PHYS 170

Lecture 02b - Motion in One Dimension (The Equations of Motion)

Introduction

Four equations for motion under constant acceleration:

Linear Equations Missing
v=v0+at s
s−s0=v0t+12at2 v
s−s0=12(v0+v)t a
v2=v20+2a(s−s0) t

There are six variables overall: s0,v0 (intial variables), and s,v,a,t (final variables).
Each of the equations is labeled on the right by the final variables that is missing. For example, the first equation has the final variables v,a,t but is missing s, so we will call it the s-equation.

The meaning of the variables:

Symbol Meaning
s0 initial displacement
v0 initial velocity
s final displacement
v final velocity
a acceleration (constant)
t time

If the acceleration is not a constant (i.e. it changes over time) then the equations are invalid. However, in this course, you can assume a is constant unless the question stated otherwise.

Simple Applications

equations_of_motion

Example - riding a bike

A bike was moving at $1.2m/s$, then the rider decided to speed up for $10s$. In the end he was traveling at $3m/s$. How far did the bike travel?

Solution

The information given:
  • $s_0=0m$
  • $v_0=1.2m/s$
  • $s=?$
  • $v=3m/s$
  • $a=$ missing
  • $t=10s$
Note that "$s=?$" above means we are looking for $s$. Out of the final variables $s, v, a, t$, only $a$ does not appear in the list above. This means we should use the $a$-equation: $$ \begin{eqnarray} s-s_0 &=& \frac{1}{2}(v_0+v)t \\ \Rightarrow s &=& \frac{1}{2}(1.2+3)(10) = 21m \end{eqnarray} $$
equations_of_motion

Example - landing a plane

A plane touches down on the runway at $75m/s$ and decelerats at $4.5m/s^2$. What is the minimal length of the runway for plane to land safely?

Solution

Note that since we want the plane to stop in the end, we know $v=0m/s$ even though the information is not explicitly given in the question. Since the plane is slowing down, its acceleration is actually negative:
  • $s_0=0m$
  • $v_0=75m/s$
  • $s=?$
  • $v=0m/s$
  • $a=-4.5m/s^2$
  • $t=$ missing
There is no $t$ above so we should use the $t$-equation: $$ \begin{eqnarray} v^2 &=& v_0^2 + 2a(s-s_0) \\ \Rightarrow s &=& s_0 + \frac{v^2 - v_0^2}{2a} = 625m \end{eqnarray} $$

Try It Yourself (click to show)

equations_of_motion

Exercise - a car speeding down a road 1

A car entered a $10m$-long road at $1.2m/s$, when it exited the road it was traveling at $2.7m/s$. We wish to find the acceleration in the end. Fill in the information below. Type in a question mark '?' for the variable you are asked to find, and type 'missing' for the missing variable.

Solution

$s_0=$
0
0.01
$m$
$v_0=$
1.2
0.02
$m/s$
$s=$
10
0.02
$m$
$v=$
2.7
0.02
$m/s$
$a=$
"?"
"not_number"
$m/s^2$
$t=$
"missing"
"not_number"
$s$
$s_0=$ $m$
$v_0=$ $m/s$
$s=$ $m$
$v=$ $m/s$
$a=$ $m/s^2$
$t=$ $s$
Feedback will appear here.
equations_of_motion

Exercise - a car speeding down a road 2

Based on your answer from the previous question, which equation below should you use to find $a$?
Write a list of the given variables, and see which of the four variables $s, v, a, t$ are missing.

Solution

  • $s_0=0m$
  • $v_0=1.2m/s$
  • $v=2.7m/s$
  • $s=-100m$
  • $a=?$
$t$ is missing, so you should use the $t$-equation.
$v = v_0 + at$
Incorrect. $s$ is given, so you should not use the equation without $s$.
$s - s_0 = v_0t + \frac{1}{2} a t^2$
Incorrect. $v$ is given, so you should not use the equation without $v$.
$s - s_0 = \frac{1}{2} (v_0+v)t$
Incorrect. You want to find $a$, but there is no $a$ in this equation!
$v^2 = v_0^2 + 2a(s-s_0)$
Correct. You are not given $t$ and not asked to find $t$, so the equation without $t$ is the one you want to use.
3
$v = v_0 + at$
$s - s_0 = v_0t + \frac{1}{2} a t^2$
$s - s_0 = \frac{1}{2} (v_0+v)t$
$v^2 = v_0^2 + 2a(s-s_0)$
Feedback will appear here.
equations_of_motion

Exercise - a car speeding down a road 3

Continue from the previous problem. Calculate the numerical value for $a$.

