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# Constant acceleration

Module by: Sunil Kumar Singh. E-mail the author

Summary: 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 r 2 = mg F = GMm r 2 = 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 r 2 g = GM r 2

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 difference 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. Significantly acceleration due to gravity is a constant irrespective of the mass "m" of the body.

The figure above shows the motion of a ball kicked from the top of a tower. The ball moves under constant acceleration of gravity during its flight to the ground. The constant acceleration, therefore, assumes significance in relation to the motion that takes place under the influence 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.

## 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,

a avg = a Δ v Δ t = đ v đ t a avg = a Δ v Δ t = đ v đ t

2: When Δt = 1 second, then

a avg = a = Δ v = v 2 - v 1 a avg = a = Δ v = v 2 - v 1

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 = 2 t 2 - 4 t + 3 ; y = - 2 t ; and z = 5 x = 2 t 2 - 4 t + 3 ; y = - 2 t ; and z = 5

Find the velocity and acceleration at t = 2 seconds from the start of motion. Also, calculate average acceleration in the first four seconds.

Solution : The component velocities in three directions are :

v x = đ x đ t = đ đ t ( 2 t 2 - 4 t + 3 ) = 4 t - 4 v y = đ y đ t = đ đ t ( - 2 t ) = - 2 v z = đ z đ t = đ đ t ( 5 ) = 0 v x = đ x đ t = đ đ t ( 2 t 2 - 4 t + 3 ) = 4 t - 4 v y = đ y đ t = đ đ t ( - 2 t ) = - 2 v z = đ z đ t = đ đ t ( 5 ) = 0

and the velocity is given by :

v = v x i + v y j + v z k v = ( 4 t - 4 ) i - 2 j v = v x i + v y j + v z k v = ( 4 t - 4 ) i - 2 j

Thus, velocity at t = 2 seconds,

v 2 = 4 i - 2 j v 2 = 4 i - 2 j

Acceleration of the particle along three axes are given as :

a x = đ 2 x đ t 2 = đ 2 đ t 2 ( 2 t 2 - 4 t + 3 ) = 4 a y = đ 2 y đ t 2 = đ 2 đ t 2 ( 4 t ) = 0 a z = đ 2 z đ t 2 = đ 2 đ t 2 ( 5 ) = 0 a x = đ 2 x đ t 2 = đ 2 đ t 2 ( 2 t 2 - 4 t + 3 ) = 4 a y = đ 2 y đ t 2 = đ 2 đ t 2 ( 4 t ) = 0 a z = đ 2 z đ t 2 = đ 2 đ t 2 ( 5 ) = 0

The resultant acceleration is given by :

a = a x i + a y j + a z k a = 4 i m / s 2 a = a x i + a y j + a z k a = 4 i m / s 2

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, a avg = 4 i a avg = 4 i . We can check the result calculating average acceleration for the first four seconds as :

a avg = v 4 - v 0 4 = 12 i - 2 j + 4 i + 2 j 4 = 4 i m / s 2 a avg = v 4 - v 0 4 = 12 i - 2 j + 4 i + 2 j 4 = 4 i m / s 2

The important fall out of a constant acceleration is that its magnitude has a constant value and its direction is fixed. 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.

It is imperative that if the initial velocity of the object is not aligned with linear constant acceleration like in the figure above, then the immediate effect of the applied force, causing acceleration, is to change the velocity. Since acceleration is defined 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.

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 fixed 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.

In the figure, see qualitatively, how the initial velocity vector, v, is modified 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 sufficient. 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.

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.

## 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 final velocity, “a” for constant acceleration and “t” for time interval of motion under consideration. This is short of convention followed by many text books and hence the convention has been retained here.

### First equation

1: v = u + a t 1: v = u + a t

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 :

At the end of 1 st second : u + a At the end of 2 nd second : u + 2 a At the end of 3 rd second : u + 3 a At the end of 4 th second : u + 4 a ------------------------------- ------------------------------- At the end of t th second : u + t a = u + a t At the end of 1 st second : u + a At the end of 2 nd second : u + 2 a At the end of 3 rd second : u + 3 a At the end of 4 th second : u + 4 a ------------------------------- ------------------------------- At the end of t th second : u + t a = u + a t

We can also derive this equation, using the defining concept of constant acceleration. We know that :

a = a avg a = Δ v Δ t = v - u t a t = v - u v = u + a t a = a avg a = Δ v Δ t = v - u t a t = v - u v = u + a t

Alternatively (using calculus), we know that :

a = đ v đ t đ v = a đ t a = đ v đ t đ v = a đ t

Integrating on both sides, we have :

Δ v = a Δ t v 2 - v 1 = a t v - u = a t v = u + a t Δ v = a Δ t v 2 - v 1 = a t v - u = a t v = u + a t

