Constant Acceleration
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1
Constant acceleration
Sunil Kumar Singh
This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License †
∗
Abstract A constant acceleration is a special case of the accelerated motion.
The constant acceleration is a special case of accelerated motion. There are vast instances of motions, which can be approximated to be under constant force and hence constant acceleration. Two of the most important forces controlling motions in our daily life are force due to gravity and friction force. Incidentally, these two forces are constant in the range of motions of bodies in which we are interested. For example, force due to gravity is given by :
F =
GMm r2
= mg
where "G" is the universal constant, "M" is the mass of earth, "m" is the mass of the body and "r" is the distance between center of earth and the body. The resulting acceleration due to gravity, g, is a constant in the immediate neighborhood of the earth surface and is given by :
g =
GM r2
The only variable for a given mass is "r", which changes with the position of the body. The distance "r", however, is equal to Earth's radius for all practical purposes as any dierence arising from the position of body on earth can be ignored. Therefore, acceleration due to gravity can safely be considered to be constant for motions close to the surface of the earth. Signi cantly acceleration due to gravity is a constant irrespective of the mass "m" of the body.
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A ball kicked from a height
Figure 1: The ball follows a parabolic path as it approaches ground
The gure above shows the motion of a ball kicked from the top of a tower.
The ball moves under
constant acceleration of gravity during its ight to the ground. The constant acceleration, therefore, assumes signi cance in relation to the motion that takes place under the in uence of gravity. In the same manner, motion on a rough plane is acted upon by the force of friction in the direction opposite to the motion. The force of friction is a constant force for the moving body and characteristic of the surfaces in contact. As a result, the object slows down at a constant rate.
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Motion in uenced by friction force
Figure 2: The block decelerates due to friction.
1 Understanding constant acceleration
The constant acceleration means that the acceleration is independent of time and is equal to a constant value. The implication of a constant acceleration is discussed here as under : 1: As acceleration is same at all instants during the motion, it follows that average acceleration is equal to instantaneous acceleration during the motion. Mathematically,
aavg = a ⇒
2: When
∆v ∆t
=
v t
∆t
= 1 second, then
aavg = a = ∆ v = v2 − v1
This means that initial velocity , on an average, is changed by the acceleration vector after every second.
Example 1: Constant acceleration Problem : The position of a particle, in meters, moving in the coordinate space is described by
the following functions in time.
x = 2t2 − 4t + 3 ; y = − 2t ; and z = 5
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Find the velocity and acceleration at t = 2 seconds from the start of motion. Also, calculate average acceleration in the rst four seconds.
Solution :
The component velocities in three directions are :
vx = vy = vz =
and the velocity is given by :
x t y t z t
= = =
t t t
2t2 − 4t + 3 ( − 2t ) = − 2 (5) = 0
= 4t − 4
v = vx i + vy j + vz k ⇒ v = ( 4t − 4 ) i − 2j
Thus, velocity at t = 2 seconds,
v2 = 4i − 2j
Acceleration of the particle along three axes are given as :
ax = ay = az =
x t2 2 y t2 2 z t2
2
= = =
2
t2
2
2t2 − 4t + 3 ( 4t ) = 0 (5) = 0
= 4
t2
2
t2
The resultant acceleration is given by :
a = ax i + ay j + az k ⇒ a = 4i m / s2
which is a constant and is independent of time. The accelerations at all time instants are, therefore, same. We know that the average and instantaneous accelerations are equal when acceleration is constant. Hence,
aavg = 4i
. We can check the result calculating average acceleration for the
rst four seconds as :
aavg =
v4 − v0 4
=
12i − 2j + 4i + 2j 4
= 4i m / s2
The important fall out of a constant acceleration is that its magnitude has a constant value and its direction is xed. A change in either of the two attributes, constituting acceleration, shall render acceleration variable. This means that acceleration is along a straight line. But does this linear nature of acceleration mean that the associated motion is also linear? Answer is no. Reason is again the disconnect between acceleration and velocity. We know that magnitude and direction of acceleration are solely determined by the mass of the object and net external force applied on it. Thus, a constant acceleration only indicates that the force i.e the cause that induces change in motion is linear. It does not impose any restriction on velocity to be linear.
