2
Motion in One Dimension
2.1
Definition of a Particle
Particle: a body whose dimensions may be neglected in describing its motion.
Example: The planets may be regarded as particles in considering their motion about
the sun, but not in considering their motion about their own axes.
2.2
Possible Types of Motion in One Dimension
No motion - particle at rest. If the position of the particle is denoted by x, then at rest
the particle will stay at a constant value of x for all time. Written as an equation this is
x(t) = A
where A is a constant and t denotes time.
Motion at a constant speed. The rate of motion of a particle is called its velocity.
For motion in 1 dimension, the velocity can be either positive (if the particle is moving
in the direction of increasing x) or negative (if the particle is moving in the direction of
decreasing x). A particle moving with constant velocity will change its position at a fixed
rate throughout time, written as an equation this is
x(t) = A + Bt
where A and B are constants.
The speed of the particle is the magnitude of the velocity. The rate of change of the
position is the velocity. Hence, the slope of the graph is a measure of the velocity. The
greater the slope, the greater the rate of change of position.
7
Figure 1: A true scale image of the size of the Sun relative to the size of the Earth. Sun
diameter is approx 1,390,000 kilometers, while the Earth has a diameter of approx
13,000 kilometers. The distance between the Sun and the Earth is approximately 150
Million kilometers.
Accelerated motion. If the speed of the particle is changing with time we say that the
particle is accelerating. Position vs time plots of accelerating particles are curves rather
than straight lines.
8
x
x
x(t) = A
x(t) = A + Bt
t
t
(a) Plot of the position as a function of (b) Plot of the position as a function of
time for a particle at rest.
time for a particle moving with constant
velocity.
Figure 2: Position as a function of time plots for a stationary particle and a particle
moving with constant velocity.
Example 2.2.1 (Figure ?? ).
Notice in the figure that the slope of the tangent line to the curve at t2 is larger than the
slope at t1 . This informs us that the particle is changing position at a faster rate at t2
than at t1 . In other words the velocity of the particle is greater at t2 than at t1 .
Example 2.2.2 (Figure ?? ).
In this case the particle oscillates between x = −A and x = +A. One can see that the
particle changes velocity from position to position (here the particle’s velocity changes
both in magnitude and direction).
9
x
x
A
x(t) = A + Bt + Ct2
t1
t2
t
x(t) = A sin(ωt)
t1
t2
-A
t
(a) Figure for Example 1 (accelerated
motion).
(b) Figure for Example 2 (accelerated
motion).
Figure 3: Two examples of the postition as a function of time for a particle moving with
accelerated motion.
x
Resting
Braking
(decelerating)
Constant
speed
Accelerating
Resting
t1
t3
t2
t4
t
Figure 4: Typical (idealised) motion of a car.
Figure ?? shows the position as a function of time for a typical car journey. The car is
at rest in traffic until t1 , the car accelerates until reaching the speed limit at time t2 and
maintains a constant speed. At t3 the driver sees a traffic light turn orange and begins
braking until at t4 the car sits at rest at the lights.
10
2.3
Mean Velocity
Suppose a particle in motion is at the point x1 at time t1 and then it moves to the point
x2 at the time t2 .
0
x1
x2
x
Figure 5: The particle is at the point x1 at time t = t2 and at the point x2 at the time
t = t2 .
The average velocity is defined to be
Important Formula 2.1 (Average Velocity).
v̄ =
x2 − x1
∆x
≡
t2 − t1
∆t
where ∆x = x2 − x1 and ∆t = t2 − t1 . The quantity ∆x is also an important quantity
and is known as the displacement.
2.3.1
Average velocity worked example.
Figure 6: Sticky running the 100m dash.
11
Example 2.3.1. Sticky (the stickperson) runs the 100m dash in 10s. What’s sticky’s
average speed?
Solution:
Sticky covers a displacement of ∆x = 100 m in a time ∆t = 10 s. The average velocity is
hence,
v̄ =
100 m
= 10 m s−1
10 s
12
2.4
Instantaneous Velocity
A more useful concept is that of instantaneous velocity, i.e. the velocity at a given time.
∆x
∆t→0 ∆t
v = lim
which we write as
Important Formula 2.2 (Instantaneous velocity).
