• t = 0 mins–30 mins: the negative gradient indicates that the car moves
with uniform velocity to the left, past the origin to a distance 20 km away
from the origin. Calculating the gradient of the line, the displacement
∆s = −30 km in ∆t = 30 mins suggests a uniform velocity of
vs =
−30 × 103 m
= −16.67 m s−1 .
30 × 60 s
• t = 30 mins–40 mins: the straight line again implies uniform motion, but
the horizontal line indicates zero displacement during the 10 mins. The
car is thus at rest for 10 mins.
• t = 40 mins–80 mins: the positive gradient indicates the car is moving
with uniform velocity to the right - i.e. towards the origin. The displacement of ∆s = +20 km in ∆t = 40 mins suggests a uniform velocity
of
vs =
+20 × 103 m
= +8.33 m s−1 .
40 × 60 s
• t = 80 mins: car has returned to the origin.
For an additional example, see Additional Materials.
3.4
5
Acceleration
We define the acceleration, a, to be the rate of change of velocity
with respect to time.
• Since it involves v s , as is a vector quantity.
• SI unit of acceleration is m s−2 or m/s2 .
Note: although in common parlance, ”slowing down” is known as ”deceleration”, we still view it as a change in velocity with time, and is thus acceleration.
5
Examples of Motion-Time Graphs, http://www.nanotech.uwaterloo.ca/∼ne131/
18
The velocity of an object can be changed in two possible ways:
• Change in magnitude - i.e. a change in the speed.
• Change in direction - e.g. motion in a circle – although vs may be
constant, its direction – i.e. the tangential to the circle – continuously
changes.
Average acceleration, aavg : if the velocity changes by ∆v during a time
interval ∆t, then
aavg =
∆v
,
∆t
(3.12)
where aavg points in the same direction as ∆v .
Instantaneous acceleration, as : in (3.12), taking the limit ∆t → 0, i.e.
so the gradient of the line joining two points on the velocity-time curve approaches a tangent to the curve at an instant t, then
dv
v (t1 + ∆t) − v (t1 )
= s .
∆t→0
dt
∆t
as ≡ lim
(3.13)
So if v s = ds/dt, then
d
as =
dt
ds
dt
=
d2 s
,
dt2
(3.14)
i.e. as is also the second derivative of the displacement vector with respect
to time. In terms of its components,
as =
d2 z
d2 y
d2 x
k̂ .
+
ĵ
+
î
dt2
dt2
dt2
19
(3.15)
Example 1: for 1-dimensional motion, the displacement of a particle is found
to vary with time as ∆s(t) = 2t2 . Calculate the velocity at t = 3 seconds, and
the acceleration of the particle.
Solution 1: recalling that vs (t) = ds/dt, we have
vs (t) =
ds
= 4t
dt
v (t = 3) = 12 m s−1 .
−→
Using (3.14), the acceleration of the particle is given by
as =
d
d2 s
(4t) = 4 m s−2 .
=
2
dt
dt
Example 2: for 3-dimensional motion, the displacement of a particle is found
to vary as ∆s = 3t î + 2 ĵ + t2 k̂ . Calculate the velocity and acceleration at
t = 2 seconds.
Solution 2: since v s = ds/dt, then v s = 3 î + 2t k̂ , and so v s (t = 2) = 3 î + 4 k̂
(i.e. the speed is 5 m s−2 ). Furthermore, using (3.15), we have
d d2 s
3 î + 2t k̂ = 2 k̂ m s−2 .
as = 2 =
dt
dt
3.5
Area Under a Velocity-Time Graph
An object’s velocity varies with time as in figure 11.
It can be shown (see Additional Materials 6 ) that
∆s = lim
∆t→0
N
X
(vs )n ∆t =
n=1
Z
tf
vs dt ,
(3.16)
ti
i.e., the integral of vs , with respect to dt, between times ti and tf ,
gives the displacement of the object in question.
6
Analysis of a Velocity-Time Graph, http://www.nanotexh.uwaterloo.ca/∼ne131/
20
Figure 11: Area under velocity-time graph.
Generalise equation (3.16) for motion in 3-dimensions:
∆s =
Z
tf
v s dt .
(3.17)
ti
Example: consider an object moving with a velocity v s = 3t î + 2 ĵ + 3 k̂ ,
such that after 2 seconds, its displacement is 2 î + 3 k̂ . Find an expression for
the object’s displacement at an arbitrary time t.
Solution: recalling equation (3.17), the displacement vector is given by the
indefinite integral
∆s =
=
Z
v s (t) dt =
Z 3t î + 2 ĵ + 3 k̂ dt
3 2
t î + 2t ĵ + 3t k̂ + C ,
2
(3.18)
where C is a constant of integration. We evaluate C using the given boundary
condition, namely that
∆s (t = 2) = 2 î + 3 k̂ .
