Kinematics is a branch of physics that models the motion of objects using position, velocity, acceleration, and time.
We've already learned about the equations for average velocity and average acceleration. Those equations are useful, but they don't give exact values for velocity or acceleration, just an average.
In the position vs. time graph below, each path from A to B has a different acceleration, but they all have some properties in common.
Example: How much time does each path from A to B take?solution
They all take the same time.
$$\Delta t = t_f - t_i$$ $$\Delta t = 50\,\mathrm{s}-5\,\mathrm{s}$$ $$\Delta t = 45\,\mathrm{s}$$Example: What is the displacement for each path.
solution
Each path travels a different distance, but they all have the same displacement.
$$\Delta x = x_f - x_i$$ $$\Delta x = 25\,\mathrm{m}-5\,\mathrm{m}$$ $$\Delta x = 20\,\mathrm{m}$$Example: What is the average velocity for each path?
solution
Each path ends up with the same total displacement over the same period of time. This means they all have the same average velocity.
The average velocity equals the displacement divided by the time period.
$$v_{\mathrm{avg}} = \frac{\Delta x}{\Delta t}$$ $$v_{\mathrm{avg}} = \frac{25\, \mathrm{m}-5\, \mathrm{m}}{50\, \mathrm{s}-5\, \mathrm{s}}$$ $$v_{\mathrm{avg}} = \frac{20\, \mathrm{m}}{45\, \mathrm{s}}$$ $$v_{\mathrm{avg}} = 0.\overline{44} \, \mathrm{\frac{m}{s}}$$Question: What is different about each path?
answer
Each path has a different:
There are many ways to move from A to B, but if acceleration is limited to a constant value only one path works. When graphed as position vs time the path is a parabola.
Question: Set the acceleration to 0.05 m/s². Write a short description of the object's velocity as it moves from A to B.answer
(at the default A and B positions)
Constant Acceleration
When acceleration is constant we can predict position and velocity at any point in time. These predictions come from the equations of motion.
derivation of the equations of motion
The first equation is the average acceleration equation rearranged with acceleration as a constant value.
$$a_{\mathrm{avg}} = \frac{\Delta v}{\Delta t}$$ $$a = \frac{v - u}{\Delta t}$$ $$a \Delta t = v - u$$ $$\large \boxed{v = u + a \Delta t}$$When acceleration is constant, velocity changes at a constant rate. This means that the average velocity equals half of the sum of initial and final velocities.
$$v_{\mathrm{avg}} = \tfrac{1}{2}(v+u)$$ $$\frac{\Delta x}{\Delta t} = \tfrac{1}{2}(v+u)$$ $$ \boxed{ \Delta x = \tfrac{1}{2}(v+u)\Delta t}$$The next one also starts with the average velocity for constant motion equation form above. We can plug our previous equation into this to remove the final velocity.
$$v_{\mathrm{avg}} = \tfrac{1}{2}(v+u)$$ $$v_{\mathrm{avg}} = \tfrac{1}{2}((a \Delta t+u)+u)$$ $$v_{\mathrm{avg}} = u + \tfrac{1}{2}a \Delta t$$ $$\frac{\Delta x}{\Delta t} = {u + \tfrac{1}{2}a \Delta t}$$ $$ \boxed{ \Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^2 }$$The last equation comes from eliminating time.
$$v = u + a \Delta t$$ $$\Delta t = \frac{v-u}{a}$$ $$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^2 $$ $$\Delta x = u \left(\frac{v-u}{a}\right) + \tfrac{1}{2}a \left(\frac{v-u}{a}\right)^2 $$ $$a\Delta x = u(v-u)+ \tfrac{1}{2}(v-u)^2 $$ $$2a\Delta x = 2u(v-u)+ (v-u)^2 $$ $$2a\Delta x = (2uv-2u^2)+ (v^2 - 2uv + u^2) $$ $$2a\Delta x = v^2 - u^2 $$ $$u^2 = v^2 -2a\Delta x$$ $$ \boxed{v^2 = u^2 +2a\Delta x}$$$$v = u+a \Delta t$$ $$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$ $$\Delta x = \tfrac{1}{2}(v+u)\Delta t$$ $$v^{2} = u^{2}+2a \Delta x$$
\(\Delta x\) = displacement [m] vector
\(\Delta t\) = time period [s]
\(v\) = final velocity [m/s] vector
\(u\) = initial velocity [m/s] vector
\(a\) = acceleration [m/s²] (constant) vector
Each equation is missing one of the five variables. Looking for the missing variable can help you choose the right equation for each situation.
