Loading....
Coupon Accepted Successfully!

 

Accelerated Motion

The velocity of a vehicle on a road is never uniform. Its velocity increases and decreases at random. As the velocity of the vehicle on the road keeps on changing, it is said to have accelerated motion.

The time rate of change of velocity of an object is called acceleration of the object.

Thus, acceleration is the change in velocity in each unit of time. In case, the change in velocity each unit of time is constant, the object is said to be moving with constant acceleration and such a motion is called uniformly accelerated motion.

On the other hand, if the change in velocity in each unit of time is not constant, the object is said to be moving with variable acceleration and such a motion is called non-uniformly accelerated motion.

In general, the motion of a vehicle on a road is non-uniformly accelerated as it speeds up and slows down at random. However, we shall restrict ourselves to the study of only uniformly accelerated motion in a straight line.

Uniformly Accelerated Motion

Consider an object in uniformly accelerated motion along a straight line OX with O as the origin.
Suppose the object reaches the point A at time t and reaches the point B at time t’, such that are the displacements of the object at times t and t’ respectively as shown in the given figure.

Average velocity

The average velocity of the object in the time interval between t and t’ is given by
……(3.1)
 
The equation (3.1) gives only the average velocity during the time interval between t and t’. When the object has accelerated motion, it may not be equal to actual velocity of the object.

However, if the object is in uniform motion (not the uniformly accelerated motion), the average velocity given by equation (3.1) will correspond to its actual velocity.

Acceleration

Consider that an object in uniformly accelerated motion along OX possesses velocities and ’ at t and t’ respectively. Then, the constant acceleration in the motion of the object is given by
       ……(3.2)
Acceleration is a vector quantity. It may be positive or negative. If the velocity of the object is steadily increasing, it has positive acceleration. On the other hand, if the velocity is decreasing steadily, the object has negative acceleration. Negative acceleration is sometimes called as retardation or deceleration. The unit of acceleration is ms-2 in SI.

The physical meaning of acceleration is that by how much the velocity of the object is changing in each unit of time.

Formulae for Uniformly Accelerated Motion

When an object is in uniformly accelerated motion, the constant acceleration is given by equation (3.2). As said in previous chapter, the displacements such as and the velocities represent vectors having the same direction and consequently the acceleration is also in the same direction.

Therefore, for the sake of convenience, in the equations (3.1) and (3.2) arrowheads over displacement, velocity and acceleration vectors may be dropped. Hence, the equation (3.2) may be written as
                        ……(3.3)
 
Since motion is uniformly accelerated,
a = constant
  1. Velocity-Time Relation
    From the equation (3.3), we have
    V’ = v + a(t’ – t)                    ……(3.4)
    The equation (3.4) is velocity-time relation, which gives the velocity of an object at any time t’ in terms of its velocity at any time t and the uniform acceleration 'a' of the object.

    Let u be the velocity of the object at t = 0 (initially) and v be its velocity at time t. Then, from the definition of acceleration, we have....(3.5)

    The equation (3.5) is velocity-time relation, which gives the velocity of an object at any time t, when moving with initial velocity u (velocity at t = 0) and uniform acceleration 'a'.
  2. Position-Time Relation
    Consider an object moving with a constant acceleration a along a straight line. Let v be the velocity and x be the displacement of the object at any time t and v’ and x’ be the corresponding values at time t’. Now, from equation (3.1), the average velocity of the object in the time interval between t and t’ may be written as

    x’ = x + (t’ – t) uav
    Also, vav = (v + v’)…..(3.6)
    Substituting the value of vav in equation (3.6), we have
    x’ = x + (t’ – t) (v + v’)               ……(3.7)
    As given by equation (3.4), the velocity of an object at any time t’ is related to its velocity at time t the equation
    v’ = v + a(t’ – t)
    where a is the constant acceleration of the motion of the object. Substituting the value of v’ in equation (3.7), we get
    x’ = x + (t’ – t) [v + v + a(t’ – t)]
    or x’ = x + v(t’ – t) + a(t’ – t)2         ……(3.8)
    The equation (3.8) relates the position of the object at time t’ to its position and velocity at any time t, when the object moves with constant acceleration a along a straight line. Let u be the velocity, x0 be the displacement of the object at t = 0 (initially) and x be its position at any time t. Then, equation (3.8) reduces to
    x = x0 + u(t-0) + a (t – 0)2
    or x = x0 + ut + at2                 ……(3.9)

    Thus, position of an object at any time t has quadratic dependence on time, when the object moves with constant acceleration along a straight line. The co-ordinate geometry tells that position-time graph for such a uniformly accelerated motion will be parabolic in nature.