(3.19)
This is the displacement after 2 seconds. However, according to our derived
21
equation (3.18), an expression for the displacement after 2 seconds is simply
obtained by substituting for t = 2 s; we have
∆s (t = 2) = 6 î + 4 ĵ + 6 k̂ + C .
(3.20)
Therefore, by equating equations (3.19) and (3.20),
6 î + 4 ĵ + 6 k̂ + C = 2 î + 3 k̂ ,
we find the expression
(3.21)
C = −4 î − 4 ĵ − 3 k̂
for C. Thus, substituting equation (3.21) into (3.18) gives the displacement
∆s =
3 2
t −4
2
î + 2 (t − 2) ĵ + 3 (t − 1) k̂
as a function of time, which is the required solution.
For an additional example, see Additional Materials.
3.6
7
Area Under an Acceleration-time Graph
Since the area under a velocity-time graph (figure 12) gives the total displacement, if as = dv s /dt, we may expect the area under an acceleration-
time graph to give a velocity.
Using a similar argument to that of the velocity (see Additional Materials 8 ),
we find
∆v s =
Z
tf
as dt ,
(3.22)
ti
i.e. the area under the acceleration-time graph between ti and tf gives the
change in velocity during the interval.
7
8
Examples of Motion-Time Graphs, http://www.nanotexh.uwaterloo.ca/∼ne131/
Analysis of a Velocity-Time Graph, http://www.nanotech.uwaterloo.ca/∼ne131/
22
Figure 12: Acceleration-time graph.
3.7
Uniform Acceleration
An object is uniformly accelerated: e.g. suppose a car accelerates continuously from rest for 6 seconds, where
• After ∆t = 2 s, car’s velocity is v = 40 mph.
• After ∆t = 4 s, car’s velocity is v = 80 mph.
• After ∆t = 6 s, car’s velocity is v = 120 mph.
• In each second, the velocity of the car has changed by the same amount,
i.e. 20 mph.
• Resulting velocity-time graph is linear in t, i.e. its gradient is constant
for all times in the range 0 s < t ≤ 6 s.
• Equal changes in velocity occur during any successive equal-time intervals - the car accelerates uniformly at 1/180 miles s−2 .
23
Figure 13: Velocity-time graph for uniform acceleration.
• Figure 13 shows a velocity-time graph graph (linear, uniform acceleration
of 1/180 m s−2 ).
• Figure 14 shows the corresponding acceleration-time graph.
Figure 14: Acceleration-time graph for uniform acceleration.
24
3.8
The Kinematic Equations of Uniform Acceleration
In 1-dimension, an object with an initial velocity us accelerates uniformly
with an acceleration as during a time interval ∆t, by which time its velocity
is vf (see figure 15).
Figure 15: Motion at uniform acceleration.
3.8.1
First Kinematic Equation
Since the acceleration is the gradient of the velocity-time graph, we have
as =
vf − us
∆vs
,
=
∆t
∆t
(3.23)
or rearranging for vf ,
vf = us + as ∆t ;
(3.24)
this equation is the first kinematic equation.
3.8.2
Second Kinematic Equation
Suppose that we desire to find the displacement ∆s of the object during the
25
time interval ∆t. If ∆s is the area under the velocity-time graph between t = 0
and t = t1 , figure 15 shows that
∆s = area of rectangle + area of triangle
1
= u∆t + (vf − us ) ∆t .
2
(3.25)
However, from equation (3.24), (vf − us ) = as ∆t; substituting this in equation
(3.25) yields
∆s = u∆t +
1
(as ∆t) ∆t ,
2
(3.26)
i.e.
1
∆s = u∆t + as (∆t)2 .
2
This equation is the second kinematic equation.
Figure 16: Motion at uniform acceleration.
Note:
26
(3.27)
• The velocity-time graph (figure 15) described by equation (3.24) is linear
in t.
• The corresponding displacement-time relationship (figure 16), given by
equation (3.27), is a parabolic curve, due to the quadratic dependence of
∆s on ∆t (given by the (∆t)2 term).
3.8.3
Third Kinematic Equation
Using (3.24) in (3.27) to eliminate ∆t, then from (3.24), if
∆t =
v f − us
,
as
(3.28)
substituting in equation (3.27) gives
∆s = u
vf − us
as
1
+ as
2
It is then straightforward to show that
v f − us
as
vf2 = u2s + 2as ∆s ,
2
.
(3.29)
(3.30)
which is the third kinematic equation.
3.9
Free Fall
Why bother deriving the three kinematic equations? Motion with uniform
acceleration is important in numerous systems.
• e.g. motion under the influence of gravity – when the object is said to
be undergoing free fall.
3.9.1
Galileo’s Free-fall Experiment (c. 1590)
• Late 16th century - several European scientists became critical of the
generally accepted theories of motion proposed by the Ancient Greeks.
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