Working with multiple equations can be complicated. It helps to follow steps:
- List the known and unknown variables.
- Choose an equation with only one unknown variable.
- Plug the known numbers into the equation.
- Simplify the numbers, then use algebra to isolate the unknown variable.
- Check if your answer agrees with your intuition and expected units.
solution
list known and unknown variables
$$u = 30 \, \mathrm{\tfrac{m}{s} }$$ $$v = 0 \, \mathrm{\tfrac{m}{s}}$$ $$\Delta x = 10 \, \mathrm{m}$$ $$a =\, ?$$identify a matching equation
$$v^{2} = u^{2}+2a \Delta x$$plug in values
$$0^2 = 30^2 + 2a(10)$$solve for the unknown
$$0 = 900 + 20a$$ $$-900 = 20a$$ $$\frac{-900}{20} = a$$ $$-45 \mathrm{\tfrac{m}{s^{2}}}=a$$check intuition and units
A car coming to a quick stop should have a large negative acceleration. The units are correct for acceleration. Our answer looks good!
You can tell what variable a number is referencing from the units. So 21 s is a time period, 21 m/s is velocity, and 21 m/s² is acceleration.
Example: A bullet aimed straight up leaves the barrel of a gun at 400 m/s. It accelerates down at 9.8 m/s². If the bullet travels for 40.8 s before stopping how far up did it go? Ignore air friction.solution
$$u = 400 \, \mathrm{\tfrac{m}{s}}$$ $$a = -9.8 \, \mathrm{\tfrac{m}{s^{2}}}$$ $$\Delta t = 40.8 \, \mathrm{s}$$ $$\Delta x =\, ?$$$$\Delta x = u\Delta t + \tfrac{1}{2} a \Delta t^{2}$$ $$\Delta x = (400\, \mathrm{\tfrac{m}{s}}) (40.8\, \mathrm{s}) + \tfrac{1}{2} (-9.8 \, \mathrm{\tfrac{m}{s^{2}}})(40.8 \, \mathrm{s})^{2}$$ $$\Delta x = 16320\, \mathrm{m} - 8157\, \mathrm{m}$$ $$\Delta x = 8163 \, \mathrm{m}$$
answer
- free fall with no air friction (a = 9.8 m/s²)
- just being at rest (v = 0 m/s) (a = 0 m/s²)
- moving at a constant speed (a = 0 m/s²)
Question: What are some situations when acceleration is NOT constant?
answer
- speeding up (accelerating) (a > 0)
- hitting the ground (a = -9.8 → a > 0 → a = 0)
- dancing (a = ?)
Assume constant acceleration from the plane's engines.
unit conversion: 1m = 3.3 ft , 1 mile = 1609 m
solution
$$\begin{aligned} v &= \mathrm{170\left(\frac{mile}{hour}\right)\left(\frac{1609 \, m}{1 \, mile}\right)\left(\frac{1 \, hour}{3600\,s}\right) = 76\,\mathrm{\tfrac{m}{s}} } \\ u &= \mathrm{rest} = 0 \\ \Delta &x = 6000\,\mathrm{ft}\scriptsize \left(\frac{1 \, \mathrm{m}}{3.3 \, \mathrm{ft} }\right)\normalsize = 1818 \, \mathrm{m} \\ a &= \,? \end{aligned}$$$$v^{2} = u^{2}+2a \Delta x$$ $$76^2=0^2+2a(1818)$$ $$5776=3636a$$ $$\frac{5776}{3636}=a$$ $$1.589 \, \mathrm{\tfrac{m}{s^{2}} }=a$$
Acceleration of Gravity
On Earth's surface everything is pulled down at 9.8 m/s². The rate is the same for cars, birds, puppies, apples, balloons, and everything.
Effects like air friction, thrust, and buoyancy can change the perceived acceleration of gravity. To keep things simple I will ignore these effect for the example problems.
solution
The acceleration of gravity comes from massive objects. Every planet, star, moon, and asteroid has a different surface gravity.
| name | g (m/s²) | |
|---|---|---|
|
|
Sun | 275 |
| Mercury | 3.7 | |
| Venus | 8.9 | |
| Earth | 9.8 | |
| Moon | 1.6 | |
| Mars | 3.7 | |
| Ceres | 0.27 | |
| Jupiter | 25.8 | |
|
|
Saturn | 10.4 |
|
|
Uranus | 8.7 |
| Neptune | 11.2 |
Example: Imagine a meteor 4000 m above the Moon falls from rest. What is the impact velocity of the meteor? How long does it take for the meteor to hit the surface?
solution
$$v = u+a \Delta t$$ $$-113=0+(-1.6)\Delta t$$ $$\frac{-113}{-1.6}=\Delta t$$ $$70.6\, \mathrm{s} = \Delta t $$
It seems like we could set the final velocity to 0 when an object hits the ground, but colliding with the ground changes the acceleration of the object and it breaks our equations of motion.