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Velocity and acceleration
Figure 3: Force and velocity are in two dierent directions
It is imperative that if the initial velocity of the object is not aligned with linear constant acceleration like in the gure above, then the immediate eect of the applied force, causing acceleration, is to change the velocity. Since acceleration is de ned as the time rate of change in velocity, the resulting velocity would be so directed and its magnitude so moderated that the change in velocity (not the resulting velocity itself ) is aligned in the direction of force.
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Change in velocity
Figure 4: The change in velocity has the same direction as that of acceleration.
As the resulting velocity may not be aligned with the direction of force (acceleration), the resulting motion may not be linear either. For motion being linear, it is essential that the initial velocity and the force applied (and the resulting acceleration) are aligned along a straight line. Examples of motions in more than one dimension with constant acceleration abound in nature. We have already seen that motion of a projectile in vertical plane has constant acceleration due to gravity, having constant magnitude, g, and xed downward direction. If we neglect air resistance, we can assume that all non- propelled projectile motions above ground are accelerated with constant acceleration. In the nutshell, we can say that constant acceleration is unidirectional and linear, but the resulting velocity may not be linear. Let us apply this understanding to the motion of a projectile, which is essentially a motion under constant acceleration due to gravity.
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Parabolic motion
Figure 5: The velocity of the ball is moderated by acceleration.
In the gure, see qualitatively, how the initial velocity vector, v, is modi ed by the constant acceleration vector, g, at the end of successive seconds. Note that combined change in both magnitude and direction of the velocity is taking place at a constant rate and is in vertically downward direction. In the context of constant acceleration, we must also emphasize that both magnitude and direction are constant. A constant acceleration in magnitude only is not su cient. For constant acceleration, the direction of acceleration should also be same (i.e constant). We can have a look at a uniform circular motion in horizontal plane, which follows a horizontal circular path with a constant speed.
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Uniform circular motion
Figure 6: The object moves along a circular path constant speed.
Notwithstanding the constant magnitude, acceleration of uniform circular motion is a variable acceleration in the horizontal plane, because direction of radial centripetal acceleration (shown with red arrow) keeps changing with time. Therefore, the acceleration of the motion keeps changing and is not independent of time as required for acceleration to be constant.
2 Equation of motion
The motion under constant acceleration allows us to describe accelerated motion, using simple mathematical construct. Here, we set out to arrive at these relationships in the form of equations. In these equations,
u
stands for initial velocity,
v
for nal velocity, for constant acceleration and t for time interval
a
of motion under consideration.
This is short of convention followed by many text books and hence the
convention has been retained here.
2.1 First equation
1 : v = u + at
This relation can be established in many ways. One of the fundamental ways is to think that velocity is changed by constant acceleration (vector) at the end of successive seconds. Following this logic, the velocity at the end of t successive seconds are as under :
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At the end of At the end of At the end of At the end of
1st 2nd 3 4
rd th
second second second second
: u + a : u + 2a : u + 3a : u + 4a
At the end of
tth
second
: u + ta = u + at
We can also derive this equation, using the de ning concept of constant acceleration. We know that :
a = aavg ⇒ a =
∆v ∆t
=
v−u t
⇒ at = v − u ⇒ v = u + at
Alternatively (using calculus), we know that :
a = ⇒
Integrating on both sides, we have :
v t
v = a t
∆v = a∆t ⇒ v2 − v1 = at ⇒ v − u = at ⇒ v = u + at
2.2 Second equation
2 : vavg =
u+v 2 The equation v
= u + at
is a linear relationship. Hence, average velocity is arithmatic mean of initial
and nal velocities :
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Average velocity
Figure 7: Average velocity is equal to mean velocity.