Solution

  • $s_0=0m$
  • $v_0=1.2m/s$
  • $s=10m$
  • $v=2.7m/s$
  • $a=?$
  • $t=$ missing
From the $t$-equation: $$ \begin{eqnarray} v^2 &=& v_0^2 + 2a(s-s_0) \\ \Rightarrow a &=& \frac{v^2-v_0^2}{2(s-s_0)} = \frac{2.7^2-1.2^2}{2(10)} = 0.29m/s^2 \end{eqnarray} $$
0.29
0.02
Select unit:
$m$
$m/s$
$m/s^2$
$s$
2
Select unit:
$m$
$m/s$
$m/s^2$
$s$
Feedback will appear here.
equations_of_motion

Exercise - roller coaster in a tunnel 1

A roller coaster was traveling at $30m/s$ when it entered a tunnel $50m$ long. While it was in the tunnel, it decelerated at the rate of $8m/s^2$. What was the velocity when it came out of the tunnel? Fill in the information below. Type in a question mark '?' for the variable you are asked to find, and type 'missing' for the missing variable.

Solution

$s_0=$
0
0.01
$m$
$v_0=$
30
0.02
$m/s$
$s=$
50
0.02
$m$
$v=$
"?"
"not_number"
$m/s$
$a=$
-8
0.02
$m/s^2$
$t=$
"missing"
"not_number"
$s$
$s_0=$ $m$
$v_0=$ $m/s$
$s=$ $m$
$v=$ $m/s$
$a=$ $m/s^2$
$t=$ $s$
Feedback will appear here.
equations_of_motion

Exercise - roller coaster in a tunnel 2

Based on your answer from the previous question, which equation below should you use?
Write a list of the given variables, and see which of the four variables $s, v, a, t$ are missing.

Solution

$t$ is missing, so you should use the $t$-equation.
$v = v_0 + at$
Incorrect. $s$ is given, so you should not use the equation without $s$.
$s - s_0 = v_0t + \frac{1}{2} a t^2$
Incorrect. $v$ is given, so you should not use the equation without $v$.
$s - s_0 = \frac{1}{2} (v_0+v)t$
Incorrect. You want to find $a$, but there is no $a$ in this equation!
$v^2 = v_0^2 + 2a(s-s_0)$
Correct. You are not given $t$ and not asked to find $t$, so the equation without $t$ is the one you want to use.
3
$v = v_0 + at$
$s - s_0 = v_0t + \frac{1}{2} a t^2$
$s - s_0 = \frac{1}{2} (v_0+v)t$
$v^2 = v_0^2 + 2a(s-s_0)$
Feedback will appear here.
equations_of_motion

Exercise - roller coaster in a tunnel 3

Use the information from the last Try It Yourself problem to find the velocity when it came out of the tunnel?

Solution

  • $s_0=0m$
  • $v_0=30m/s$
  • $s=50m$
  • $a=-8m/s^2$
  • $v=?$
$t$ is missing. From the $t$-equation: $$ \begin{eqnarray} v^2 &=& v_0^2 + 2a(s-s_0) \\ \Rightarrow v &=& \sqrt{v_0^2+2a(s-s_0)} = \sqrt{30^2+2(-8)(50)} = 10m/s \end{eqnarray} $$
10
0.02
Select unit:
$m$
$m/s$
$m/s^2$
$s$
1
Select unit:
$m$
$m/s$
$m/s^2$
$s$
Feedback will appear here.
equations_of_motion

Example - roller coaster in a tunnel 4

Same situation as the last problem, restated here for clarity. A roller coaster was traveling at $30m/s$ when it entered a tunnel $50m$ long. While it was in the tunnel, it decelerated at the rate of $8m/s^2$. How long did it take for the roller coaster to come out of the tunnel?