We have to end problems before they collide with the ground to keep acceleration constant at 9.8 m/s².
Example: You drop a rock from 2.0 meters above the ground. It hits the ground after 1.03 seconds. What planet, moon, or asteroid are you on?solution
We can solve for the acceleration and compare it to the surface gravities on the chart.
People often use words to communicate numbers. Words like "rest" or "stop" tell you velocity is equal to 0.
Also, if you throw an object up and gravity slows it down the velocity will be 0 at the peak of it's path, at the maximum height.
Example: An object is thrown straight up from ground level at 19.6 m/s. Calculate its maximum height.solution
$$u = +19.6\,\mathrm{m/s}$$ $$v = 0\,\mathrm{m/s}$$ $$a = -9.8\,\mathrm{m/s^2}$$ $$v^2 = u^2 + 2a\Delta x$$ $$0^2=19.6^2+2(-9.8)\Delta x$$ $$0=384.16-19.6\Delta x$$ $$19.6\Delta x=384.16$$ $$\Delta x=\frac{384.16}{19.6}$$ $$\Delta x = 19.6\,\mathrm{m}$$2-D Motion (for constant acceleration)
Solving for 2-Dimensional motion can reuse the methods from 1-Dimension if we divide the problem up into 2 directions. We also should set the two directions to be perpendicular so that they are independent from each other. This gives each direction unrelated positions, velocities, and accelerations, but they still share the same time period.
| Δt = | |
| Δx = | Δy = |
| u = | u = |
| v = | v = |
| a = | a = |
You can organize the variables in columns for x and y with shared time.
setup
Let's make the x-direction east-west, and the y-direction north-south, like a compass. East and north will be positive, while south and west will be negative.
| Δt = 10 s | |
| Δx = ? | Δy = ? |
| u = 5 m/s | u = 3 m/s |
| v | v |
| a = -2 m/s² | a = 1 m/s² |
solution
We'll start with the north-south, y direction.
$$\Delta y = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$ $$\Delta y = (3)(10) + \tfrac{1}{2}(1)(10)^{2}$$ $$\Delta y = 30 + 50$$ $$\Delta y = 80 \, \mathrm{m}$$The east-west, x direction can use the same equation.
$$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$ $$\Delta x = (5)(10) + \tfrac{1}{2}(-2)(10)^{2}$$ $$\Delta x = 50 + -100$$ $$\Delta x = -50 \, \mathrm{m}$$The object moved 80 m north and 50 m west. We can also calculate the displacement with the Pythagorean theorem.
$$d^2 = x^2 + y^2$$ $$d^2 = (-50)^2 + (80)^2$$ $$d^2 = 8900$$ $$d = 94 \, \mathrm{m}$$2-D Projectile Motion
When solving for objects in free fall on the surface of the Earth you can let the x-direction be horizontal and the y-direction be vertical. This choice puts the acceleration from gravity in only the y-direction.
| Δt = | |
| ↔ horizontal | ↕ vertical |
| Δx = | Δy = |
| u = | u = |
| v = | v = |
| a = 0 | a = -9.8 m/s² |
Vectors directed down or left are negative and vectors directed up or right are positive.
setup
| Δt = | |
| Δx = ? | Δy = -1.5 m |
| u = 2 m/s | u = 0 |
| v = | v = |
| a = 0 | a = -9.8 m/s² |
solution
$$\Delta y = u\Delta t + \tfrac{1}{2}a \Delta t^2$$ $$-1.5 = 0\Delta t + \tfrac{1}{2}(-9.8)\Delta t^2$$ $$-1.5 = -4.9\Delta t^2$$ $$-4.9\Delta t^2 = -1.5$$ $$\sqrt{\Delta t^2} = \sqrt{0.31}$$ $$\Delta t = \pm 0.55$$The negative time answer is valid, but not what we are looking for.
$$\Delta t = 0.55 \, \mathrm{s}$$With this new information we can use time to solve on the horizontal side.
| Δt = 0.55 s | |
| Δx = ? | Δy = 1.5 m |
| u = 2 m/s | u = 0 |
| v = 2 m/s | v = -5.42 m/s |
| a = 0 | a = 9.8 m/s² |
Aim and click to fire. Please be careful not to hit each other.
| Δt = | |
| Δx = | Δy = |
| u = | u = |
| v = | v = |
| a = 0 | a = -9.8 m/s² |
answer
The accelerations and initial velocities stay constant. Also the final horizontal velocity is constant.
