} + a_xt \\
\hookrightarrow v_{ix} &= \frac{x_f – \frac{1}{2}a_xt^2}{t} & &= (26.6\ \textrm{m/s}) + \left(-5.6\ \textrm{m/s}^2\right)(4.2\ \textrm{s}) \\
&= \frac{x_f}{t} – \frac{a_xt}{2} & &= 3.1\ \textrm{m/s} \\
&= \frac{62.4\ \textrm{m}}{4.2\ \textrm{s}} – \frac{\left(-5.6\ \textrm{m/s}^2\right)(4.2\ \textrm{s})}{2} \\
&= 26.6\ \textrm{m/s}
\end{align*}
\]
Practice 12.1
12.1.2 Free Fall
An object in free fall is subject only to the influence of gravity. This definition applies whether the object is traveling up or down—as long as gravity is the only force acting, the object is in free fall. Using this constraint, we can determine the acceleration of any object in free fall. Start with a free body diagram:
Then apply Newton’s second law:
\[
\begin{align*}
F_\textit{net,y} &= ma_y \\
-F^G &= ma_y \\
-mg &= ma_y \\
a_y &= -g
\end{align*}
\]
The magnitude of the object’s acceleration is \(g = 9.8\) m/s2, and the acceleration points towards the ground. (If you want to do a quick in-your-head approximation, you can use \(g \approx 10\) m/s2.) Note that this can apply equally well to an object that is traveling up, as for one that is traveling down. So, an object can be in “free fall” while still traveling up—this is different than how we use the term in everyday language—as long as gravity is the only force acting on it.
In the late 1500s and early 1600s, scientists showed that all falling objects have the same constant acceleration when air resistance can be neglected. (Contrary to popular belief, Galileo did not actually drop things off of the Leaning Tower of Pisa to show this. He instead used inclined planes instead to carry out controlled experiments.)
In 1971, Apollo 15 astronauts dropped a feather and hammer on the moon, where there is basically no air resistance. You can see footage of that here. While it is very cool, the footage from the moon is rather grainy. Recently, a similar demonstration was carried out on Earth, in a vacuum chamber. This video uses slow motion filming so that you can really see what’s going on.
Example 12.4
A person throws a ball straight up. The plot below shows a position vs time graph of the ball’s motion.
Draw an acceleration vs. time graph and a velocity vs. time graph for period of time starting just after the ball leaves the person’s hand, to the time just before it is caught again.
Velocity vs time
Velocity is the rate of change (slope) of position. The position vs time graph begins with a very steep positive slope which gradually becomes shallower. At the apex of the parabola, the slope is zero. It is then negative, and gradually becomes steeper.
Acceleration vs time
Acceleration is the rate of change (slope) of velocity. The slope of the velocity graph is constant; there is no curvature to the line. The slope of the line is negative.
A note on air resistance
Air resistance mostly depends on two factors: the speed and cross-sectional area of the object. For all of the situations that we will consider, air resistance is very small compared to the other forces acting on the object. In fact, this is the case for the vast majority of free fall situations you’ll encounter in everyday life! You almost need to go out of your way and set up some specific parameters for air resistance to not be negligible.
Example 12.5
A cat steps off of a 2.0 m tall ledge. How much time does it take to reach the ground? What is its velocity when it lands?
Following convention, I’ll use a \(+y\) up coordinate system. This means “towards the ground” is the \(-y\) direction. Since the cat simply steps off the ledge (instead of jumping up) there is no initial velocity.
| Known | Unknown |
|---|---|
| \(y_i = 2.0\ \textrm{m}\) | \(v_{fy}\) |
| \(y_f = 0\) | \(t\) |
| \(v_{iy} = 0\) | |
| \(a_y = -g\) |
Use equation (12.1) to determine time. Then use equation (12.2) to find the final velocity. We will use \(y\) instead of \(x\), to represent motion along the \(y) axis.