Solution

  • $s_0=0m$
  • $v_0=30m/s$
  • $s=50m$
  • $a=-8m/s^2$
  • $v=$ missing
  • $t=?$
$v$ is missing. From the $v$-equation: $$ \begin{eqnarray} s - s_0 &=& v_0 t + \frac{1}{2}at^2 \\ \Rightarrow 50 &=& 30t -4t^2 \\ \Rightarrow 4t^2-30t+50 &=& 0 \\ \Rightarrow 2t^2-15t+25 &=& 0 \\ \Rightarrow t &=& \frac{15\pm \sqrt{15^2-4(2)(25)}}{2(2)} = 2.5s \text{ or } 5s \text{(rejected)} \end{eqnarray} $$ We reject the second solution ($t=5s$) to the quadratic equation because the equation assumes constant negative acceleration. When taken literally, means the roller coaster will continue to decelerate even after it comes out of the tunnel at $2.5s$. In this case, the roller coaster will eventually reverse direction and move backward, and finally arriving at the exit of the tunnel for a second time (in the reversed direction) at $5s$. In real life, of course the roller coaster will at some point change its acceleration (perhaps to go at constant velocity after it comes out of the tunnel), therefore the mathematical assumption that $a$ is constant forever is not really valid. As a result the second solution for the return of the roller coaster should be ignored.
equations_of_motion

Example - roller coaster in a tunnel 4 (alternative method)

The last example we used the quadratic equation to solve for $t$. However, there is in fact an easier way to solve this problem.

Solution

Recall that we found $v=10m/s$ in the Try It Yourself problem "roller coaster in a tunnel 3", putting all the information together:
  • $s_0=0m$
  • $v_0=30m/s$
  • $s=50m$
  • $a=-8m/s^2$
  • $v=10m/s$
  • $t=?$
With this extra information, there is no variable missing from the list above. When no variable is missing, it means you have more information than you need to solve the problem. You can then use any of the four equations of motion that contains $t$ to solve it. The shortest equation, namely the $s$-equation, will do nicely: $$ \begin{eqnarray} v &=& v_0 + at \\ \Rightarrow t &=& \frac{v-v_0}{a} \\ &=& \frac{10-30}{-8} = 2.5s \end{eqnarray} $$ This alternative approach gives the same answer as before. While this method is faster, it assummes (dangerously) that your own calculation of $v=10m/s$ is correct. In the exam, it is perhaps unwise to make such a bet. I personally prefer the quadratic method despite it being more tedious because that way I could have gotten $v$ wrong and yet I would still calculate the time correctly.

Acceleration due to Gravity

Gravity pulls everything downward in such a way that the vertical acceleration of an object falling on the surface of the Earth is given by: a=−9.8m/s2 The negative sign denotes the fact that gravity always ponts downward. The value of a does not depend on the mass.

The magnitude of the acceleration due to gravity is defined to be g: g=|a|=9.8m/s2 Note that g is by definition positive (because it is the magnitude), so never write something like g=−9.8m/s2. Be aware of the difference between the notations of a and g, the former is negative and the latter is negative:
  • a=−g=−9.8m/s2
  • g=|a|=9.8m/s2
  • a=9.8m/s2
  • g=−9.8m/s2
Below is a picture of an apple and a piece of feather under free fall inside a vacuum chamber (to eliminate the air resistance). Even though the feather is much lighter than the apple, they both reach the ground at exactly the same time because their downward acceleration is exactly the same.
two simple vectors

Video of Free Fall in Air and in Vacuum (click to show)

equations_of_motion

Example - ball falling from a building

A ball is being dropped from the top of a $100m$-tall building. You want the ball to reach the ground in $4s$. How fast and in what direction should you throw the ball?