⇒ vavg =
u+v 2
2.3 Third equation
3 : s = ∆ r = r2 − r1 = ut +
1 2 2 at This equation is derived by combining the two expressions available for the average velocity.
vavg =
and
∆r ∆t
=
r2 − r1 t
vavg =
( v1 + v2 ) 2
=
(v+u) 2
Combining two expressions of average acceleration and rearranging, we have :
⇒ ( r2 − r1 ) =
Using the relation,
(v+u)t 2
v = u + at
and substituting for v, we have :
⇒ ( r2 − r1 ) =
( u + at + u ) t 2 1 2 2 at
⇒ s = ∆ r = r2 − r1 = = ut +
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Alternatively, we can derive this equation using calculus. Here,
v = ⇒
Integrating between the limits on both sides,
r t
r = v t
r =
Now substituting
v t
v,
⇒ r = ( u + at ) t ut + at t
1 2 2 at
⇒ ∆r =
⇒ s = ∆ r = r2 − r1 = ut +
(
The expression for the displacement has two terms : one varies linearly ( t) with the time and the other 1 2 2 at ) varies with the square of time. The rst term is equal to the displacement due to non-accelerated motion i.e the displacement when the particle moves with uniform velocity, . The second term represents
u
u
the contribution of the acceleration (change in velocity) towards displacement. This equation is used for determining either displacement(∆ ) or position(
r
r1 ).
A common simpli cation,
used widely, is to consider beginning of motion as the origin of coordinate system so that
r1 = 0 ⇒ ∆ r = r2 − r1 = r2 = r ( say )
In this case, both nal position vector and displacement are equal. This simpli cation, therefore, allows us to represent both displacement and position with a single vector variable
r.
3 Graphical interpretation of equations of motion
The three basic equations of motion with constant acceleration, as derived above, are :
1 : v = u + at 2 : vavg =
(u+v) 2 1 2 2 at
3 : s = ∆ r = r2 − r1 = ut +
These three equations completely describe motion of a point like mass, moving with constant acceleration. We need exactly ve parameters to describe the motion under constant acceleration :
u , v , r1 , r2 and t
.
It can be emphasized here that we can not use these equations if the acceleration is not constant. We should use basic dierentiation or integration techniques for motion having variable acceleration (nonuniform acceleration). These equations serve to be a ready to use equations that avoids dierentiation and integration. Further, it is evident that equations of motion are vector equations, involving vector addition. We can evaluate a motion under constant acceleration, using either graphical or algebraic method based on components. Here, we interpret these vector equations, using graphical technique. For illustration purpose, we apply these equations to a motion of an object, which is thrown at an angle of acceleration is "g", which is directed vertically downward. corresponding bold faced symbol
θ
from the horizontal. The magnitude
g.
Let acceleration vector be represented by
Let
and corresponding position vectors are The nal velocity at time instant
v1 and v2 r1 and r2 .
be the velocities at time instants
t1
and
t2
respectively
t2
, is given by :
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v = u + at ⇒ v2 = v1 + g ( t2 − t1 )
Graphically, the nal velocity is obtained by modifying initial vector
v1
by the vector
g ( t2 − t1 )
.
Graphical representation of rst equation
Figure 8
Now, we discuss graphical representation of second equation of motion. The average velocity between two time instants or two positions is given by :
vavg = ⇒
(u+v) 2 v2 ) vavg = ( v1 + 2
The vector addition involved in the equation is graphically represented as shown in the gure. Note that average velocity is equal to half of the vector sum
v1 + v2
.