\[
\begin{align*}
y_f &= y_i + v_{iy}t + \frac{1}{2}a_yt^2 & v_{fy} &= v_{iy} + a_yt \\
0 &= y_i + (0)t – \frac{1}{2}gt^2 & &= 0-gt \\
\hookrightarrow t &= \sqrt{\frac{2y_i}{g}} & &= -\left(9.81\ \textrm{m/s}^2\right)(0.639\ \textrm{s}) \\
&= \sqrt{\frac{2(2.0\ \textrm{m})}{9.81\ \textrm{m/s}^2}} & &= -6.26\ \textrm{m/s} \\
&= 0.639\ \textrm{s}
\end{align*}
\]
Note that the velocity is negative, since the cat is traveling in the negative direction.
Example 12.6
When a volcano erupts, there are often blobs of lava that are shot up into the air, called “lava bombs.” Consider a lava bomb that takes 4.75 s to go up and return to the caldera (the same place it was launched from). What was the initial velocity?
I’ll use a \(+y\) up coordinate system.
| Known | Unknown |
|---|---|
| \(y_i = y_f =0 \) | \(v_{iy}\) |
| \(a_y = -g\) | |
| \(t = 4.75\ \textrm{s}\) |
We have everything we need to use equation (12.1).
\[
\begin{align*}
y_f &=y_i + v_{i,y}t + \frac{1}{2}at^2 \\
0 &= 0 + v_{i,y}t – \frac{1}{2}gt^2 \\
\hookrightarrow v_{i,y} &= \frac{1}{2}gt \\
&= \frac{1}{2}\left(9.81\ \textrm{m/s}^2\right)(4.75\ \textrm{s}) \\
&= 23.3\ \textrm{m/s}
\end{align*}
\]
Example 12.7
A parachutist is descending at 10 m/s and accidentally drops their camera from an altitude of 50 m. How much time does it take the camera to reach the ground? What is the camera’s velocity when it hits the ground?
I’ll use a \(+y\) up coordinate system.
| Known | Unknown |
|---|---|
| \(y_i = 50\ \textrm{m} \) | \(t\) |
| \(y_f = 0 \) | \(v_{fy}\) |
| \(v_{iy} = -10\ \textrm{m/s}\) | |
| \(a_y = -g\) |
Use equation (12.1) to find time.
\[
\begin{align*}
y_f &= y_i + v_{iy} t + \frac{1}{2}a_yt^2 \\
0 &= y_i + v_{iy} t – \frac{1}{2}gt^2
\end{align*}
\]
Note that this is quadratic in time. Let’s rearrange it slightly to be in a standard form:
\[
0 = \underbrace{-\frac{1}{2}g}_{A}t^2 + \underbrace{v_{iy}}_{B}t + \underbrace{y_i}_{C}
\]
Use the quadratic formula:
\[
\begin{align*}
t &= \frac{-B \pm \sqrt{B^2 – 4AC}}{2A} \\
&= \frac{-v_{iy} \pm \sqrt{(-v_{iy})^2 – 4\left(-\frac{1}{2}g\right)(y_i)}}{2\left(-\frac{1}{2}g\right)} \\
&= \frac{-v_{iy} \pm \sqrt{ v_{iy}^2 + 2gy_i}}{-g} \\
&= \frac{-(-10\ \textrm{m/s}) \pm \sqrt{(-10\ \textrm{m/s})^2 + 2\left(9.81\ \textrm{m/s}^2\right)(50\ \textrm{m})}}{-9.81\ \textrm{m/s}^2}
\end{align*}
\]
This gives two solutions: \(t = -4.37\ \textrm{s}\) and \(t = 2.33\ \textrm{s}\). We know the camera did not travel backwards in time, so we take the positive solution (2.33 s). Now that we know time, we can use equation (12.2) to find the velocity just before it hits the ground:
\[
\begin{align*}
v_{fy} &= v_{iy} + a_yt \\
&= v_{iy} – gt \\
&= (-10\ \textrm{m/s}) – \left(9.81\ \textrm{m/s}^2\right)(2.33\ \textrm{s}) \\
&= -32.9\ \textrm{m/s}
\end{align*}
\]
Example 12.8
A cement block “accidentally” falls from rest from the ledge of a 54 m high building. When the block is 14 m above the ground, a 2 m tall man looks up and notices the block directly above him. How much time does the man have to get out of the way? On average, it takes a person about 300 ms (0.3 s) to react—that is, to notice and get out of the way. Will this man be able to move out of the way in time?