Solution

  • $s_0=0m$
  • $v_0=?$
  • $s=-100m$
  • $v=$ missing
  • $a=-g=-9.8m/s^2$
  • $t=4s$
Note that the displacement $s$ is negative because you want the ball to fall down, not up! Since $v$ is missing, we use the $v$-equation: $$ \begin{eqnarray} s - s_0 &=& v_0t + \frac{1}{2}at^2 \\ \Rightarrow v_0 &=& \frac{s-s_0}{t} - \frac{1}{2}at \\ &=& \frac{-100}{4} - \frac{1}{2} (-9.8)(4) = -5.4m/s \end{eqnarray} $$ The negative sign means the initial velocity should be pointing downward.
equations_of_motion

Example - ball falling from a building

Continue from the last example. What is the final velocity of the ball?

Solution

  • $s_0=0m$
  • $v_0=-5.4m/s$
  • $s=-100m$
  • $v=?$
  • $a=-g=-9.8m/s^2$
  • $t=4s$
None of the variables are missing, so we can use any of the equations that contains $v$. We choose the shortest one, namely the $s$-equation: $$ v = v_0 + at = -5.4 + (-9.8) (4) = -44.6m/s $$ The negative sign means the final velocity points downward, as expected.

Try It Yourself (click to show)

equations_of_motion

Exercise - falling from a cliff 1

A ball was dropped downward from the top of a cliff with an intial speed of $1m/s$. The ball later hit the ground at $10m/s$. How much time did it take for the ball to fall to the ground? Fill in the information below. Type in a question mark '?' for the variable you are asked to find, and type 'missing' for the missing variable.
In Physics, the word speed' means the magnitude of the velocity. So when someone say the speed is $10m/s$, it means $v$ could be $+10m/s$ or $-10m/s$. You have to read the question carefully to figure out the sign using the direction.

Solution

Note that the initial and final velocity are both negative because they both point downward according to the question.
$s_0=$
0
0.01
$m$
$v_0=$
-1
0.02
$m/s$
$s=$
"missing"
"not_number"
$m$
$v=$
-10
0.02
$m/s$
$a=$
-9.8
0.02
$m/s^2$
$t=$
"?"
"not_number"
$s$
$s_0=$ $m$
$v_0=$ $m/s$
$s=$ $m$
$v=$ $m/s$
$a=$ $m/s^2$
$t=$ $s$
Feedback will appear here.
equations_of_motion

Exercise - falling from a cliff 2

Based on your answer from the previous question, which equation below should you use?
Write a list of the given variables, and see which of the four variables $s, v, a, t$ are missing.

Solution

$s$ is missing, so you should use the $s$-equation.
$v = v_0 + at$
Correct. You are not given $s$ and not asked to find $s$, so the equation without $s$ is the one you want to use.
$s - s_0 = v_0t + \frac{1}{2} a t^2$
Incorrect. $v$ is given, so you should not use the equation without $v$.
$s - s_0 = \frac{1}{2} (v_0+v)t$
Incorrect. $a$ is given, so you should not use the equation without $a$.
$v^2 = v_0^2 + 2a(s-s_0)$
Incorrect. You want to find $t$, but there is no $t$ in this equation!
0
$v = v_0 + at$
$s - s_0 = v_0t + \frac{1}{2} a t^2$
$s - s_0 = \frac{1}{2} (v_0+v)t$
$v^2 = v_0^2 + 2a(s-s_0)$
Feedback will appear here.
equations_of_motion

Exercise - falling from a cliff 3

Based on your answer to the previous two problems, how much time did it take for the ball to fall to the ground?

Solution

  • $s_0=0m$
  • $v_0=-1m/s$
  • $s=$ missing
  • $a=-g=-9.8m/s^2$
  • $v=-10m/s$
  • $t=?$
$s$ is missing, so we choose the $s$-equation: $$ \begin{eqnarray} v &=& v_0 +at \\ \Rightarrow t &=& \frac{v-v_0}{a} \\ &=& \frac{(-10)-(-1)}{-9.8} = 0.92s \end{eqnarray} $$

In the exam, if you miss any of the negative signs in the calculation, you will lost most of the points of that question even if you have done everything else correctly (and even if you have written down the right answer!). The signs here are of paramount importance and must be handled very carefully!