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Graphical representation of second equation
Figure 9
Third equation of motion provides for displacement in terms of two vector quantities - initial velocity and acceleration. The displacement,
s, is equal to addition of two vector terms :
1 2 2 at 1 2g
s = ∆ r = r2 − r1 = ut +
⇒ s = ∆ r = r2 − r1 = v1 ( t2 − t1 ) +
( t2 − t1 )
2
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Graphical representation of third equation
Figure 10
4 Equations of motion in component form
The application of equations of motion graphically is tedious. In general, we use component representation that allows us to apply equations algebraically. We use equations of motion, using component forms of vector quantities involved in the equations of motion. The component form of the various vector quantities are :
r = xi + y j + z k ∆ r = ∆ xi + ∆ y j + ∆ z k u = ux i + uy j + uz k v = vx i + vy j + vz k v = vavgx i + vavgy j + vavgz k a = ax i + ay j + az k
Using above relations, equations of motion are :
1 : vx i + vy j + vz k = ( ux i + uy j + uz k ) + ( ax i + ay j + az k ) t 2 : vavgx i + vavgy j + vavgz k = 3: ∆ xi + ∆ y j + ∆ z k = (
( ux i + uy j + uz k ) + ( vx i + vy j + vz k ) 2 1 ux i + uy j + uz k ) t + 2 ( ax i +
ay j + az k ) t2
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Example 2: Acceleration in component form Problem : A particle is moving with an initial velocity
( 0.4i + 0.3j ) m / s2
( 8i + 2j ) m / s , having an acceleration
, which is a constant
Solution :
−1
. Calculate its speed after 10 seconds.
The particle has acceleration of
( 0.4i + 0.3j ) m / s2 = 0.5 m / s
2
acceleration. Its magnitude is
0.4 + 0.3
2
2
making an angle with x-direction.
θ = tan
3 4
. Time interval is 10 seconds. Thus, applying equation of motion, we have :
vx i + vy j + vz k = ( ux i + uy j + uz k ) + ( ax i + ay j + az k ) t ⇒ v = ( 8i + 2j ) + ( 0.4i + 0.3j ) x 10 = 12i + 5j
Speed i.e. magnitude of velocity is :
⇒ v =
122 + 52
= 13 m / s
note: This example illustrates the basic nature of the equations of motion. If we treat them as
scalar equations, we may be led to wrong answers. For example, magnitude of initial velocity i.e. speed is
( 82 + 22 ) = 8.25 m / s , Whereas magnitude of acceleration is
2
. Now, using equation of motion as scalar equation, we have :
0.42 + 0.32
=
0.5 m / s
v = u + at = 8.25 + 0.5 x 10 = 13.25 m / s
5 Equivalent scalar system of equations of motion
We have discussed earlier that a vector quantity in one dimension can be conveniently expressed in terms of an equivalent system of scalar representation. The advantage of linear motion is that we can completely do away with vector notation with an appropriate scheme of assigning plus or minus signs to the quantities involved. The equivalent scalar representation takes advantage of the fact that vectors involved in linear motion has only two possible directions. The one in the direction of chosen axis is considered positive and the other against the direction of the chosen axis is considered negative. At the same time, the concept of component allows us to treat a motion into an equivalent system of the three rectilinear motions in the mutually perpendicular directions along the axes. perpendicular directions. Once we follow the rules of equivalent scalar representation, we can treat equations of motion as scalar equations in the direction of an axis, say x - axis, as : The two concepts, when combined together, renders it possible to treat equations of motion in scalar terms in mutually three
1x : vx = ux + ax t 2x : vavgx = 3x :
( ux + vx ) 2
=
∆ x = x2 − x1 =
( x2 − x1 ) t 1 ux t + 2 ax t2
We have similar set of equations in the remaining two directions. We can obtain the composite interpretation of the motion by combing the individual result in each direction. In order to grasp the method, we rework the earlier example.
Example 3: Acceleration in scalar form Problem : A particle is moving with an initial velocity
( 0.4i + 0.3j ) m / s2
( 8i + 2j ) m / s , having an acceleration
. Calculate its speed after 10 seconds.
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Solution :
The motion in x direction :
ux = 8 m / s ; ax = 0.4 m / s2 ; 10 s and, vx = ux + ax t ⇒ vx = 8 + 0.4 x 10 = 12 m / s
The motion in y direction :
uy = 2 m / s ; ay = 0.3 m / s2 ; 10 s and, vy = uy + ay t ⇒ vy = 2 + 0.3 x 10 = 5 m / s
Therefore, the velocity is :
⇒ v = 2i + 5j ⇒ v = 122 + 52 = 13m / s2
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Constant acceleration
Sunil Kumar Singh
This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License †
∗
Abstract A constant acceleration is a special case of the accelerated motion.
