Kinematic-Motion and time

First of all we will study what exactly is rest and motion.

An object is said to be in motion when it changes it's position with respect to an observer along the passage of time. An object is said to be at rest when it does not change it position wih respect to am observer along the passage of time.

If a person is standing on the Earth then he would be in state of ________ with respect to a person standing on the Moon.

Solution: Motion. This is because the Moon is revolving around the Earth and so the person will be seen in motion.

In the above example the earth is rotating around it's axis and the moon is revolving around the earth. Then why does the moon appear in motion to us and not in rest?

Solution: The speed of the rotation of the earth and that of the revolution of the moon is different. Hence the resultant speed is not zero and the moon appears to be in motion to us.

Now we will come to what is distance and displacement. Distance-It is the total path length travelled between the initial and final point Displacement-It is the shortest distance between the initial and final point. Both have same S.I. Unit that is m. Distance is a scalar quantity while displacement is a vector quantity.

A man started from point A moved 3m left and then 4m south reaching point B. Find the total distance and the displacement he travelled between point A and B.

Solution: As the distance is the total path length travelled it will be (3+4)m = 7m.Therefore the distance is 7m. Now to find the displacement we will refer to the given figure. As we know that the displacement is the shortest length between 2 points it will be the diagonal distance between A an C. It is shown by the dotted lines. Now to find this distance we will use the Pythagoras Theorem. It states that \[{ (Hypotenuse) }^{ 2 }\quad =\quad { Perpendicular }^{ 2 }+\quad { Base }^{ 2 }\].Therefore, we get that \[{ AC }^{ 2 }\quad =\quad { 3 }^{ 2 }\quad +\quad { 4 }^{ 2 }\\ therefore\quad AC\quad =\quad \sqrt { 25 } \quad =\quad 5\] Hence the displacement is 5m

Further we will know about speed and velocity. Speed is the ratio of the distance and time. \[Speed\quad =\quad \frac { Distance }{ Time } \] Velocity is the ratio of displacement. \[Velocity\quad =\quad \frac { Displacement }{ Time } \] Both speed and velocity have S.I. Unit m/s. Speed is a scalar quantity and vector is a vector quantity. Now what if the speed keeps changing throughout a journey. We will simply find the average speed. The average speed is the ratio of the total distance travelled and the total time taken. \[Average\quad Speed\quad =\quad \frac { Total\quad distance\quad travelled }{ Total\quad time\quad taken } \]

A person travels from A to B with a speed of 5m/s. He returns from B to A with a speed of 6 m/s. Find the average speed and average velocity.

Solution: Let the distance from A to B be x meters. So time taken from A to B is \[\frac { distance }{ speed } \] = \[\frac { x }{ 5 } \]. Now time taken from B to A is \[\frac { x }{ 6} \] So, Average speed is \[\frac { total\quad distance\quad }{ total\quad time } \quad \] = \[\frac { x+x\quad }{ \frac { x }{ 5 } +\frac { x }{ 6 } } \quad \\ =\quad \frac { 2x }{ \frac { 6x+5x }{ 30 } } \\ =\quad \frac { 2x\quad \times \quad 30 }{ 11x } \\ =\frac { 60 }{ 11 } \\ =5.45\] Therefore the average speed is 5.45m/s. Now, we know that Average Velocity = \[\frac { Total\quad displacement }{ Total\quad time\quad } \quad \] =0, because total displacement is o.

Now acceleration will be studied. Acceleration is a vector quantity that is defined as a rate at which an object changes its velocity. An object is accelerating if it is changing its velocity. So,\(Acceleration\quad =\quad \frac { change\quad in\quad velocity }{ time\quad taken } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { v-u }{ t } \\ where,\quad t\quad =\quad time\quad taken\\ v\quad =\quad final\quad velocity\quad and\\ u\quad =\quad initial\quad velocity\)