Given that the position of an object as a function of time is x(t) the velocity as a function
of time is given by
v(t) =
dx(t)
dt
or more compactly,
v=
2.4.1
dx
dt
Instantaneous velocity worked examples.
Note that the velocity is the rate of change of position with time
Example 2.4.1 ( Figure ?? ).
Suppose a particles position is governed by the equation
x = A + Bt
what is the particles velocity?
Solution:
dx
d
= (A + Bt)
dt
dt
=0+B
=B
v=
see Figure ?? for a plot of the function.
13
ASIDE 2.1 (Derivative of a polynomial).
d n
(t ) = nt(n−1)
dt
Example 2.4.2 ( Figure ?? ).
Take the particle’s position function to be
x(t) = A + Bt + Ct2 .
Find the particles velocity as a function of time.
Solution:
d
(A + Bt + Ct2 )
dt
= 0 + B + 2Ct
v=
thus we have,
v(t) = B + 2Ct
If C > 0, the velocity increases through time. See Figure ?? for a plot of the position
function.
14
Example 2.4.3. The position of an object moving in a straight line is given by
x = 3t − 4t2 + t3
where x is the particle’s position in metres and t denotes time in seconds.
(a) What is the position of the object at t = 0, 1 s and 3 s?
(b) What is the object’s displacement between t = 0 and t = 2 s?
(c) What is the average velocity of the object between t = 2 s and t = 4 s?
(d) What is the velocity of the object at t = 3 s?
Solution:
(a) At t = 0 s,
x = [3(0) − 4(0)2 + (0)3 ] m
= 0 m.
At t = 2 s
x = [3(2) − 4(2)2 + (2)3 ] m
= [6 − 16 + 8] m
= −2 m.
At t = 4 s
x = [3(4) − 4(4)2 + (4)3 ] m
= [12 − 64 + 64] m
= 12 m.
(b) We know that at t = 0 s the object is at x = 0 m, while at t = 2 s the object is at
x = −2 m. Thus the displacement is
∆x = [−2 − 0] m
= −2 m.
Thus the object has travelled to the left of the origin.
15
. . . example continued
(c) At t = 2 s, x = −2 m and at t = 4 s, x = 12 m. Thus the average velocity of the
object over this time interval is
x2 − x1
t2 − t1
12 m − (−2) m
14 m
=
=
4s − 2s
2s
−1
= 7ms
v̄ =
(d) We have the position function
x = 3t − 4t2 + t3
and can find the instantaneous velocity of the object (in m s−1 ) by taking the derivative
dx
d
= (3t − 4t2 + t3 )
dt
dt
= 3 − 8t + 3t2 .
v=
This tells us how the velocity varies with time. We are interested in the velocity at
a particular instant in time, namely t = 3 s,
v = [3 − 8(3) + 3(3)2 ] m s−1
= 6 m s−1 .
16
2.5
Acceleration
In our pervious set of examples we have seen that the velocity of an object can vary in
time, i.e. v ≡ v(t). The rate of change of the velocity is defined as the acceleration
Important Formula 2.3 (Instantaneous acceleration).
Given that the velocity of an object as a function of time is v(t), the acceleration of the
object is defined as the rate of change of the velocity with respect to time and is given
by
a(t) =
dv(t)
dt
or more compactly
a=
2.5.1
dv
dt
Acceleration worked examples.
Example 2.5.1. We have already found that an object governed by the position function
x = 3t − 4t2 + t3
has a velocity
v = 3 − 8t + 3t2 .
What is the object’s acceleration?
Solution:
To find the acceleration one need only differentiate the expression for the object’s velocity
a=
d
dv
= (3 − 8t + 3t2 )
dt
dt
= −8 + 6t
Note that in this case the acceleration varies in time. When t = 0 s the acceleration
a = −8 m s−2 and when t = 4 s the object has an acceleration a = 16 m s−2
Note 2.1. The units of acceleration are
units of velocity
m s−1
=
= m s−2
units of time
s
17
Example 2.5.2. A particle is moving in a straight line. Its position, x, from the origin
O on the line, at time t, is given by
x = t − 5t2
where x is in meters and t is in seconds.
Obtain
(1) the average velocity over the interval t = 2 s to t = 5 s,
(2) the velocity at t = 4 s and
(3) the time at which the velocity is zero.
(4) Obtain the acceleration at t = 2 s.
