2.4 Kinematic Equations
On pp. 46-47 of the text, four kinematic equations are derived:
𝑣 = 𝑣0 + 𝑎𝑡
1
𝑥 = 𝑥0 + 𝑣 + 𝑣0 𝑡
2
1 2
𝑥 = 𝑥0 + 𝑣0 𝑡 + 𝑎𝑡
2
𝑣 2 = 𝑣02 + 2𝑎(𝑥 − 𝑥0 )
These all carry the assumptions that (1) acceleration is constant,
and (2) t0 = 0. If the latter assumption is not the case, you can
still use these equations, making the substitution t → t – t0.
You will use these equations on countless occasions throughout
PC141. Make sure to include them on your exam cheat sheets!
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 1
2.4 Kinematic Equations
We have seen that the slope of a velocity –
vs – time graph is equal to the
instantaneous acceleration.
We can gain more information by
examining the area under the line. In the
top graph on the right, 𝑣0 = 0, and 𝑣 = 𝑎𝑡.
The area is that of a triangle with base t
1
1
and height v. This area is 𝐴 = 𝑣𝑡 = 𝑎𝑡 2 .
2
2
But (from the 3rd equation on the last slide)
this is equal to 𝑥 − 𝑥0 = ∆𝑥.
The same result is obtained even when
𝑣0 ≠ 0 (bottom plot), as you should check
for yourself.
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 2
2.4 Kinematic Equations
As it turns out, the area under the v(t)
curve is equal to ∆𝑥 even when
acceleration is not constant. However, to
prove this (or to apply it in problem solving)
requires calculus, so we won’t worry about
it.
The same approach can be used to
determine ∆𝑣 using an acceleration – vs –
time graph. Later in today’s class, we will
tackle a problem that involves piecewise
continuous acceleration.
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 3
Problem #1: Average Speed
WBL LP 2.15
An object accelerates uniformly from rest for t seconds.
The object’s average speed for this time interval is
1
A 2 𝑎𝑡
1 2
B 2 𝑎𝑡
PC141 Intersession 2013
C
2𝑎𝑡
D
2𝑎𝑡 2
Day 4 – May 13 – WBL 2.4-2.5
Slide 4
Problem #2: From the Earth to the Moon
In his 1865 novel From the Earth to the Moon, Jules Verne proposed sending
people to the moon via a rocket cannon that accelerates a capsule along a length
of 220 m, attaining a final speed of 11.2 km/s. What acceleration would the
passengers experience?
Fun(ish) fact: 11.2 km/s is the “escape velocity” from
Earth. An object must attain this velocity if it is to
break free from the gravitational pull of Earth.
Solution: In class
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 5
Problem #3: A Blow to the Head
A soccer ball is headed with and without the use of a helmet. In both cases, the
acceleration of the player’s head is plotted as a function of time. At time t = 7 ms,
what is the difference in the speed acquired by the bare head and the speed
acquired by the helmeted head?
Solution: In class
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 6
Problem #4: Sprinting
The current world record for the 100 m dash is 9.58 seconds, set by Usain Bolt in
2009. Analysis of that run shows that he reached his top speed around the 70
metre mark, at which point his speed was 12.42 m/s. What was his average
speed during the race? Assuming that his acceleration was constant over the first
70 metres, what was the magnitude of his acceleration during that portion of the
race?
Solution: In class
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 7
Problem #5: Worst. Car. Ever.
The “slowest” car ever produced is the 2009 Tata Nano, which accelerates from 0
to 60 miles per hour in a blistering 29.4 seconds. What is the magnitude of this
acceleration, assuming that it is constant over this duration? How much distance
does it cover during this time?
Solution: In class
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 8
Problem #6: Sprinter vs. Car
It seems that the 2009 Tata Nano has a much weaker acceleration that Usain
Bolt. How much of a head start would the car require in order to “beat” Usain
Bolt over 100 metres?
