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

# Constant acceleration (application)

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

Summary: Solving problems is an essential part of the understanding

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

## Hints on solving problems

1. Though acceleration is constant and hence one – dimensional, but the resulting motion can be one, two or three dimensional – depending on the directional relation between velocity and acceleration.
2. Identify : what is given and what is required. Establish relative order between given and required attribute.
3. Use differentiation method to get a higher order attribute in the following order : displacement (position vector) → velocity → acceleration.
4. Use integration method to get a lower order attribute in the following order : acceleration → velocity → displacement (position vector).

## Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the motion with constant acceleration. The questions are categorized in terms of the characterizing features of the subject matter :

• Average velocity
• Differentiation and Integration method
• Components of constant acceleration
• Rectilinear motion with constant acceleration
• Equations of motion

## Average velocity

### Example 1

Problem : A particle moves with an initial velocity “u” and a constant acceleration “a”. What is average velocity in the first “t” seconds?

Solution : The particle is moving with constant acceleration. Since directional relation between velocity and acceleration is not known, the motion can have any dimension. For this reason, we shall be using vector form of equation of motion. Now, the average velocity is given by :

v a = Δ r Δ t v a = Δ r Δ t

The displacement for motion with constant acceleration is given as :

Δ r = u t + 1 2 a t 2 Δ r = u t + 1 2 a t 2

Thus, average velocity is :

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

## Differentiation and Integration methods

### Example 2

Problem : A particle is moving with a velocity 2 i + 2 t j 2 i + 2 t j in m/s. Find (i) acceleration and (ii) displacement at t = 1 s.

Solution : Since velocity is given as a function in “t”, we can find acceleration by differentiating the function with respect to time.

a = đ đ t ( 2 i + 2 t j ) = 2 j a = đ đ t ( 2 i + 2 t j ) = 2 j

Thus, acceleration is constant and is directed in y-direction. However, as velocity and acceleration vectors are not along the same direction, the motion is in two dimensions. Since acceleration is constant, we can employ equation of motion for constant acceleration in vector form,

Δ r = u t + 1 2 a t 2 Δ r = ( 2 i + 2 t j ) t + 1 2 x 2 j x t 2 Δ r = u t + 1 2 a t 2 Δ r = ( 2 i + 2 t j ) t + 1 2 x 2 j x t 2

For t = 1 s

Δ r = ( 2 i + 2 x 1 j ) x 1 + 1 2 x 2 j x 1 2 Δ r = 2 i + 3 j Δ r = ( 2 i + 2 x 1 j ) x 1 + 1 2 x 2 j x 1 2 Δ r = 2 i + 3 j

Note 1 : We should remind ourselves that we obtained displacement using equation of motion for constant acceleration. Had the acceleration been variable, then we would have used integration method to find displacement.

Note 2 : A constant acceleration means that neither its magnitude or direction is changing. Therefore, we may be tempted to think that a constant acceleration is associated with one dimensional motion. As we see in the example, this is not the case. A constant acceleration can be associated with two or three dimensional motion as well. It is the relative directions of acceleration with velocity that determines the dimension of motion – not the dimension of acceleration itself.

### Example 3

Problem : The coordinates of a particle (m/s) in a plane at a given time “t” is 2 t , t 2 2 t , t 2 . Find (i) path of motion (ii) velocity at time “t” and (iii) acceleration at time “t”.

Solution : Clearly, the position of the particle is a function of time and the particle moves in a two dimensional xy - plane. Here,

x = 2 t y = t 2 x = 2 t y = t 2

In order to find the relation between “x” and “y”, we substitute “t” from the first equation in to second as :

y = ( x 2 ) 2 = x 2 4 x 2 = 4 y y = ( x 2 ) 2 = x 2 4 x 2 = 4 y

Hence, path of motion is parabolic. Now, the position vector is :

r = 2 t i + t 2 j r = 2 t i + t 2 j

Differentiating with respect to time, the velocity of the particle is :

v = đ r đ t = 2 i + 2 t j m / s v = đ r đ t = 2 i + 2 t j m / s

Further differentiating with respect to time, the acceleration of the particle is :

a = đ v đ t = 2 j m / s 2 a = đ v đ t = 2 j m / s 2

## Components of acceleration

### Example 4

Problem : At a certain instant, the components of velocity and acceleration are given as :

v x = 4 m / s ; v y = 3 m / s ; a x = 2 m / s 2 ; a y = 1 m / s 2 . v x = 4 m / s ; v y = 3 m / s ; a x = 2 m / s 2 ; a y = 1 m / s 2 .

What is the rate of change of speed?

Solution : Here, phrasing of question is important. We are required to find the rate of change of speed – not the rate of change of velocity or magnitude of rate of change of velocity. Let us have a look at the subtle differences in the meaning here :

1: The rate of change of velocity

The rate of change of velocity is equal to acceleration. For the given two dimensional motion,

a = đ v đ t = đ đ t ( a x i + a y j ) a = đ v đ t = 2 i + 1 j a = đ v đ t = đ đ t ( a x i + a y j ) a = đ v đ t = 2 i + 1 j

2: The magnitude of rate of change of velocity

The magnitude of rate of change of velocity is equal to magnitude of acceleration. For the given two dimensional motion,

| a | = | đ v đ t | = ( 2 2 + 1 2 ) = 5 m / s 2 | a | = | đ v đ t | = ( 2 2 + 1 2 ) = 5 m / s 2

3: The rate of change of speed

The rate of change of speed (dv/dt) is not equal to the magnitude of acceleration, which is equal to the absolute value of the rate of change of velocity. It is so because speed is devoid of direction, whereas acceleration consists of both magnitude and direction.