(5) Obtain the displacement and the distance travelled in the first 5 seconds of motion.
Solution:
(1)
At t = 2 s
At t = 5 s
x = [2 − 5(2)2 ] m = [2 − 20] m
= −18 m.
x = [2 − 5(5)2 ] m = [2 − 125] m
= −120 m.
We are looking for the average velocity over this interval,
−120 m − (−18) m
x2 − x1
=
t2 − t1
5s − 2s
(−120 + 18) m
=
3s
= −34 m s−1 .
v̄ =
(2) The instantaneous velocity (in m s−1 ) is found by differentiating the position function
v≡
dx
d
= (t − 5t2 )
dt
dt
= 1 − 10t.
The question asks for the velocity at t = 4 s,
v = [1 − 10(4)] m s−1 = [1 − 40] m s−1
= −39 m s−1
18
. . . example continued
(3) We have an expression for the velocity
v = 1 − 10t.
The question asks for the time t for which the velocity is zero v = 0,
0 = 1 − 10t
10t = 1 ⇒ t = 1/10 s
Note 2.2. Points where v = 0 are often very important, so important in
fact that they have their own name, turning points of the object’s motion.
(4) The acceleration is found by simply differentiating the object’s velocity,
a≡
dv
d
= (1 − 10t)
dt
dt
= [0 − 10] m s−2
= −10 m s−2 .
The acceleration is constant i.e. it takes the same value throughout time. Thus at
t = 2 s, or any other time for that matter, a = −10 m s−2 .
(5) The displacement (often denoted by a scripted letter ‘s’; here we will use S) between
t = 0 and t = 5 s. To find the displacement we require the position of the object at
these times:
At t = 0 s
At t = 5 s
x = [0 − 5(0)2 ] m
= 0 m.
x = [5 − 5(5)2 ] m
= −120 m.
The displacement is thus,
S = ∆x = x2 − x1
= −120 m − 0 m
= −120 m
19
. . . example continued
(5) The displacement between t = 0 and t = 5 s
ASIDE 2.2. The distance an object travels is distinct to its displacement.
The distance travelled is the length of the path the object takes when getting
from a point a to a point b. The displacement is the difference between an
objects final position and its initial position, i.e. the shortest path from
point a to point b. As an example, you can walk around the entire north
campus of Maynooth University starting from outside the science building
and return to same spot you started from. Here your starting position and
final position are the same and hence your displacement is 0 m; however the
distance travelled is approx 1.5km.
we know that the object has a turning point when t = 1/10 s, and at t = 5 s the
object is at x = −120 m. Begin by finding the object’s position at the its turning
point,
when t = 0.1 s,
x = [0.1 − 5(0.1)2 ] m
= [0.1 − 0.05] m
= +0.05 m
This is the point at which the velocity changes direction.
-120
0.05
0
x
From the figure one can see that the total distance travelled D is given by,
D = [0.05 + 0.05 + 120] m
= 120.1 m
20
3
Motion with Constant Acceleration
A special class of motion in physics is one where the acceleration is constant. For example,
objects falling near the Earth’s surface fall with a constant acceleration.
(a) The fateful apple?
(b) Home.
Figure 7: An apple falling from a tree to the Earth’s surface is an example of motion
with constant acceleration.
This case is special as one can derive a number of useful equations of motion under these
conditions, that are readily useable for a number of physical situations.
3.1
The equations of motion
3.1.1
The first equation of motion.
Consider an object moving with constant acceleration a. If at t = 0 the object has a
velocity v0 and at a time t it has a velocity v, then we have
a=
v − v0
∆v
=
∆t
t−0
Hence, at = v − v0
Important Formula 3.1 (Constant acceleration — first equation of motion).
v = v0 + at
21
(1)
Note 3.1 (plot of the first equation of motion).
If one makes a plot of Equation ?? with velocity v against against time t, one will find a
straight line with slope equal to a. Recall that the equation of a line in the xy-plane is
y = mx + c
which is the same form as Equation ??, let v = y, t = x, m = a nad c = v0 .
a
v
a = const.
v = v0 + at
a
v0
t
t
(a) Object has constant acceleration.
(b) The slope of the velocity vs. time
plot is equal to the object’s acceleration.
Figure 8: plot of the first equation of motion for constant acceleration.