0.92
0.01
Select unit:
$m$
$m/s$
$m/s^2$
$s$
3
Select unit:
$m$
$m/s$
$m/s^2$
$s$
Feedback will appear here.
equations_of_motion

Exercise - falling from a cliff 4 (reversed direction)

A ball was thrown upward from the top of a cliff with an intial speed of $1m/s$. The ball later hit the ground at $10m/s$. How much time did it take for the ball to fall to the ground? Fill in the information below. Type in a question mark '?' for the variable you are asked to find, and type 'missing' for the missing variable.
This problem is the same as 'falling from a cliff 1' except that the initial velocity now points upward instead of down. The direction change could be specified by choosing the appropriate sign. Up is positive, down is negative.

Solution

Since the inital velocity was upward, $v_0$ is positive. Final velocity $v$ is negative because it points downward.
$s_0=$
0
0.01
$m$
$v_0=$
1
0.02
$m/s$
$s=$
"missing"
"not_number"
$m$
$v=$
-10
0.02
$m/s$
$a=$
-9.8
0.02
$m/s^2$
$t=$
"?"
"not_number"
$s$
$s_0=$ $m$
$v_0=$ $m/s$
$s=$ $m$
$v=$ $m/s$
$a=$ $m/s^2$
$t=$ $s$
Feedback will appear here.
equations_of_motion

Exercise - falling from a cliff 5

Continue from the last problem (the ball was thrown upward). Find the time it took for the ball to fall down the cliff.

Solution

  • $s_0=0m$
  • $v_0=+1m/s$
  • $s=$ missing
  • $v=-10m/s$
  • $a=-g=-9.8m/s^2$
  • $t=?$
$s$ is missing, so we choose the $s$-equation: $$ \begin{eqnarray} v &=& v_0 +at \\ \Rightarrow t &=& \frac{v-v_0}{a} \\ &=& \frac{(-10)-1}{-9.8} = 1.12s \end{eqnarray} $$

In the exam, if you miss any of the negative signs in the calculation, you will lost most of the points of that question even if you have done everything else correctly (and even if you have written down the right answer!). The signs here are of paramount importance and must be handled very carefully!

1.12
0.01
Select unit:
$m$
$m/s$
$m/s^2$
$s$
3
Select unit:
$m$
$m/s$
$m/s^2$
$s$
Feedback will appear here.
equations_of_motion

Exercise - falling from a cliff 6

Continue from the last problem 'falling from a cliff 5'. What is the final displacment of the ball? How tall is the cliff?

Solution

  • $s_0=0m$
  • $v_0=1m/s$
  • $s=?$
  • $v=-10m/s$
  • $a=-g=-9.8m/s^2$
  • $t=$ missing
Even though you calculated $t$ from the last problem, it is a good idea to not use it because you might have made a mistake in your calculation of $t$. Without using $t$ from the last problem, we treat $t$ as missing, so we choose the $t$-equation: $$ \begin{eqnarray} v^2 &=& v_0^2 + 2 a (s-s_0) \\ \Rightarrow s &=& \frac{v^2-v_0^2}{2a} \\ &=& \frac{(-10)^2-1^2}{2(-9.8)} = -5.05m \end{eqnarray} $$ The negative sign means the ball moved downward by $5.05m$. This implies that the height of the cliff is $5.05m$. While the displacement of the ball is negative, you should not give a negative answer as the height of the cliff (how could a cliff be negatively tall?).

In the exam, if you miss any of the negative signs in the calculation, you will lost most of the points of that question even if you have done everything else correctly (and even if you have written down the right answer!). The signs here are of paramount importance and must be handled very carefully!

$s=$
-5.05
0.02

height of cliff =
5.05
0.02
Select unit:
$m$
$m/s$
$m/s^2$
$s$
0
$s=$
height of cliff =
Select unit:
$m$
$m/s$
$m/s^2$
$s$
Feedback will appear here.