Solution: In class
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 9
2.5 Free Fall
The most obvious example of constant acceleration occurs when
an object moves near the Earth’s surface (by “near” we mean
anywhere within ±100 km or so of sea level). Here, there is a
constant acceleration directed downward, with a magnitude of
𝑔 = 9.80 m/s 2
The symbol “g” indicates that gravity is responsible for this
acceleration…but you didn’t hear that from me, since we’re not
discussing dynamics (the why of motion) until chapter 4.
In chapter 7 we will learn where the number “9.80” comes from.
For now, just know that it is affected not only by altitude, but
also by latitude and longitude, for various reasons. Therefore, it
is not useful to quote g with any more significant figures than
that which is shown above.
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 10
2.5 Free Fall
When an object’s motion is influenced only by g, we say that it is
in free fall. Although this term seems to imply that the object is
moving downward, it in fact refers to any motion that is subject
only to g, even if it projected upward or at an angle (such as a
drive of a golf ball). Remember, even when an object is traveling
upward, it is still accelerating downward. This is what causes it
to slow down, momentarily stop, then begin its downward
journey.
Notice that g doesn’t depend on any
properties (mass, shape…) of the
moving object. This is only true in a
vacuum. Otherwise, air resistance
comes into play, and the object isn’t
truly in free fall.
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 11
2.5 Free Fall
When analyzing objects as they move vertically, it is customary
to label the vertical axis as y, and to let y increase in the upward
direction. In this case:
1. We can still use all of the 1D kinematics equations from this
chapter, replacing x with y.
2. When applying these equations, use an acceleration 𝑎 =
− 𝑔 = −9.80 m/s 2 . The negative sign indicates that
acceleration is in the negative y direction. If you forget the
negative sign, you’re essentially claiming that your object is
𝑣 = 𝑣0 − 𝑔𝑡
“falling up”!
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
1 2
𝑦 = 𝑦0 + 𝑣0 𝑡 − 𝑔𝑡
2
1
𝑦 = 𝑦0 + 𝑣 + 𝑣0 𝑡
2
𝑣 2 = 𝑣02 − 2𝑔(𝑦 − 𝑦0 ) Slide 12
2.5 Free Fall
Here are a few facts about free fall motion for an object thrown
vertically upward. These may be of benefit in solving more
difficult problems later on…
1. If the object returns to its original elevation, the upward and
downward portions of the journey take the same time.
2. If the object returns to its original elevation, it does so with
the same speed as that at which it was launched. The
velocity isn’t the same, since the direction has changed. This
concept will be beneficial when we study kinetic and
potential energy.
3. At the top of the object’s trajectory, its velocity is
momentarily zero.
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 13
Problem #7: Dropping Rocks
You drop a rock off a bridge. When the rock has fallen 4 m, you
drop a second rock. As the two rocks continue to fall, what
happens to their separation?
A
The separation increases as they fall
B
The separation stays constant at 4 m
C
The separation decreases as they fall
D
It is impossible to answer without more information
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 14
Problem #8: Your Idiot Upstairs Neighbour
WBL EX 2.73
A particular apartment building has floors which are vertically separated by 4.0
m. Somebody from an upper floor drops an apple from their window. A resident
on the 1st floor measures the time that the apple takes to pass across his entire
window, which measures 1.35 m vertically. From what floor was the apple
dropped?
Solution: In class
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 15
Problem #9: Vertical Rocket
WBL EX 2.75
A model rocket is fired vertically upward, with a constant acceleration of 12.0
m/s2. When it reaches a height of 1000 m, the engines immediately cut off,
leaving the model rocket in free fall. What is the rocket’s speed at the moment
the engines cut off? What is the maximum altitude of the rocket? What is the
time it takes to reach this maximum altitude, measured from the moment of
launch from the ground?
Solution: In class
PC141 Intersession 2013
Day 4 – May 13 – WBL 2.4-2.5
Slide 16