### Second equation

2: v avg = u + v 2 2: v avg = u + v 2

The equation v = u + a t v = u + a t is a linear relationship. Hence, average velocity is arithmatic mean of initial and final velocities :

v avg = u + v 2 v avg = u + v 2

### Third equation

3: s = Δ r = r 2 - r 1 = u t + 1 2 a t 2 3: s = Δ r = r 2 - r 1 = u t + 1 2 a t 2

This equation is derived by combining the two expressions available for the average velocity.

v avg = Δ r Δ t = r 2 - r 1 t v avg = Δ r Δ t = r 2 - r 1 t

and

v avg = ( v 1 + v 2 ) 2 = ( v + u ) 2 v avg = ( v 1 + v 2 ) 2 = ( v + u ) 2

Combining two expressions of average acceleration and rearranging, we have :

( r 2 - r 1 ) = ( v + u ) t 2 ( r 2 - r 1 ) = ( v + u ) t 2

Using the relation, v = u + a t v = u + a t and substituting for v, we have :

( r 2 - r 1 ) = ( u + a t + u ) t 2 s = Δ r = r 2 - r 1 = = u t + 1 2 a t 2 ( r 2 - r 1 ) = ( u + a t + u ) t 2 s = Δ r = r 2 - r 1 = = u t + 1 2 a t 2

Alternatively, we can derive this equation using calculus. Here,

v = đ r đ t đ r = v đ t v = đ r đ t đ r = v đ t

Integrating between the limits on both sides,

đ r = v đ t đ r = v đ t

Now substituting v,

đ r = ( u + a t ) đ t Δ r = u t + a t đ t s = Δ r = r 2 - r 1 = u t + 1 2 a t 2 đ r = ( u + a t ) đ t Δ r = u t + a t đ t s = Δ r = r 2 - r 1 = u t + 1 2 a t 2

The expression for the displacement has two terms : one varies linearly (ut) with the time and the other ( 1 2 a t 2 1 2 a t 2 ) varies with the square of time. The first term is equal to the displacement due to non-accelerated motion i.e the displacement when the particle moves with uniform velocity, u. The second term represents the contribution of the acceleration (change in velocity) towards displacement.

This equation is used for determining either displacement(Δr) or position( r 1 r 1 ). A common simplification, used widely, is to consider beginning of motion as the origin of coordinate system so that

r 1 = 0 Δ r = r 2 - r 1 = r 2 = r ( say ) r 1 = 0 Δ r = r 2 - r 1 = r 2 = r ( say )

In this case, both final position vector and displacement are equal. This simplification, therefore, allows us to represent both displacement and position with a single vector variable r.

## Graphical interpretation of equations of motion

The three basic equations of motion with constant acceleration, as derived above, are :

1: v = u + a t 2: v avg = ( u + v ) 2 3: s = Δ r = r 2 - r 1 = u t + 1 2 a t 2 1: v = u + a t 2: v avg = ( u + v ) 2 3: s = Δ r = r 2 - r 1 = u t + 1 2 a t 2

These three equations completely describe motion of a point like mass, moving with constant acceleration. We need exactly five parameters to describe the motion under constant acceleration : u , v , r 1 , r 2 and t u , v , r 1 , r 2 and t .

It can be emphasized here that we can not use these equations if the acceleration is not constant. We should use basic differentiation or integration techniques for motion having variable acceleration (non-uniform acceleration). These equations serve to be a ready to use equations that avoids differentiation 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 θ from the horizontal. The magnitude of acceleration is "g", which is directed vertically downward. Let acceleration vector be represented by corresponding bold faced symbol g. Let v 1 v 1 and v 2 v 2 be the velocities at time instants t 1 t 1 and t 2 t 2 respectively and corresponding position vectors are r 1 r 1 and r 2 r 2 .

The final velocity at time instant t 2 t 2 , is given by :

v = u + a t v 2 = v 1 + g ( t 2 - t 1 ) v = u + a t v 2 = v 1 + g ( t 2 - t 1 )

Graphically, the final velocity is obtained by modifying initial vector v 1 v 1 by the vector g ( t 2 - t 1 ) g ( t 2 - t 1 ) .

Now, we discuss graphical representation of second equation of motion. The average velocity between two time instants or two positions is given by :

v avg = ( u + v ) 2 v avg = ( v 1 + v 2 ) 2 v avg = ( u + v ) 2 v avg = ( v 1 + v 2 ) 2

The vector addition involved in the equation is graphically represented as shown in the figure. Note that average velocity is equal to half of the vector sum v 1 + v 2 v 1 + v 2 .