3.1.2
The second equation of motion.
We begin by examining the average velocity, v̄, of an object that is moving with constant
acceleration a.
v
v
v - v0
-
v
v0
t
22
From Equation ?? we have that
v = v0 + at.
Examining the average velocity of the object,
1
1
v̄ = (v0 + v) = (v0 + v0 + at)
2
2
1
= (2v0 + at)
2
thus we have
1
v̄ = v0 + at.
2
(2)
If at t = 0 the object is at the position x0 and at time t the object is at x, the from the
definition of the average velocity we have,
v̄ ≡
∆x
x − x0
x − x0
=
=
∆t
t−0
t
from Equation ??
1
x − x0
v + at =
2
t
1
⇒ x − x0 = v0 t + at2 .
2
Now we note that x−x0 is just the displacement of the object, we let S = x−x0 so we have
Important Formula 3.2 (Constant acceleration — second equation of motion).
1
S = v0 t + at2
2
23
(3)
ASIDE 3.1. A graph of S vs. t is typically not a straight line for constant acceleration
problems. We have
1
x = x0 + v0 t + at2
2
The velocity at time t can be found by differentiating this expression for x,
v≡
dx
= v0 + at
dt
which is the the same equation as Equation ??
S
Slope = v
t
Figure 9: An example of a displacement vs time graph of an object moving with
constant acceleration.
3.1.3
The third equation of motion.
One can also obtain an expression relating the initial and final velocities of the displacement. To do this note that from Equation ??
t=
v − v0
.
a
Substituting this expression for time into Equation ?? S = v0 t + 12 at2 , we obtain,
24
(4)
(v − v0 ) 1 a(v − v0 )2
+
a
2
a2
(v − v0 ) 1 (v − v0 )2
= v0
+
a
2
a
S = v0
multiplying both sides of the equation by 2a,
2aS = 2v0 (v − v0 ) + (v − v0 )2
2
2
+ v2
=
2v
2vv
0 v −2v0 +v − 0
0
| {z }
|{z}
= v 2 − v02 .
We have now arrived at the third equation of motion
Important Formula 3.3 (Constant acceleration — third equation of motion).
v 2 = v02 + 2aS
3.1.4
(5)
Summary.
The main results of this section are that for a particle moving with constant acceleration
a the following relations hold
v = v0 + at
1
S = v0 t + at2
2
v 2 = v02 + 2aS
where S = x − x0 is the displacement.
25
3.1.5
Constant acceleration examples.
Example 3.1.1.
A particle is moving in a straight line with a constant deceleration of 3.0 m s−2 . If initially
it has a velocity of 10 m s−1 find the time when the velocity is zero.
Solution:
Given:
v0 = 10 m s−1 ,
v = 0 m s−1 ,
a = −3.0 m s−2 ,
t =?
From Equation ?? we have
v = v0 + at
filling in the given information yields,
0 = 10 m s−1 + (−3.0 m s−2 )t
0 = 10 s − 3.0t
3t = 10 s ⇒ t = 3.3 s
Example 3.1.2.
A particle starts from a point 0 with an initial velocity of 2 m s−1 and travels along a
straight line with a constant acceleration of 2 m s−2 . Two seconds later a second particle
starts from a rest at 0 and travels along the same line with an acceleration of 6 m s−2 .
Find how far from 0 the second particle overtakes the first.
Solution:
Suppose the first particle has been traveling for T seconds when it is overtaken
t=0
a1 = 2 ms-2
a2 = 6 ms-2
x
0
x
d
26
t=2
x
0
t=T
0
a1 = 2 ms-2
. . . example continued
we have,
First Particle: After a time T s,
S = d m,
v0 = 2 m s−1 ,
t = T s,
a = 2 m s−2
From
1
S = v0 t + at2
2
⇒
d = 2T + T 2
(6)
Second Particle: Is travelling for a time (T − 2) s,
S = d m,
v0 = 0,
t = (T − 2) s,
a = 6 m s−2
From
1
S = v0 t + at2
2
⇒
1
d = 0(T − 2) + 6(T − 2)2
2
2
= 3(T − 2)
From Equation ?? and Equation ??:
2T + T 2 = 3(T − 2)2
2T + T 2 = 3(T 2 − 4T + 4).
