Introduction to Mechanics
Kinematics Equations (Part 2)
Lana Sheridan
De Anza College
Jan 21, 2016
Last time
• introduced the kinematics equations
Overview
• rest of the kinematics equations
• derivations
• using the equations
The Kinematics Equations
For constant acceleration:
v = v0 + at
1
∆x = v0 t + at 2
2
1
∆x = vt − at 2
2
v0 + v
∆x =
t
2
v 2 = v02 + 2 a ∆x
For zero acceleration:
∆x = vt
The Kinematics Equations: the “no-initial-velocity”
equation
We can build a very similar equation to that last one.
This time we rearrange v = v0 + at to give:
v0 = v − at
And put that into the equation
v0 + v
∆x =
t
2
(v − at) + v
∆x =
2
1 2
= vt − at
2
For constant acceleration:
1
x(t) = x0 + vt − at 2
2
t
(4)
The Kinematics Equations: the “no-time” equation
The last equation we will derive is a scaler equation.
The Kinematics Equations: the “no-time” equation
The last equation we will derive is a scaler equation.
v0 + v
∆x =
t
2
We could also write this as:
(∆x) i =
v0 + v
t i
2
where ∆x, vi , and vf could each be positive or negative.
The Kinematics Equations: the “no-time” equation
The last equation we will derive is a scaler equation.
v0 + v
∆x =
t
2
We could also write this as:
(∆x) i =
v0 + v
t i
2
where ∆x, vi , and vf could each be positive or negative.
We do the same for equation (1):
v i = (v0 + at) i
The Kinematics Equations: the “no-time” equation
The last equation we will derive is a scaler equation.
v0 + v
∆x =
t
2
We could also write this as:
(∆x) =
v0 + v
t
2
where ∆x, vi , and vf could each be positive or negative.
We do the same for equation (1):
v = (v0 + at)
Rearranging for t:
t=
v − v0
a
The Kinematics Equations: the “no-time” equation
v − v0
;
t=
a
∆x =
v0 + v
2
t
Substituting for t in our ∆x equation:
v0 + v
v − v0
∆x =
2
a
2a∆x = (v0 + v )(v − v0 )
so,
v 2 = v02 + 2 a ∆x
(5)
Example 2-7, pg 34
Jets at JFK International Airport accelerate from rest at one end
of a runway, and must attain takeoff speed before reaching the
other end of the runway.
(a) Plane A has acceleration a and takeoff speed vto . What is the
minimum length of runway, ∆xA , required for this plane? Give a
symbolic answer.
(b) Plane B has the same acceleration as plane A, but requires
twice the takeoff speed. Find ∆xB and compare with ∆xA .
(c) Find the minimum runway length for plane A if a = 2.20 m/s2
and vto = 95.0 m/s. (These values are typical for a 747 jetliner.)
nt-Acceleration Equation of Motion: Velocity in Terms of Displacement
0
2
Example 2-7,2 pg 34
+ 2a1x - x02 = v0 + 2a¢x
2–12
Jets at JFK International Airport accelerate from rest at one end
of a runway,
and the
must
attainattakeoff
speed to
before
reaching
the
ation allows
us to relate
velocity
one position
the velocity
at ansition, other
without
how much time is involved. The next Example
endknowing
of the runway.
ow Equation
2–12Acan
used.
(a) Plane
hasbeacceleration
a and takeoff speed vto . What is the
minimum length of runway, ∆xA , required for this plane? Give a
symbolic answer.
ner (b) Plane B has the same acceleration as plane A, but requires
celeratetwice
from rest
one endspeed.
of a runway,
takeoffwith
speed
before
theattakeoff
Findand
∆xBmust
andattain
compare
∆x
A.
way. (a) Plane
A
has
acceleration
a
and
takeoff
speed
What
is
the
minimum
.
v
to plane A if a = 2.20 m/s2
(c) Find the minimum runway length for
r this plane? Give a symbolic answer. (b) Plane B has the same acceleration as
and vto = 95.0 m/s. (These values are typical for a 747 jetliner.)
keoff speed. Find ¢xB and compare with ¢xA. (c) Find the minimum runway
2
and vto = 95.0 m/s. (These values are typical for a 747 jetliner.)