Let “v” be the instantaneous speed, which is given in terms of its component as :

v 2 = v x 2 + v y 2 v 2 = v x 2 + v y 2

We need to find rate of change of speed i.e dv/dt, using the values given in the question. Therefore, we need to differentiate speed with respect to time,

2 v đ v đ t = 2 v x ( đ v x đ t ) + 2 v y ( đ v y đ t ) 2 v đ v đ t = 2 v x ( đ v x đ t ) + 2 v y ( đ v y đ t )

If we ponder a bit, then we would realize that when we deal with component speed or magnitude of component velocity then we are essentially dealing with unidirectional motion. No change in direction is possible as components are aligned to a fixed axis. As such, equating rate of change in speed with the magnitude of acceleration in component direction is valid. Now, proceeding ahead,

đ v đ t = v x a x + v y a y v đ v đ t = v x a x + v y a y v

đ v đ t = v x a x + v y a y ( v x 2 + v y 2 ) đ v đ t = v x a x + v y a y ( v x 2 + v y 2 )

Putting values, we have :

đ v đ t = 4 x 2 + 3 x 1 ( 4 2 + 3 2 ) đ v đ t = 11 5 = 2.2 m / s đ v đ t = 4 x 2 + 3 x 1 ( 4 2 + 3 2 ) đ v đ t = 11 5 = 2.2 m / s

Note: This is an important question as it brings out differences in interpretation of familiar terms. In order to emphasize the difference, we summarize the discussion as hereunder :

i: In general (i.e two or three dimensions),

đ v đ t | đ v đ t | đ v đ t | a | đ v đ t a đ v đ t | đ v đ t | đ v đ t | a | đ v đ t a

ii: In the case of one dimensional motion, the inequality as above disappears.

đ v đ t = a đ v đ t = a

## Rectilinear motion with constant acceleration

### Example 5

Problem : A block is released from rest on a smooth inclined plane. If S n S n denotes the distance traveled by it from t = n - 1 second to t = n seconds, then find the ratio :

A = S n S n + 1 A = S n S n + 1

Solution : It must be noted that the description of linear motion is governed by the equations of motion whether particle moves on a horizontal surface (one dimensional description) or on an inclined surface (two dimensional description). Let us orient our coordinates so that the motion can be treated as one dimensional unidirectional motion. This allows us to use equations of motion in scalar form,

S n = u + a 2 ( 2 n - 1 ) S n = u + a 2 ( 2 n - 1 )

Here, u = 0, thus

S n = a 2 ( 2 n - 1 ) S n = a 2 ( 2 n - 1 )

Following the description of term Sn as given by the question, we can define Sn+1 as the linear distance from t = n second to t = n + 1 seconds. Thus, substituting “n” by “n+1” in the formulae, we have :

S n + 1 = a 2 ( 2 n + 2 - 1 ) S n + 1 = a 2 ( 2 n + 1 ) S n + 1 = a 2 ( 2 n + 2 - 1 ) S n + 1 = a 2 ( 2 n + 1 )

The required ratio is :

A = S n S n + 1 = ( 2 n - 1 ) ( 2 n + 1 ) A = S n S n + 1 = ( 2 n - 1 ) ( 2 n + 1 )

## Equations of motion

### Example 6

Problem : A force of 2 N is applied on a particle of mass 1 kg, which is moving with a velocity 4 m/s in a perpendicular direction. If the force is applied all through the motion, then find displacement and velocity after 2 seconds.

Solution : It is a two dimensional motion, but having a constant acceleration. Notably, velocity and accelerations are not in the same direction. In order to find the displacement at the end of 2 seconds, we shall use algebraic method. Let the direction of initial velocity and acceleration be along “x” and “y” coordinates (they are perpendicular to each other). Also, let “A” be the initial position and “B” be the final position of the particle. The displacement between A (position at time t =0) and B (position at time t = 2 s) is given as :

AB = u t + 1 2 a t 2 AB = u t + 1 2 a t 2

For time t = 2 s,

AB t=2 = 2 u + 1 2 a x 2 2 = 2 ( u + a ) AB t=2 = 2 u + 1 2 a x 2 2 = 2 ( u + a )

This is a vector equation involving sum of two vectors at right angles. According to question,

u = 4 m / s ; a = F m = 2 1 = 2 m / s 2 u = 4 m / s ; a = F m = 2 1 = 2 m / s 2

Since u and a perpendicular to each other, the magnitude of the vector sum (u + a) is :

| u + a | = ( u 2 + a 2 ) = ( 4 2 + 2 2 ) = 2 5 | u + a | = ( u 2 + a 2 ) = ( 4 2 + 2 2 ) = 2 5

Hence, magnitude of displacement is :

AB t=2 = 2 | u + a | = = 2 x 2 5 = 4 5 m AB t=2 = 2 | u + a | = = 2 x 2 5 = 4 5 m

Let the displacement vector makes an angle “θ” with the direction of initial velocity.

tan θ = a u = 2 4 = 1 2 tan θ = a u = 2 4 = 1 2

Let the direction of initial velocity and acceleration be along “x” and “y” coordinates (they are perpendicular to each other). Then,

u = 4 i a = 2 j u = 4 i a = 2 j

Using equation of motion for constant acceleration, the final velocity is :

v = u + a t v = 4 i + 2 j x 1 = 4 ( i + j ) v = u + a t v = 4 i + 2 j x 1 = 4 ( i + j )

The magnitude of velocity is :

v = | v | = 4 ( 1 2 + 1 2 ) = 4 2 m v = | v | = 4 ( 1 2 + 1 2 ) = 4 2 m

Let the final velocity vector makes an angle “θ” with the direction of initial velocity.

tan θ = 1 1 = 1 θ = 45 ° tan θ = 1 1 = 1 θ = 45 °

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