27
(7)
. . . example continued
Rearranging terms we have
2T 2 − 14T + 12 = 0
T 2 − 7T + 6 = 0
(T − 6)(T − 1) = 0
Hence the meeting times are mathematically found to be T = 1 or T = 6. We disregard
the T = 1 s solution, although mathematically correct, as it is unphysical (particle 2
starts 2 seconds after particle 1!).
Thus we have
t = T s = 6 s.
We can find where the particle 1 meets particle 2 by returning to Equation ??
d = 2(6) + (6)2 = 48
The particles have travelled 48 m before the second overtakes the first.
ASIDE 3.2. It’s not always possible to so simply factorise the a quadratic equation,
however one can always find the roots of a quadratic
ax2 + bx + c = 0
with the general formula
x=
−b ±
28
√
b2 − 4ac
.
2a
3.2
The Velocity-Time Graph for Constant Acceleration
For an object moving with constant acceleration we have see that
v = v0 + at.
This is the equation of a straight line of slope a and intercept on the velocity axis of v0 .
v
v
v - v0
A
v0
B
t
Figure 10: A velocity vs time graph for an object moving with the constant acceleration.
The are under the curve of the velocity-time graph is made up of the area of the triangle
and the area of the rectangle.
Total Area = Area of triangle + Area of rectangle
= A + B.
We know the area of a triangle is given by A = 12 (Base)(height) and the area of a rectangle
is simply given by B = (Base)(height) and thus,
1
Total Area = (t)(v − v0 ) + (t)(v0 )
2
1
1
= vt − v0 t + v0 t
2
2
1
= (v + v0 )t.
2
Recalling the first equation of motion (??),
v = v0 + at,
29
and substituting it into our expression for the area yields
1
Total Area = (v0 + at + v0 )t
2
1
= (2v0 + at)t
2
1
= v0 t + at2
2
= S.
We have thus found that the area under the velocity time graph of an object moving with
constant acceleration is the displacement.
Note 3.2.
Area under a velocity time graph for constant acceleration = S.
3.2.1
Worked example.
Example 3.2.1.
A car is moving along a straight line. It is taken from rest to a velocity of 20.0 m s−1 by
constant acceleration of 5.00 m s−2 . It maintains a constant velocity of 20.0 m s−1 for 5.00
seconds and then is brought to rest by a constant acceleration of −2.00 m s−2 . Draw a
velocity time graph and find the distance covered by the car.
Solution:
There are three stages of motion in this example
Stage 1. Accelerating
Stage 2. Constant velocity
Stage 3. Decelerating.
In order to sketch a velocity time graph it is necessary to know how much time the car
spends in each of these three stages of motion.
30
. . . example continued
Stage 1. We have the following pieces of information
v0 = 0 m s−1 ,
v = 20.0 m s−1
and a = 5.00 m s−2
using Equation ??
v = v0 + at
20.0 m s = 0 + (5.00 m s−2 )t
⇒ t = 4.00 s
−1
Stage 2. Constant velocity of 20.0 m s−1 for 5.00 s
Stage 3.
v0 = 20.0 m s−1 ,
v = 0 m s−1
and a = −2.00 m s−2
using Equation ??
v = v0 + at
0 = 20.0 − 2.00t
⇒ t = 10.0 s
v
20
2
1
0
t=4
3
t=9
31
t = 19 t
. . . example continued
The displacement is simply the sum of all three indicated areas on the velocity time
graph.
1
S1 = (4.00 s)(20.0 m s−1 ) = 40.0 m
2
S2 = (5.00 s)(20.0 m s−1 ) = 100 m
1
S3 = (10.0 s)(20.0 m s−1 ) = 100 m
2
Now the total displacement is given by
STotal = S1 + S2 + S3
= 40.0 m + 100 m + 100 m
= 240 m
We are however asked for the total distance travelled by the car which in this example is
equal to the total displacement
Total Distance = 240 m
Note 3.3.
Since all the areas of the velocity time graph are positive, the distance travelled
and the car’s total displacement are equal.
32
3.3
Deriving the equations of motion using calculus
We have seen the equations of motion for a particle moving with constant acceleration
using geometric arguments and a little algebra. We can however, derive the equations of
motion using a little calculus. The only integral we require is given by
Note 3.4. Given the function f (x) = xn , its integral is given by
Z
Z
f (x) dx =
xn dx =
xn+1
+C
n+1
where C is a constant of integration.