    If we call x – x0 = S, displacement of the object in time t, then equation (3.9) becomes
    S = ut + at2                      ------(3.10)
  3. Velocity-Displacement Relation
    From equation (3.4) we have
    t’ – t =
    Substituting for t’ – t in equation (3.8), we have
    x’ = x + v+ a
    Or x’ – x = [2v (v’-v) + (v’-v)2]
    = [2vv’ – 2v2 + v’2 + v2 – 2cc’] =
    or v’2 – v2 = 2a (x’ – x)          ……(3.11)

    The equation (3.11) is velocity-displacement relation between velocities of the object at times t’ and t and the corresponding positions at these instants.

    Also, from equation (3.5) we have
    t =

    Substituting for t in equation (3.9) we have
    x = x0 + u + a 2
    Or x – x0 = [2u(v-u)+(v-u)2]
    = [2uv – 2u2 + v2 + u2 – 2uv] =
    of v2 – u2 = 2a(x-x0)              ……(3.12)

    The equation (3.12) is velocity-displacement relation between velocities of the object at time t and t = 0 and the corresponding positions of these instants. The equation (3.12) can be obtained from equation (3.11) by merely changing v’ to v; v to u ;x’ to x and x to x0.

    If we denote x – x0 = S, the distance covered by the object in time t, then equation (3.12) becomes
    V2 – u2 = 2aS                        ……(3.13)

    The equations (3.3), (3.4), (3.5), (3.8), (3.9), (3.11), (3.12) and (3.13) are the formulae for kinematics of uniformly accelerated motion.

Displacement in nth Second of Motion

Consider a particle starting with initial velocity u and moving with uniform acceleration a. Let the displacement of the particle in the nth second of its motion be Snth.
 
Then,
Snth = Sn – Sn-1             ……(3.14)
 
Where Sn and Sn-1 are displacements of the particle in n and n-1 seconds. The values of Sn and Sn-1 can be obtained by putting t = n and t = n – 1 in the equation.
S = ut + at2
Therefore, from equation (3.14) we have
Snth = -

      = -

      = u(n-n+1) + a (n2 – n2 –1 + n)

or Snth = u + a(2n-1) -----(3.15)
 
Example

An electron is emitted with a velocity of 5 × 106 ms-2. If its final velocity is 7 × 106 ms-1, calculate the distance discovered by the electron.


Solution
Here, u = 5
× 106 ms-1; v = 7 × 106 ms-1 ; a = 3 × 1014 ms-2

We know v2 – u2 = 2aS

 (7 × 106)2 – (5 × 10)6 = 2 × 3 1014 × S or 49 × 1012 – 25 × 1012 = 6 × 1014S

or S = = 0.04 m
 

 
Example

A car moving along a straight highway with speed of 126 km h-1 is brought to a stop within a distance of 200 m then how long did it take for the car to go?


Solution

Here, distance covered, S = 200 m

Initial speed of the car, u = 126 km h-1 = 126 × (1000m) × (60 × 60s)-1 = 35 ms-1  

Final speed of the car, v = 0

Let a be the retardation of the car

We know, v2 – u2 = 2aS

(0)2 – (35)2 = 2a × 200 or a = - = 3.06 ms-2

v = u + at or 0 = 35 + t

or t = = 11.43 s
 

Velocity-Time Graph (Uniformly Accelerated Motion)

To study the velocity-time graph for a uniformly accelerated motion, consider an object moving with an initial velocity of 10ms-1 and a constant acceleration of ms-2. Thus,
 
u = 10ms-1 ; a = 5ms-2
 
The equation (3.5) can be used to find velocity of the object at time t = 1s, 2s, 3s …………. The following table gives the velocity of the object at times t = 0, 1s, 2s, ……… If we draw the velocity-time graph the object in uniformly accelerated motion, is shown in the figure.

