# 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 tâ€™, such that are the displacements of the object at times t and tâ€™ respectively as shown in the given figure.

# Average velocity

The average velocity of the object in the time interval between t and tâ€™ is given byâ€¦â€¦(3.1)

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 â€™ at t and tâ€™ respectively. Then, the constant acceleration in the motion of the object is given byâ€¦â€¦(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

â€¦â€¦(3.3)

a = constant

**Velocity-Time Relation**

From the equation (3.3), we have

Vâ€™ = v + a(tâ€™ â€“ t) â€¦â€¦(3.4)

The equation (3.4) is velocity-time relation, which gives the velocity of an object at any time tâ€™ 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'.**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 vâ€™ and xâ€™ be the corresponding values at time tâ€™. Now, from equation (3.1), the average velocity of the object in the time interval between t and tâ€™ may be written as

xâ€™ = x + (tâ€™ â€“ t) u_{av}

Also, vav = (v + vâ€™)â€¦..(3.6)

Substituting the value of vav in equation (3.6), we have

xâ€™ = x + (tâ€™ â€“ t) (v + vâ€™) â€¦â€¦(3.7)

As given by equation (3.4), the velocity of an object at any time tâ€™ is related to its velocity at time t the equation

vâ€™ = v + a(tâ€™ â€“ t)

where a is the constant acceleration of the motion of the object. Substituting the value of vâ€™ in equation (3.7), we get

xâ€™ = x + (tâ€™ â€“ t) [v + v + a(tâ€™ â€“ t)]

or xâ€™ = x + v(tâ€™ â€“ t) + a(tâ€™ â€“ t)^{2}â€¦â€¦(3.8)

The equation (3.8) relates the position of the object at time tâ€™ 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 â€“ 0)^{2 }

or x = x0 + ut + at2 â€¦â€¦(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 â€“ x_{0}= S, displacement of the object in time t, then equation (3.9) becomes

S = ut + at^{2 }------(3.10)**Velocity-Displacement Relation**

From equation (3.4) we have

tâ€™ â€“ t =

Substituting for tâ€™ â€“ t in equation (3.8), we have

xâ€™ = x + v+ a

Or xâ€™ â€“ x = [2v (vâ€™-v) + (vâ€™-v)^{2}]

= [2vvâ€™ â€“ 2v^{2}+ vâ€™^{2}+ v^{2}â€“ 2ccâ€™] =

or vâ€™^{2}â€“ v^{2}= 2a (xâ€™ â€“ x) â€¦â€¦(3.11)

The equation (3.11) is velocity-displacement relation between velocities of the object at times tâ€™ 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 = x_{0}+ u + a^{2 }

Or x â€“ x_{0}= [2u(v-u)+(v-u)2]

= [2uv â€“ 2u^{2}+ v^{2}+ u^{2}â€“ 2uv] =

of v^{2}â€“ u^{2}= 2a(x-x_{0}) â€¦â€¦(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 vâ€™ to v; v to u ;xâ€™ to x and x to x_{0}.

If we denote x â€“ x_{0}= S, the distance covered by the object in time t, then equation (3.12) becomes

V^{2}â€“ u^{2}= 2aS â€¦â€¦(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 S_{nth}.

S

_{nth}= S

_{n}â€“ S

_{n-1}â€¦â€¦(3.14)

_{n}and S

_{n-1 }are displacements of the particle in n and n-1 seconds. The values of S

_{n}and S

_{n}-1 can be obtained by putting t = n and t = n â€“ 1 in the equation.

S = ut + at

^{2}

Therefore, from equation (3.14) we have

S

_{nth}= -

= -

= u(n-n+1) + a (n^{2} â€“ n^{2} â€“1 + n)

_{nth}= u + a(2n-1) -----(3.15)

*An electron is emitted with a velocity of 5 Ã— 10 ^{6} ms^{-2}. If its final velocity is 7 Ã— 10^{6} ms^{-1}, calculate the distance discovered by the electron.*

**Solution**

Here, u = 5 Ã— 10^{6} ms^{-1}; v = 7 Ã— 10^{6} ms^{-1} ; a = 3 Ã— 10^{14} ms^{-2}

We know v^{2} â€“ u^{2} = 2aS

âˆ´ (7 Ã— 10^{6})^{2} â€“ (5 Ã— 10)^{6} = 2 Ã— 3 10^{14} Ã— S or 49 Ã— 10^{12} â€“ 25 Ã— 10^{12} = 6 Ã— 10^{14}S

or S = = 0.04 m

*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 Ã— (1000m) Ã— (60 Ã— 60s)^{-1} = 35 ms^{-1}

Final speed of the car, v = 0

Let a be the retardation of the car

We know, v^{2} â€“ u^{2} = 2aS

(0)^{2} â€“ (35)^{2} = 2a Ã— 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,

^{-1}; a = 5ms

^{-2}

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 |

_{1}till it attains a velocity v moves with this velocity till time t

_{2}i.e. for a time interval (t

_{2}â€“ t

_{1}) and then uniformly decelerates for a time interval (t

_{3}â€“ t

_{2}) and finally, it again comes to rest. In the time interval between t = 0 to t = t

_{1}, the object has positive acceleration and its velocity goes on increasing, while in time interval between t = t

_{2}and t = t

_{3}, the object has negative acceleration and its velocity goes on decreasing.

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

# 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,

_{0}= 10m ; u = 5ms

^{-1}; a = 5ms

^{-2}

Time (s) |
Velocity (m) |

0 |
10.0 |

1 |
17.5 |

2 |
30.0 |

3 |
47.5 |

4 |
70.0 |

5 |
97.5 |

In this x â€“ 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

- in time interval between t = 0 and t = t
_{1}, the object remains at rest; - in time interval between t = t
_{1}to t = t_{2}, the object speeds up* with a constant acceleration; - in time interval between t = t
_{3}to t = t_{3 }the object slows down with constant velocity. - in time interval between t = t
_{3}to t = t_{4}, the object slows down with a constant acceleration and - in time interval between t = t
_{4}to t = t_{5}, 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 tâ€™, consider two points A and B on the velocity time graph. Then, the object at time t and tâ€™ respectively. Also, Aâ€™Bâ€™ = tâ€™ â€“ t= area of trapezium ABBâ€™Aâ€™

= (AAâ€™ + BBâ€™) Ã— Aâ€™Bâ€™

= (v + vâ€™) Ã— (tâ€™ â€“ t)

a =

or tâ€™ â€“ t = â€¦â€¦(3.16)

Area under vâ€“t graph between t and tâ€™ = (v +vâ€™) Â´ = = ---- (3.17)

But from the equation vâ€™

^{2}â€“ v

^{2}= 2a (xâ€™ â€“ x), we have

= xâ€™ â€“ x â€¦.(3.18)

Therefore, from the equations (3.17) and (3.18), it follows that

xâ€™- x = area under the v â€“ t graph between t and tâ€™