A further concept that we require is that of the instantaneous rate of change, an idea
that we will return to later in the course. At this point it suffices to note that one can
write the the acceleration and velocity as
The velocity of an object is the rate of change of position in time,
v=
dx
dt
and the acceleration of an object is the rate of change of velocity
a=
dv
dt
We are now ready to derive the equations of motion.
3.3.1
The first equation of motion
We begin by examining the motion of an object, which begins its motion at time t = 0
with an initial velocity v0 . As was the previous case we look at motion with constant
acceleration,
a = constant
Using the framework of calculus we have
33
dv
≡ a = constant
dt
and hence we have
dv = a dt
(8)
Now we have the particle stating with an initial velocity of v0 and we take the velocity
at time t to be v, we integrate (??) using these limits2
Zv
Zt
dv =
Zt
a dt = a
0
v0
dt
0
In the last step, we have factored out a as it is a constant. Performing the integration
we have
v
t
v
=a t
v0
0
filling in the limits we have
v − v0 = a (t − 0)
which is the first equation of motion for constant acceleration
v = v0 + at.
2
In fact on a technical note one should be more careful with the integration variable we are using.
Since we are integrating with respect to t and v one should not have these dummy variables appearing in
our limits, as we cannot integrate the limits before the integrand itself! This is a relatively minor issue
in all practicality, all that is required is that we relabel the “dummy variables”, for example
Zv
0
Zt
dv =
v0
a dt0
0
would suffice. As stated earlier this is minor technicality and if you found the original integral disturbing then a career as a mathematician may be worth considering!
34
Note 3.5.
Z
Z
dv = v
3.3.2
and
dt = t
The second equation of motion
To derive the second equation of motion we recall that the velocity can be written as
v≡
dx
dt
and hence our first equation of motion can be written as
dx
= v0 + at.
dt
we can see that an infinitesimally small change in position is equivalent to
dx = v0 dt + at dt
(9)
If at t = 0 we have the object starting at the position x = x0 and take the particle to be
at position x after time t, integrating (??) gives us
Zx
Zt
dx =
x0
Zt
v0 dt +
0
at dt
0
Zt
= v0
Zt
dt + a
0
t dt
0
since v0 and a are constants. Hence
x
t
2 t
t
.
x
= v0 t + a
2 0
x0
0
Filling in the limits we have
35
x − x0 = v0 t + a
t2 − 02
2
,
which is the second equation of motion for a particle moving with constant acceleration
1
S = x − x0 = v0 t + at2
2
3.3.3
The third equation of motion
The third equation of motion is found in the same manner as before, in fact there is no
new information offered by the third equation that the first two do not have as it is a
combination of the first two.
One begins with the two equations of motion already derived
v = v0 + at
1
S = v0 t + at2 .
2
(10)
(11)
The aim is to use the first equation (??) to find an expression for t and to substitute this
expression into the second equation (??). From (??) we have
t=
v − v0
a
and substituting this expression into (??) we have
v − v0
S = v0 ·
+
a
v − v0
+
= v0 ·
a
2
1
v − v0
a·
2
a
1 (v − v0 )2
2
a
multiplying both sides by 2a we have
2aS = 2v0 (v − v0 ) + (v − v0 )2
= 2v0 v − 2v02 + v 2 − 2v0 v + v02
= v 2 − v02
36
which is the third equation
v 2 = v02 + 2aS
3.4
Freely Falling Bodies
(a) Nastia Liukin, Olympic
silver medalist in the beam
event.
(b) Lin Yue and
Huo Liang, Olympic
gold medalists.
(c) Torah Bright, Olympic gold
medalist in the Halfpipe event.
All bodies, regardless of their size shape or composition, fall with the same acceleration
at the point near the Earth’s surface. This acceleration is called the free fall acceleration
or acceleration due to gravity and is denoted by g.
Near the Earth’s surface g = 9.8 m s−2 . The direction of g is toward the centre of the
Earth. It varies from 9.832 m s−2 at the North Pole to 9.761 m s−2 on the top of Mount
Chimborazo in Ecuador. In Maynooth the value is approximately g = 9.814 m s−2 .
y
a = -9.8ms-2
= -g
0
x
Figure 11: The particle falls in the negative y direction toward the center of the Earth
(the origin). Thus the acceleration due to gravity has a negative sign.