Question: Why do all the shots that land on the ground end with a negative Δy?
answer
$$\Delta y = y_f - y_i$$Δy is the vertical displacement. It measures the difference between the starting height and the ending height. It doesn't matter how high the object goes, if it starts on the ground and ends on the ground the vertical displacement is going to be zero.
The Δy is negative because the tank's turret is above ground. When the projectile ends up on the ground, it is lower then the starting point on the turret.
Question: Where on the projectile's arc is its vertical velocity zero?
answer
The vertical velocity is zero at the top of the arc, the highest point.
solution
strategy
Warning: This is a long complicated example problem. Take your time. Get organized. Some equations will be dead ends. Other equations will lead to the quadratic equation. You can avoid using the quadratic equation if you solve for the final vertical velocity before you solve for time.
Start with the 2 column structure. You already know the accelerations.
| Δt = | |
| Δx = | Δy = |
| u = | u = |
| v = | v = |
| a = 0 | a = -9.8 m/s² |
You can find the x and y part of the initial velocity with trigonometry.
You also know Δy. It's just the difference between the starting and ending point in the vertical direction. The path of the ball doesn't matter, just look at the difference.
solution
| Δt = ? | |
| Δx = ? | Δy = -2.0 m |
| u = 10 cos(60) = 5.0 m/s | u = 10 sin(60) = 8.66 m/s |
| v = ? | v = ? |
| a = 0 | a = -9.8 m/s² |
$$v = u+a \Delta t$$ $$-10.7 = 8.66+(-9.8)\Delta t$$ $$1.97 \, \mathrm{s} = \Delta t$$
$$\Delta x = u\Delta t + \tfrac{1}{2}a \Delta t^{2}$$ $$\Delta x = 5(1.97) + \tfrac{1}{2}(0)(1.97)^{2}$$ $$\Delta x = 9.85 \, \mathrm{m}$$
| Δt = 1.97 s | |
| Δx = 9.85 m | Δy = -2.0 m |
| u = 10 cos(60) = 5.0 m/s | u = 10 sin(60) = 8.66 m/s |
| v = 5.0 m/s | v = -10.7 m/s |
| a = 0 | a = -9.8 m/s² |
strategy
At first it might seem like there isn't enough information to find the vertical distance. The trick is to end the problem at the highest point on the arc which makes the final velocity 0.
solution
| Δt | |
| Δx | Δy = ? |
| u = 5 m/s | u = 8.7 m/s |
| v = 5 m/s | v = 0 |
| a = 0 | a = -9.8 m/s² |
We don't need to use any horizontal information.
$$v^2 = u^2 + 2 a \Delta y$$ $$0^2=8.7^2+2(-9.8)\Delta y$$ $$0=75.69-19.6\Delta y$$ $$19.6\Delta y=75.69$$ $$\Delta y=\frac{75.69}{19.6}$$ $$\Delta y = 3.86 \, \mathrm{m}$$solution
| Δt = ? | |
| Δx = | Δy = 0 m |
| u = | u = 150 sin(35) = 86.0 m/s |
| v = | v = |
| a = 0 | a = -9.8 m/s² |
We don't have to use the quadratic equation if we factor out time and solve two separate equations.
$$0 = \Delta t(86.0 + \small\frac{1}{2}(-9.8)\Delta t)$$ $$0=\Delta t \quad \quad 0 = 86.0 + \small\frac{1}{2}(-9.8)\Delta t$$The solution of zero time is technically correct, but not interesting.
$$0 = 86.0 + \small\frac{1}{2}(-9.8)\Delta t$$ $$-86.0 = -4.9\Delta t$$ $$\Delta t = 17.6 \, \mathrm{s}$$In this simulation we can fire boxes at a wall. Use 2-D kinematics calculations to predict what height the gap in the wall should be to let a box through.
We need to send a box through the gap in the wall again. This time we can change the initial vertical velocity to get it through.