KINEMATICS

When an object changes its position relative to a chosen reference object (Frame of reference) it is in motion relative to that `frame'. Description of motion requires knowledge of the following terms :

2.1 Definitions

The motion of an object is represented in Figure 9. The object moves from the point P at t = 0 sec to the point Q in interval dt sec.

Consider,

If the time interval D t is sufficiently large then the velocity obtained is called average velocity. For as accurate description of motion it is necessary to choose D t as small as possible i.e. D t ® 0; then the velocity obtained is called instantaneous velocity.

As the displacements must be directly proportional to the interval of time hence the ratio of displacement to time interval will be constant whether Dt is sufficiently small or arbitrarily large.

Similar considerations also apply to acceleration also

Even if acceleration keeps on changing with time, we do not find it necessary to consider the rate of change in acceleration for the description of motion, and so on. The reason for this will be found when we study dynamics. The causes of acceleration are forces according to Newton's laws of motion.

Motion in one dimension


Motion under uniform acceleration

Let a particle move along X-axis (chosen for convenience as the motion is one dimensional) with uniform acceleration a .

For uniformly accelerated motion, velocity and position can be determined at any instant. Thus motion is accurately and completely described.

Consider
The equations (1),(3) (5) are called linear kinematical equations of motion under uniform acceleration.

The vector form of the equation has an advantage as the conventional +ve & -ve sign for vectors pointing to the right & left respectively need not be kept in mind. The equations may also be represented in the scalar form also as

Problem
An object moving with a velocity of 25 m/s is uniformly retarded for 5 seconds and is brought to rest. Find the retardation & the displacement. Solution
Using vector form

Area ABCD = Area of Rectangle ABED + Area of DEC
= AB ´ AD + 1 DE ´ EC
                      2
= (3 - 1) ´ (30 - 0) + 1 (3 - 1) ´ (70 - 30)
                                  2
= 60 + 40
= 100 m.

Now, displacement between 1st to 3rd sec is
= 10 ´ 3 + 1 ´ 10 ´ 32 - {10 ´ 1 + 1 ´ 20 ´ 12 }
                  2                                    2
= 10 [3 + 9] - { 10 (1 + 1) }
= 10 {12 - 2}
= 10 { 10}
= 100 m
Thus displacement in a given interval is
= Area under velocity-time graph for the given interval.
This result is obvious, as can be seen from
v = ds (in scalar form)
      dt
\ ds = v dt

 

The above results can be generalized when the acceleration is not uniform.
Thus, to summarize
i) v = slope of tangent to s-t graph
ii) Ds = Area under v-t graph                                     ........(8)
iii) a = slope of tangent to v-t graph.
iv)Dv = Area under a-t graph.





Motion in Two Dimensions
Projectile motion

The motion of an object projected at an angle to the horizontal level and subjected to acceleration due to the gravity of the Earth alone, is called projectile motion. The air resistance is assumed to be absent and the horizontal level is considered to be a plane and the curvature of Earth's surface is neglected.

Locally, the projectile motion can be analyzed as a two component motion : Horizontal component with uniform horizontal velocity and vertical component with uniform acceleration due to gravity

of Earth.

With given velocity V_o there are two angles of projection q & 90^o - q for which the horizontal range remains same; the vertical heights however will be different. Also if q = 45^o then sin 2q = sin 90 ^o= 1 (maximum)

possible at q = 45^o ; then a = 90^o - q = 90^o - 45^o= 45^o