Thus, velocity-time graph of a uniformly accelerated motion is a straight line inclined to the time-axis. When the object has positive constant acceleration (as in the present example), the graph slopes upward (positive slope).

In case the object has negative constant acceleration, the velocity-time graph will slope downward (negative slope).
Time (s) Velocity (ms-1)
0 10
1 15
2 20
3 25
4 30
5 35

               

To highlight the fact that what useful information can be drawn from (v – t) graph of an object, consider the v – t graph of an object as shown in the figure. It shows that the object which is initially at rest, first uniformly accelerates for time t1 till it attains a velocity v moves with this velocity till time t2 i.e. for a time interval (t2 – t1) and then uniformly decelerates for a time interval (t3 – t2) and finally, it again comes to rest. In the time interval between t = 0 to t = t1, the object has positive acceleration and its velocity goes on increasing, while in time interval between t = t2 and t = t3, the object has negative acceleration and its velocity goes on decreasing.
It may be pointed out that it is wrong to think that velocity of an object increases or decreases depending on whether it has positive or negative acceleration. It is the value of time t in the velocity-time relation.
 
V = u + at
 
which affects the change in velocity i.e. whether the body speeds up or speeds down. It may be pointed out that an object always slows down for the time before its velocity becomes zero and speeds up after that instant. The time at which the velocity of the object will become zero can be obtained by setting v = 0 in the equation given above. Setting v = 0, we have
t =

Therefore, for the object moving with initial velocity of 10 ms-1 and a constant acceleration of 5 ms-2, we have

t = - = -2s
 

The object, whose motion is under consideration, will slow down for t < -2 s and will speed up for t > - 2s. It can be ascertained by producing the velocity-time graph of the figure, in backward direction.

Position-Time Graph (Uniformly Accelerated Motion)

Consider an object moving along a straight path with initial velocity of 5 ms-1 and a constant acceleration of 5ms-2. Suppose that the object is at a distance of 10m from the origin of position-axis time t = 0. Thus,
 
x0 = 10m ; u = 5ms-1 ; a = 5ms-2
 
The equation (3.9) can be used to find the position of the object at t = 1s, 2s, 3s,…….. The following table gives the position of the object at times t = 0, 1s, 2s…………. The figure, represents the position-time graph of the object. It may be pointed out that the position-time graph of an object having accelerated motion is parabolic in nature.

        Time (s)

           Velocity (m)

           0

               10.0

           1

               17.5

           2

               30.0

           3

               47.5

           4

               70.0

           5

               97.5

 

                

To know what information the x – t graph of an object can give, let us consider the x – t graph of an object as shown in the figure.

In this x – t graph of the object, the part AB is horizontal straight line, part BC is parabolic in nature, part CD is an oblique straight line, part DE is again parabolic (but curving in opposite direction) and finally part EF is again a horizontal straight line. It follows that
  1. in time interval between t = 0 and t = t1, the object remains at rest;
  2. in time interval between t = t1 to t = t2, the object speeds up* with a constant acceleration;
  3. in time interval between t = t3 to t = t3 the object slows down with constant velocity.
  4. in time interval between t = t3 to t = t4, the object slows down with a constant acceleration and
  5. in time interval between t = t4 to t = t5 , the object again remains at rest.

Distance covered as Area Under Velocity Time graph

The velocity-time graph of an object may be used to calculate the distance covered by the object in certain time interval. Figure represents the velocity-time graph of a uniformly accelerated motion. Corresponding to time t and t’, consider two points A and B on the velocity time graph. Then, the object at time t and t’ respectively. Also, A’B’ = t’ – t
Area under velocity-time graph between t and t’
= area of trapezium ABB’A’
= (AA’ + BB’) × A’B’
= (v + v’) × (t’ – t)
 
Now, from definition of the acceleration,
a =
or t’ – t = ……(3.16)
 
Substituting for t’ – t in equation (3.16), we have
Area under v–t graph between t and t’ = (v +v’) ´ = = ---- (3.17)
But from the equation v’2 – v2 = 2a (x’ – x), we have
= x’ – x ….(3.18)
Therefore, from the equations (3.17) and (3.18), it follows that
x’- x = area under the v – t graph between t and t’




Test Your Skills Now!
Take a Quiz now
Reviewer Name