The constant acceleration is a special case of accelerated motion. There are vast instances of motions, which can be approximated to be under constant force and hence constant acceleration. Two of the most important forces controlling motions in our daily life are force due to gravity and friction force. Incidentally, these two forces are constant in the range of motions of bodies in which we are interested. For example, force due to gravity is given by :
F =
GMm r2
= mg
where "G" is the universal constant, "M" is the mass of earth, "m" is the mass of the body and "r" is the distance between center of earth and the body. The resulting acceleration due to gravity, g, is a constant in the immediate neighborhood of the earth surface and is given by :
g =
GM r2
The only variable for a given mass is "r", which changes with the position of the body. The distance "r", however, is equal to Earth's radius for all practical purposes as any dierence arising from the position of body on earth can be ignored. Therefore, acceleration due to gravity can safely be considered to be constant for motions close to the surface of the earth. Signi cantly acceleration due to gravity is a constant irrespective of the mass "m" of the body.
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A ball kicked from a height
Figure 1: The ball follows a parabolic path as it approaches ground
The gure above shows the motion of a ball kicked from the top of a tower.
The ball moves under
constant acceleration of gravity during its ight to the ground. The constant acceleration, therefore, assumes signi cance in relation to the motion that takes place under the in uence of gravity. In the same manner, motion on a rough plane is acted upon by the force of friction in the direction opposite to the motion. The force of friction is a constant force for the moving body and characteristic of the surfaces in contact. As a result, the object slows down at a constant rate.
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Motion in uenced by friction force
Figure 2: The block decelerates due to friction.
1 Understanding constant acceleration
The constant acceleration means that the acceleration is independent of time and is equal to a constant value. The implication of a constant acceleration is discussed here as under : 1: As acceleration is same at all instants during the motion, it follows that average acceleration is equal to instantaneous acceleration during the motion. Mathematically,
aavg = a ⇒
2: When
∆v ∆t
=
v t
∆t
= 1 second, then
aavg = a = ∆ v = v2 − v1
This means that initial velocity , on an average, is changed by the acceleration vector after every second.
Example 1: Constant acceleration Problem : The position of a particle, in meters, moving in the coordinate space is described by
the following functions in time.
x = 2t2 − 4t + 3 ; y = − 2t ; and z = 5
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Find the velocity and acceleration at t = 2 seconds from the start of motion. Also, calculate average acceleration in the rst four seconds.
Solution :
The component velocities in three directions are :
vx = vy = vz =
and the velocity is given by :
x t y t z t
= = =
t t t
2t2 − 4t + 3 ( − 2t ) = − 2 (5) = 0
= 4t − 4
v = vx i + vy j + vz k ⇒ v = ( 4t − 4 ) i − 2j
Thus, velocity at t = 2 seconds,
v2 = 4i − 2j
Acceleration of the particle along three axes are given as :
ax = ay = az =
x t2 2 y t2 2 z t2
2
= = =
2
t2
2
2t2 − 4t + 3 ( 4t ) = 0 (5) = 0
= 4
t2
2
t2
The resultant acceleration is given by :
a = ax i + ay j + az k ⇒ a = 4i m / s2
which is a constant and is independent of time. The accelerations at all time instants are, therefore, same. We know that the average and instantaneous accelerations are equal when acceleration is constant. Hence,
aavg = 4i
. We can check the result calculating average acceleration for the
rst four seconds as :
aavg =
v4 − v0 4
=
12i − 2j + 4i + 2j 4
= 4i m / s2
The important fall out of a constant acceleration is that its magnitude has a constant value and its direction is xed. A change in either of the two attributes, constituting acceleration, shall render acceleration variable. This means that acceleration is along a straight line. But does this linear nature of acceleration mean that the associated motion is also linear? Answer is no. Reason is again the disconnect between acceleration and velocity. We know that magnitude and direction of acceleration are solely determined by the mass of the object and net external force applied on it. Thus, a constant acceleration only indicates that the force i.e the cause that induces change in motion is linear. It does not impose any restriction on velocity to be linear.