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 :

s = Δ r = r 2 - r 1 = u t + 1 2 a t 2 s = Δ r = r 2 - r 1 = v 1 ( t 2 - t 1 ) + 1 2 g ( t 2 - t 1 ) 2 s = Δ r = r 2 - r 1 = u t + 1 2 a t 2 s = Δ r = r 2 - r 1 = v 1 ( t 2 - t 1 ) + 1 2 g ( t 2 - t 1 ) 2

## 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 = x i + y j + z k Δ r = Δ x i + Δ y j + Δ z k u = u x i + u y j + u z k v = v x i + v y j + v z k v = v avgx i + v avgy j + v avgz k a = a x i + a y j + a z k r = x i + y j + z k Δ r = Δ x i + Δ y j + Δ z k u = u x i + u y j + u z k v = v x i + v y j + v z k v = v avgx i + v avgy j + v avgz k a = a x i + a y j + a z k

Using above relations, equations of motion are :

1: v x i + v y j + v z k = ( u x i + u y j + u z k ) + ( a x i + a y j + a z k ) t 2: v avgx i + v avgy j + v avgz k = ( u x i + u y j + u z k ) + ( v x i + v y j + v z k ) 2 3: Δ x i + Δ y j + Δ z k = ( u x i + u y j + u z k ) t + 1 2 ( a x i + a y j + a z k ) t 2 1: v x i + v y j + v z k = ( u x i + u y j + u z k ) + ( a x i + a y j + a z k ) t 2: v avgx i + v avgy j + v avgz k = ( u x i + u y j + u z k ) + ( v x i + v y j + v z k ) 2 3: Δ x i + Δ y j + Δ z k = ( u x i + u y j + u z k ) t + 1 2 ( a x i + a y j + a z k ) t 2

### Example 2: Acceleration in component form

Problem : A particle is moving with an initial velocity ( 8 i + 2 j ) m / s ( 8 i + 2 j ) m / s , having an acceleration ( 0.4 i + 0.3 j ) m / s 2 ( 0.4 i + 0.3 j ) m / s 2 . Calculate its speed after 10 seconds.

Solution : The particle has acceleration of ( 0.4 i + 0.3 j ) m / s 2 ( 0.4 i + 0.3 j ) m / s 2 , which is a constant acceleration. Its magnitude is ( 0.4 2 + 0.3 2 ) = 0.5 m / s 2 ( 0.4 2 + 0.3 2 ) = 0.5 m / s 2 making an angle with x-direction. θ = tan -1 ( 3 4 ) θ = tan -1 ( 3 4 ) . Time interval is 10 seconds. Thus, applying equation of motion, we have :

v x i + v y j + v z k = ( u x i + u y j + u z k ) + ( a x i + a y j + a z k ) t v = ( 8 i + 2 j ) + ( 0.4 i + 0.3 j ) x 10 = 12 i + 5 j v x i + v y j + v z k = ( u x i + u y j + u z k ) + ( a x i + a y j + a z k ) t v = ( 8 i + 2 j ) + ( 0.4 i + 0.3 j ) x 10 = 12 i + 5 j

Speed i.e. magnitude of velocity is :

v = ( 12 2 + 5 2 ) = 13 m / s v = ( 12 2 + 5 2 ) = 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 ( 8 2 + 2 2 ) = 8.25 m / s ( 8 2 + 2 2 ) = 8.25 m / s , Whereas magnitude of acceleration is ( 0.4 2 + 0.3 2 ) = 0.5 m / s 2 ( 0.4 2 + 0.3 2 ) = 0.5 m / s 2 . Now, using equation of motion as scalar equation, we have : v = u + a t = 8.25 + 0.5 x 10 = 13.25 m / s v = u + a t = 8.25 + 0.5 x 10 = 13.25 m / s

## 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. The two concepts, when combined together, renders it possible to treat equations of motion in scalar terms in mutually three 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 :

1x: v x = u x + a x t 2x: v avgx = ( u x + v x ) 2 = ( x 2 - x 1 ) t 3x: Δ x = x 2 - x 1 = u x t + 1 2 a x t 2 1x: v x = u x + a x t 2x: v avgx = ( u x + v x ) 2 = ( x 2 - x 1 ) t 3x: Δ x = x 2 - x 1 = u x t + 1 2 a x t 2

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 ( 8 i + 2 j ) m / s ( 8 i + 2 j ) m / s , having an acceleration ( 0.4 i + 0.3 j ) m / s 2 ( 0.4 i + 0.3 j ) m / s 2 . Calculate its speed after 10 seconds.

Solution : The motion in x – direction :

u x = 8 m / s ; a x = 0.4 m / s 2 ; 10 s and, v x = u x + a x t v x = 8 + 0.4 x 10 = 12 m / s u x = 8 m / s ; a x = 0.4 m / s 2 ; 10 s and, v x = u x + a x t v x = 8 + 0.4 x 10 = 12 m / s

The motion in y – direction :

u y = 2 m / s ; a y = 0.3 m / s 2 ; 10 s and, v y = u y + a y t v y = 2 + 0.3 x 10 = 5 m / s u y = 2 m / s ; a y = 0.3 m / s 2 ; 10 s and, v y = u y + a y t v y = 2 + 0.3 x 10 = 5 m / s

Therefore, the velocity is :

v = 2 i + 5 j v = ( 12 2 + 5 2 ) = 13 m / s 2 v = 2 i + 5 j v = ( 12 2 + 5 2 ) = 13 m / s 2

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