Ultimately, we’re looking for the time it takes the block to go from 14 m to 2 m, the time from when the man sees the block to when it would reach his head. To do so, we’ll need to know the speed of the block at the time that he sees it.
We’ll break this problem into two parts: the block falling from 54 m to 14 m; and then from 14 m to 2 m. For both parts I’ll use a \(+y\) up coordinate system.
Part 1: the block falls from the ledge to a point 14 m above the ground
| Known | Unknown |
|---|---|
| \(y_i = 54\ \textrm{m} \) | \(t\) |
| \(y_f = 14\ \textrm{m} \) | \(v_{fy}\) |
| \(v_{iy} = 0\) | |
| \(a_y = -g\) |
Use equation (12.1) to find the time it takes the block to reach 14 m.
\[
\begin{align*}
y_f &= y_i +v_{iy}t + \frac{1}{2}a_yt^2 \\
y_f &= y_i + (0)t – \frac{1}{2}gt^2 \\
\hookrightarrow t &= \sqrt{\frac{2(y_i – y_f)}{g}}
\end{align*}
\]
Substitute this into equation (12.2) to find the speed of the block at this time. Since we’re not looking for the time specifically, we avoid rounding errors by doing this substitution instead of calculating a value for time.
\[
\begin{align*}
v_{fy} &= v_{iy} + a_yt \\
&= 0-g\sqrt{\frac{2(y_i – y_f)}{g}} \\
&= -\sqrt{g^2\left(\frac{2(y_i – y_f)}{g}\right)} \\
&= -\sqrt{2g(y_i – y_f)}
\end{align*}
\]
As with time, we’re not actually interested in finding a particular value for this speed. However, it will make our work easier for part 2 of the problem. Substituting the values from the table of knowns and unknowns, we find the block is traveling down (the \(-y\) direction) at 28 m/s at this time. Note that you could also find this by analyzing conservation of energy!
Part 2: the block falls from 14 m to the top of the man’s head
| Known | Unknown |
|---|---|
| \(y_i = 14\ \textrm{m} \) | \(t\) |
| \(y_f = 2\ \textrm{m} \) | |
| \(v_{iy} = -28\ \textrm{m/s}\) | |
| \(a_y = -g\) |
Use equation (12.1).
\[
\begin{align*}
y_f &= y_i + v_{iy} t + \frac{1}{2}a_yt^2 \\
\hookrightarrow 0 &= \underbrace{\frac{1}{2}g}_{A}t^2 \underbrace{- v_{iy}}_{B} t + \underbrace{(y_f – y_i)}_{C} \\
\hookrightarrow t &= \frac{-B \pm \sqrt{B^2 – 4AC}}{2A} \\
&= \frac{-(-v_{iy}) \pm \sqrt{(-v_{iy})^2 – 4\left(\frac{1}{2}g\right)(y_f – y_i)}}{2\left(\frac{1}{2}g\right)} \\
&= \frac{v_{iy} \pm \sqrt{ v_{iy}^2 – 2g(y_f – y_i)}}{g} \\
&= \frac{(-28\ \textrm{m/s}) \pm \sqrt{(-28\ \textrm{m/s})^2 – 2\left(9.81\ \textrm{m/s}^2\right)\left((2\ \textrm{m}) – (14\ \textrm{m})\right)}}{9.81\ \textrm{m/s}^2}
\end{align*}
\]
This gives two solutions: \(t = 0.400\ \textrm{s}\) and \(t = -6.11\ \textrm{s}\). We know the block did not travel backwards in time, so we take the positive solution (0.400 s). This is 400 ms, which is greater than the man’s reaction time. He will be able to move out of the way of the cement block 🙂