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Velocity and acceleration
Figure 3: Force and velocity are in two dierent directions
It is imperative that if the initial velocity of the object is not aligned with linear constant acceleration like in the gure above, then the immediate eect of the applied force, causing acceleration, is to change the velocity. Since acceleration is de ned as the time rate of change in velocity, the resulting velocity would be so directed and its magnitude so moderated that the change in velocity (not the resulting velocity itself ) is aligned in the direction of force.
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Change in velocity
Figure 4: The change in velocity has the same direction as that of acceleration.
As the resulting velocity may not be aligned with the direction of force (acceleration), the resulting motion may not be linear either. For motion being linear, it is essential that the initial velocity and the force applied (and the resulting acceleration) are aligned along a straight line. Examples of motions in more than one dimension with constant acceleration abound in nature. We have already seen that motion of a projectile in vertical plane has constant acceleration due to gravity, having constant magnitude, g, and xed downward direction. If we neglect air resistance, we can assume that all non- propelled projectile motions above ground are accelerated with constant acceleration. In the nutshell, we can say that constant acceleration is unidirectional and linear, but the resulting velocity may not be linear. Let us apply this understanding to the motion of a projectile, which is essentially a motion under constant acceleration due to gravity.
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Parabolic motion
Figure 5: The velocity of the ball is moderated by acceleration.
In the gure, see qualitatively, how the initial velocity vector, v, is modi ed by the constant acceleration vector, g, at the end of successive seconds. Note that combined change in both magnitude and direction of the velocity is taking place at a constant rate and is in vertically downward direction. In the context of constant acceleration, we must also emphasize that both magnitude and direction are constant. A constant acceleration in magnitude only is not su cient. For constant acceleration, the direction of acceleration should also be same (i.e constant). We can have a look at a uniform circular motion in horizontal plane, which follows a horizontal circular path with a constant speed.
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Uniform circular motion
Figure 6: The object moves along a circular path constant speed.
Notwithstanding the constant magnitude, acceleration of uniform circular motion is a variable acceleration in the horizontal plane, because direction of radial centripetal acceleration (shown with red arrow) keeps changing with time. Therefore, the acceleration of the motion keeps changing and is not independent of time as required for acceleration to be constant.
2 Equation of motion
The motion under constant acceleration allows us to describe accelerated motion, using simple mathematical construct. Here, we set out to arrive at these relationships in the form of equations. In these equations,
u
stands for initial velocity,
v
for nal velocity, for constant acceleration and t for time interval
a
of motion under consideration.
This is short of convention followed by many text books and hence the
convention has been retained here.
2.1 First equation
1 : v = u + at
This relation can be established in many ways. One of the fundamental ways is to think that velocity is changed by constant acceleration (vector) at the end of successive seconds. Following this logic, the velocity at the end of t successive seconds are as under :
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At the end of At the end of At the end of At the end of
1st 2nd 3 4
rd th
second second second second
: u + a : u + 2a : u + 3a : u + 4a
At the end of
tth
second
: u + ta = u + at
We can also derive this equation, using the de ning concept of constant acceleration. We know that :
a = aavg ⇒ a =
∆v ∆t
=
v−u t
⇒ at = v − u ⇒ v = u + at
Alternatively (using calculus), we know that :
a = ⇒
Integrating on both sides, we have :
v t
v = a t
∆v = a∆t ⇒ v2 − v1 = at ⇒ v − u = at ⇒ v = u + at
2.2 Second equation
2 : vavg =
u+v 2 The equation v
= u + at
is a linear relationship. Hence, average velocity is arithmatic mean of initial
and nal velocities :
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Average velocity
Figure 7: Average velocity is equal to mean velocity.