37
3.4.1
Equations of motion for a particle in the Earth’s gravitational field (at
the surface).
v = v0 − gt
1
S = v0 t − gt2
2
2
2
v = v0 − 2gS
These are the equations of motion for a particle moving with a constant acceleration
a = −g.
3.4.2
Worked examples.
Example 3.4.1. An object is dropped from rest and falls freely. Determine the position
and velocity of the body after 1.00, 2.0 and 4.0 seconds.
Solution:
One is free to choose the starting position of the object here, as such choose y0 = 0 m at
t = 0 s. From the question one is given that v0 = 0 m s−1 as the object is dropped from
rest.
y
t=0
0
t=1
t=2
38
x
. . . example continued
From the second equation of motion Equation ?? with acceleration a = −g we have
1
S = y − y0 = v0 t − gt2
2
1
y − 0 = (0)t − gt2
2
1 2
y = − gt
2
which gives the particle’s position after a time t s. Notice that the position will be a
negative value which means that the object is falling downward.
From the first equation of motion Equation ?? with a = −g one finds that
v = v0 − gt
= −gt
which gives the object’s velocity at a time t.
After t = 1.00 s
After t = 2.0 s
1
1
y = − (9.8 m s−2 )(1.00 s)2 y = − (9.8 m s−2 )(2.0 s)2
2
2
= −4.9 m
= −20 m
v = −9.8 m s−2 (1.00 s)
= −9.8 m s−1 .
v = −9.8 m s−2 (2.0 s)
= −20 m s−1 .
Complete the problem for t = 4 s.
39
Example 3.4.2. A stone is thrown vertically upward from the top of a tower and hits
the ground 10 seconds later with a speed of 51 m s−1 . Find the height of the tower.
Solution:
y
V0
x
0
h
51ms-1
We have that
v = −51 m s−1 ,
t = 10 s and a = −g.
Begin by finding the initial launch velocity of the stone using
v = v0 − gt
−51 m s = v0 − 9.8 m s−1 (10 s)
−51 m s−1 = v0 − 98 m s−1
⇒ v0 = 47 m s−1 .
−1
It is now possible to calculate the displacement
1
S = v0 t − gt2
2
1
= 47 m s−1 (10 s) − (9.8 m s−2 )(10 s)2
2
= 470 m − 490 m
= −20 m.
We are interested in the height of the tower and thus we have
h = 20 m
40
Example 3.4.3. A ball is thrown vertically upward from the ground with a speed of
25.2 m s−1 .
(a) How long does it take to reach its highest point?
(b) How high does it rise?
(c) At what times will it be 27.0 m above the ground?
Solution:
y
V0 = 25.2 ms-1
y0 = 0m
x
(a) When the ball reaches its highest point v = 0 m s−1 (we have a turning point). we
also have,
v0 = 25.2 m s−1 ,
v = 0 m s−1
and a = −g
From the first equation of motion we have
v = v0 − gt
0 = 25.2 m s−1 − (9.8 m s−2 )t
⇒ t = 2.57 s
(b) To find how high the ball rises we compute the displacement of the ball
41
. . . example continued
v 2 = v02 − 2gS
02 = (25.2 m s−1 )2 − 2(9.8 m s−2 )(S)
⇒ S = 32.4 m
Thus the highest point the ball reaches is 32.4 m above the ground.
(c) To find the time t when S = y = 27.0 m we note that
S = 27.0 m,
v0 = 25.2 m s−1
and a = −g
From the second equation of motion we have
1
S = v0 t − gt2
2
1
27.0 m = (25.2 m s−1 )t − (9.8 m s−2 )(t2 )
2
⇒ (4.9 s−2 )t2 − (25.2 s−1 )t + 27.0 = 0
which is a quadratic equation in t. This can be solved to yield
p
(−25.2)2 − 4(4.9)(27)
s
2(4.9)
= (2.57 ± 1.05) s .
t=
25.2 ±
Thus we have that the ball reaches a height of 27.0 m at t = 1.52 s (on the way
upwards) and at t = 3.62 s (on the way down).
y
y
t = 1.52s
t = 3.62s
x
x
42