Sketch:
o be the dihe acceleramilarly, the
s.
v0 = 0
a
v = vto
x
!x
¢x, the distance the plane travels in attaining takeoff speed, in terms of the
Example 2-7, pg 34
(a) Plane A has acceleration a and takeoff speed vto . What is the
minimum length of runway, ∆xA , required for this plane? Give a
symbolic answer.
Hypothesis: ∆xA will be directly proportional to vto and inversely
proportional to the acceleration, a.
Example 2-7, pg 34
(a) Plane A has acceleration a and takeoff speed vto . What is the
minimum length of runway, ∆xA , required for this plane? Give a
symbolic answer.
Hypothesis: ∆xA will be directly proportional to vto and inversely
proportional to the acceleration, a.
(b) Plane B has the same acceleration as plane A, but requires
twice the takeoff speed. Find ∆xB and compare with ∆xA .
Hypothesis: ∆xB will be twice as big as ∆xA .
Example 2-7, pg 34
(a) Plane A has acceleration a and takeoff speed vto . What is the
minimum length of runway, ∆xA , required for this plane? Give a
symbolic answer.
Hypothesis: ∆xA will be directly proportional to vto and inversely
proportional to the acceleration, a.
(b) Plane B has the same acceleration as plane A, but requires
twice the takeoff speed. Find ∆xB and compare with ∆xA .
Hypothesis: ∆xB will be twice as big as ∆xA .
(c) Find the minimum runway length for plane A if a = 2.20 m/s2
and vto = 95.0 m/s.
Hypothesis: I guess a minimum runway length would be about 2
football fields, about 200 m.
Example 2-7, pg 34
Given: a, v = vto , v0 = 0.
Asked for: ∆x.
Strategy: Use equation
v 2 = v02 + 2a(∆x)
Example 2-7, pg 34
Given: a, v = vto , v0 = 0.
Asked for: ∆x.
Strategy: Use equation
v 2 = v02 + 2a(∆x)
(a) Rearrange for ∆x:
0
2
vto
2
= v07
+ 2a(∆xA )
∆xA =
2
vto
2a
Example 2-7, pg 34
Given: a, v = vto , v0 = 0.
Asked for: ∆x.
Strategy: Use equation
v 2 = v02 + 2a(∆x)
(a) Rearrange for ∆x:
0
2
vto
2
= v07
+ 2a(∆xA )
∆xA =
2
vto
2a
(b) Now for plane B, v = 2vto :
(2vto )2 = 2a(∆xB )
2
4vto
∆xB =
2a
∆xB = 4∆xA
Example 2-7, pg 34
(c) Find the minimum runway length for plane A if a = 2.20 m/s2
and vto = 95.0 m/s.
Using our expression from (a):
∆xA =
=
2
vto
2a
(95.0 m/s)2
2(2.20 m/s2 )
= 2.05 × 103 m
Example 2-7, pg 34
Analysis: My hypotheses were not correct!
For part (a), the distance was inversely proportional to the
acceleration, but it was proportional to the square of the takeoff
velocity.
For part (b), ∆xB is four times as big as ∆xA . This makes sense
because the distance is proportional to the square of the speed.
In part (c), my guess was an order of magnitude too small. The
747 is one of the biggest commercial jets, so it makes sense that it
needs a long runway. Looking at Google Maps, the length of the
SJC airport is about 3 km, which makes sense if runways need to
be at least 2 km.
The Kinematics Equations
For constant acceleration:
v = v0 + at
1
∆x = v0 t + at 2
2
1
∆x = vt − at 2
2
v0 + v
∆x =
t
2
v 2 = v02 + 2 a ∆x
For zero acceleration:
∆x = vt
The Kinematics Equations Summary
For constant acceleration:
v = v0 + at
1
∆x = v0 t + at 2
2
1
∆x = vt − at 2
2
v0 + v
t
∆x =
2
v 2 = v02 + 2 a ∆x
For zero acceleration:
x = vt
Summary
• kinematics equations for constant acceleration
Homework Walker Physics:
• Ch 2, onward from page. Problems: 67, 97, 101, 121