⇒ vavg =
u+v 2
2.3 Third equation
3 : s = ∆ r = r2 − r1 = ut +
1 2 2 at This equation is derived by combining the two expressions available for the average velocity.
vavg =
and
∆r ∆t
=
r2 − r1 t
vavg =
( v1 + v2 ) 2
=
(v+u) 2
Combining two expressions of average acceleration and rearranging, we have :
⇒ ( r2 − r1 ) =
Using the relation,
(v+u)t 2
v = u + at
and substituting for v, we have :
⇒ ( r2 − r1 ) =
( u + at + u ) t 2 1 2 2 at
⇒ s = ∆ r = r2 − r1 = = ut +
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Alternatively, we can derive this equation using calculus. Here,
v = ⇒
Integrating between the limits on both sides,
r t
r = v t
r =
Now substituting
v t
v,
⇒ r = ( u + at ) t ut + at t
1 2 2 at
⇒ ∆r =
⇒ s = ∆ r = r2 − r1 = ut +
(
The expression for the displacement has two terms : one varies linearly ( t) with the time and the other 1 2 2 at ) varies with the square of time. The rst term is equal to the displacement due to non-accelerated motion i.e the displacement when the particle moves with uniform velocity, . The second term represents
u
u
the contribution of the acceleration (change in velocity) towards displacement. This equation is used for determining either displacement(∆ ) or position(
r
r1 ).
A common simpli cation,
used widely, is to consider beginning of motion as the origin of coordinate system so that
r1 = 0 ⇒ ∆ r = r2 − r1 = r2 = r ( say )
In this case, both nal position vector and displacement are equal. This simpli cation, therefore, allows us to represent both displacement and position with a single vector variable
r.
3 Graphical interpretation of equations of motion
The three basic equations of motion with constant acceleration, as derived above, are :
1 : v = u + at 2 : vavg =
(u+v) 2 1 2 2 at
3 : s = ∆ r = r2 − r1 = ut +
These three equations completely describe motion of a point like mass, moving with constant acceleration. We need exactly ve parameters to describe the motion under constant acceleration :
u , v , r1 , r2 and t
.
It can be emphasized here that we can not use these equations if the acceleration is not constant. We should use basic dierentiation or integration techniques for motion having variable acceleration (nonuniform acceleration). These equations serve to be a ready to use equations that avoids dierentiation and integration. Further, it is evident that equations of motion are vector equations, involving vector addition. We can evaluate a motion under constant acceleration, using either graphical or algebraic method based on components. Here, we interpret these vector equations, using graphical technique. For illustration purpose, we apply these equations to a motion of an object, which is thrown at an angle of acceleration is "g", which is directed vertically downward. corresponding bold faced symbol
θ
from the horizontal. The magnitude
g.
Let acceleration vector be represented by
Let
and corresponding position vectors are The nal velocity at time instant
v1 and v2 r1 and r2 .
be the velocities at time instants
t1
and
t2
respectively
t2
, is given by :
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v = u + at ⇒ v2 = v1 + g ( t2 − t1 )
Graphically, the nal velocity is obtained by modifying initial vector
v1
by the vector
g ( t2 − t1 )
.
Graphical representation of rst equation
Figure 8
Now, we discuss graphical representation of second equation of motion. The average velocity between two time instants or two positions is given by :
vavg = ⇒
(u+v) 2 v2 ) vavg = ( v1 + 2
The vector addition involved in the equation is graphically represented as shown in the gure. Note that average velocity is equal to half of the vector sum
v1 + v2
.
