Kinematics (Rectilinear and Curvilinear Motion)
Kinematics
Kinematics is a branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the mass of each of the forces that caused the motion.
Kinematics is the mathematical description of motion. The term is derived from the Greek word kinema, meaning movement. In order to quantify motion,
a mathematical coordinate system, called a reference frame, is used to describe space and time. Once a reference frame has been chosen, we can introduce the physical concepts of position, velocity and acceleration in a mathematically precise manner. The figure shows a Cartesian coordinate system in one dimension with unit vector ˆi pointing in the direction of increasing x -coordinate. |
| Figure: A one-dimensional Cartesian coordinate system. |
IMPORTANT TERMS:-
Velocity: The velocity of a body may be defined as its rate of change of displacement, with respect to its surroundings, in a particular direction. As the velocity is always expressed in a particular direction, therefore it is a vector quantity.
Acceleration: The acceleration of a body may be defined as the rate of change of its velocity. It is said to be positive, when the velocity of a body increases with time, and negative when the velocity decreases with time. The negative acceleration is also called retardation. In general, the term acceleration is used to denote the rate at which the velocity is changing. It may be uniform or variable.
Uniform acceleration: If a body moves in such a way that its velocity changes in equal magnitudes in equal intervals of time, it is said to be moving with a uniform acceleration
Variable acceleration: If a body moves in such a way, that its velocity changes in unequal magnitudes in equal intervals of time, it is said to be moving with a variable acceleration.
Distance traversed: It is the total distance moved by a body. Mathematically, if body is moving with a uniform velocity (v), then in (t) seconds, the distance traversed
s = v X t
In this chapter, we shall discuss the motion under uniform acceleration only.
MOTION UNDER UNIFORM ACCELERATION
Consider the *linear motion of a particle starting from O and moving along OX with a uniform
acceleration as shown in Fig. 17.1. Let P be its position after t seconds.
Let u = Initial velocity,
v = Final velocity,
t = Time (in seconds) taken by the particle to change its velocity
from u to v.
a = Uniform positive acceleration, and
s = Distance travelled in t seconds.
Since in t seconds, the velocity of the particle has increased steadily from (u) to (v) at the
rate of a, therefore total increase in velocity
= a t
∴ v = u + a t ...(i)
MOTION UNDER FORCE OF GRAVITY
But, if the motion takes place against the force of gravity, i.e., the particle is projected upwards, the corresponding equations will be :
Notes: 1. In this case, the value of u is taken as negative due to upward motion.
2. In this case, the distances in upward direction are taken as negative, while those in the
downward direction are taken as positive.
DISTANCE TRAVELLED IN THE nth SECOND
17.5. (a) and (b). Here we shall discuss the following two cases :
1. When the body is moving with a uniform velocity.
Thus we see that the area of the figure OABC (i.e., velocity × time) represents the distance,
2. When the body is moving with a variable velocity
We know that the distance traversed by a body, traversed by the body, to some scale.
From the geometry of the Fig. 17.5 (b),
GRAPHICAL REPRESENTATION OF VELOCITY, TIME AND DISTANCE TRAVELLED BY A BODY
The motion of body may also be represented by means of a graph. Such a graph may be drawn by plotting velocity as ordinate and the corresponding time as abscissa as shown in Fig.
17.5. (a) and (b). Here we shall discuss the following two cases :
1. When the body is moving with a uniform velocity.
Consider the motion of a body, which is represented by the graph OABC as shown in Fig. 17.5(a). We know that the distance traversed by the body,
s = Velocity × Time
2. When the body is moving with a variable velocity
We know that the distance traversed by a body, traversed by the body, to some scale.
S= ut + 1/2 at²
From the geometry of the Fig. 17.5 (b),
we know that area of the figure OABC = Area (OADC + ABD)
But area of figure OADC = u × t
and
and
area of figure ABD = 1/2 × t × at =1/2 at²
∴ Total area OABC = S=ut+1/2 at²
Thus, we see that the area of the OABC represents the distance traversed by the body to some scale. From the figure, it is also seen
tan α = at/t = a
Thus, tan α represents the acceleration of the body.
∴ Total area OABC = S=ut+1/2 at²
Thus, we see that the area of the OABC represents the distance traversed by the body to some scale. From the figure, it is also seen
tan α = at/t = a
Thus, tan α represents the acceleration of the body.
Example:- A train moving with a velocity of 30 km.p.h. has to slow down to 15 km.p.h.
due to repairs along the road. If the distance covered during retardation be one kilometer and that
covered during acceleration be half a kilometer, find the time lost in the journey.
∴ Total time, t = t₁ + t₂ = 2.67 + 1.33 = 4 min
If the train had moved uniformly with a velocity of 30 km/hr,
then the time required to cover 1.5 km=60/30 * 3/2 =3 min. ...(iii)
∴ Time lost = 4 – 3 = 1 min Ans.
If the train had moved uniformly with a velocity of 30 km/hr,
then the time required to cover 1.5 km=60/30 * 3/2 =3 min. ...(iii)
∴ Time lost = 4 – 3 = 1 min Ans.
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