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Graphical representation of second equation
Figure 9
Third equation of motion provides for displacement in terms of two vector quantities - initial velocity and acceleration. The displacement,
s, is equal to addition of two vector terms :
1 2 2 at 1 2g
s = ∆ r = r2 − r1 = ut +
⇒ s = ∆ r = r2 − r1 = v1 ( t2 − t1 ) +
( t2 − t1 )
2
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Graphical representation of third equation
Figure 10
4 Equations of motion in component form
The application of equations of motion graphically is tedious. In general, we use component representation that allows us to apply equations algebraically. We use equations of motion, using component forms of vector quantities involved in the equations of motion. The component form of the various vector quantities are :
r = xi + y j + z k ∆ r = ∆ xi + ∆ y j + ∆ z k u = ux i + uy j + uz k v = vx i + vy j + vz k v = vavgx i + vavgy j + vavgz k a = ax i + ay j + az k
Using above relations, equations of motion are :
1 : vx i + vy j + vz k = ( ux i + uy j + uz k ) + ( ax i + ay j + az k ) t 2 : vavgx i + vavgy j + vavgz k = 3: ∆ xi + ∆ y j + ∆ z k = (
( ux i + uy j + uz k ) + ( vx i + vy j + vz k ) 2 1 ux i + uy j + uz k ) t + 2 ( ax i +
ay j + az k ) t2
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Example 2: Acceleration in component form Problem : A particle is moving with an initial velocity
( 0.4i + 0.3j ) m / s2
( 8i + 2j ) m / s , having an acceleration
, which is a constant
Solution :
−1
. Calculate its speed after 10 seconds.
The particle has acceleration of
( 0.4i + 0.3j ) m / s2 = 0.5 m / s
2
acceleration. Its magnitude is
0.4 + 0.3
2
2
making an angle with x-direction.
θ = tan
3 4
. Time interval is 10 seconds. Thus, applying equation of motion, we have :
vx i + vy j + vz k = ( ux i + uy j + uz k ) + ( ax i + ay j + az k ) t ⇒ v = ( 8i + 2j ) + ( 0.4i + 0.3j ) x 10 = 12i + 5j
Speed i.e. magnitude of velocity is :
⇒ v =
122 + 52
= 13 m / s
note: This example illustrates the basic nature of the equations of motion. If we treat them as
scalar equations, we may be led to wrong answers. For example, magnitude of initial velocity i.e. speed is
( 82 + 22 ) = 8.25 m / s , Whereas magnitude of acceleration is
2
. Now, using equation of motion as scalar equation, we have :
0.42 + 0.32
=
0.5 m / s
v = u + at = 8.25 + 0.5 x 10 = 13.25 m / s
5 Equivalent scalar system of equations of motion
We have discussed earlier that a vector quantity in one dimension can be conveniently expressed in terms of an equivalent system of scalar representation. The advantage of linear motion is that we can completely do away with vector notation with an appropriate scheme of assigning plus or minus signs to the quantities involved. The equivalent scalar representation takes advantage of the fact that vectors involved in linear motion has only two possible directions. The one in the direction of chosen axis is considered positive and the other against the direction of the chosen axis is considered negative. At the same time, the concept of component allows us to treat a motion into an equivalent system of the three rectilinear motions in the mutually perpendicular directions along the axes. perpendicular directions. Once we follow the rules of equivalent scalar representation, we can treat equations of motion as scalar equations in the direction of an axis, say x - axis, as : The two concepts, when combined together, renders it possible to treat equations of motion in scalar terms in mutually three
1x : vx = ux + ax t 2x : vavgx = 3x :
( ux + vx ) 2
=
∆ x = x2 − x1 =
( x2 − x1 ) t 1 ux t + 2 ax t2
We have similar set of equations in the remaining two directions. We can obtain the composite interpretation of the motion by combing the individual result in each direction. In order to grasp the method, we rework the earlier example.
Example 3: Acceleration in scalar form Problem : A particle is moving with an initial velocity
( 0.4i + 0.3j ) m / s2
( 8i + 2j ) m / s , having an acceleration
. Calculate its speed after 10 seconds.
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Solution :
The motion in x direction :
ux = 8 m / s ; ax = 0.4 m / s2 ; 10 s and, vx = ux + ax t ⇒ vx = 8 + 0.4 x 10 = 12 m / s
The motion in y direction :
uy = 2 m / s ; ay = 0.3 m / s2 ; 10 s and, vy = uy + ay t ⇒ vy = 2 + 0.3 x 10 = 5 m / s
Therefore, the velocity is :
⇒ v = 2i + 5j ⇒ v = 122 + 52 = 13m / s2
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