Kinematics (Physics) Motion for high school and AP Physics 1

This lecture takes the concept of distance and displacement in detail with examples and also the difference between them.

Distance and Displacement

→ The actual path or length travelled by a object during its journey from its initial position to its final position is called the distance.

→ Distance is a scalar quantity which requires only magnitude but no direction to explain it.

Example: Ramesh travelled 65 km. (Distance is measured by odometer in vehicles.)

→ Displacement is a vector quantity requiring both magnitude and direction for its explanation.

Example: Ramesh travelled 65 km south-west from Clock Tower.

→ Displacement can be zero (when initial point and final point of motion are same)

Example: circular motion.

This lecture in continuation to previous lecture explains about displacement in more depth.

In this lecture the application of the conept of displacement is taken in more depth.

The concept of displacement have been applied on numericals for better understanding of concept.

This lecture explains the concept of speed with examples .

Speed

→ The measurement of distance travelled by a body per unit time is called speed.

• Speed (v) = Distance Travelled/Time Taken = s/t

• SI unit = m/s (meter/second)

→ If a body is executing uniform motion, then there will be a constant speed or uniform motion.

→ If a body is travelling with non-uniform motion, then the speed will not remain uniform but have different values throughout the motion of such body.

→ For non-uniform motion, average speed will describe one single value of speed throughout the motion of the body.

• Average speed = Total distance travelled/Total time taken

Conversion Factor

• Change from km/hr to m/s = 1000m/(60×60)s = 5/18 m/s

Example: What will be the speed of body in m/s and km/hr if it travels 40 kms in 5 hrs ?

Solution

Distance (s) = 40 km

Time (t) = 5 hrs.

Speed (in km/hr) = Total distance/Total time = 40/5 = 8 km/hr

40 km = 40 × 1000 m = 40,000 m

5 hrs = 5 × 60 × 60 sec.

Speed (in m/s) = (40 × 1000)/(5×60 ×60) = 80/36 = 2.22 m/s

The concept of average speed have been applied to numericals based on to and fro jorney . Also short cut trick for solving these kind of numericals is also provided in this video .

In this video student will understand about the concept of velocity and average velocity in depth.

Velocity

→ It is the speed of a body in given direction.

• Velocity = Displacement/Time

→ Velocity is a vector quantity. Its value changes when either its magnitude or direction changes.

→ For non-uniform motion in a given line, average velocity will be alculated in the same way as done in average speed.

• Average velocity = Total displacement/Total time

• For uniformly changing velocity, the average velocity can be calculated as follows :

Avg. Velocity (vavg) = (Initial velocity + Final velocity)/2 = (u+v)/2

where, u = initial velocity, v = final velocity

• SI unit of velocity = ms-1

• Velocity = Displacement/Time

→ It can be positive (+ve), negative (-ve) or zero.

Difference between speed and velocity is being taught in this video for school exams.

Differences Between Speed and Velocity

S.No.            SPEED                                                               VELOCITY

1.         It is defined as the rate of change of distance.   It is defined as the rate of change of displacement.

2.          It is a scalar quantity.                                                It is a vector quantity.

3.        It can never be negative or zero.                             It can be negative,zero or positive.

4.        Speed is velocity without direction.                          Velocity is directed speed.

5. Speed may or may not be equal to velocity.               A body may possess different velocities but the same speed.

6. Speed never decreases with time. For a moving body, . Velocity can decrease with time. For a moving body , it 

    it is never zero                                                                                                   can be zero.

7. Speed in SI is measured in ms-1                                             Velocity in SI, is measured in ms-1

This video takes numerical based on average speed given in NCERT example 8.1

The student learn about the concept of acceleration fundamental concept in this video.

Acceleration

→ Acceleration is seen in non-uniform motion and it can be defined as the rate of change of velocity with time.

• Acceleration (a) = Change in velocity/Time = (v-u)/t

where, v = final velocity, u = initial velocity

→ If v > u, then ‘a’ will be positive (+ve).

Retardation/Deaceleration

→ Deaceleration is seen in non-uniform motion during decrease in velocity with time. It has same definition as acceleration.

• Deaceleration (a') = Change in velocity/Time = (v-u)/t

Here, v < u, ‘a’ = negative (-ve).

Student learn about the concept of acceleration in more depth in this video.

Example 1: A car speed increases from 40 km/hr to 60 km/hr in 5 sec. Calculate the acceleration of car.

Solution

u = 40km/hr = (40×5)/18 = 100/9 = 11.11 m/s

v = 60 km/hr = (60×5)/18 = 150/9 = 16.66 m/s

t = 5 sec

a = (v-u)/t = (16.66 - 11.11)/5

= 5.55/5 = 1.11 ms-2

Example 2: A car travelling with a speed of 20 km/hr comes into rest in 0.5 hrs. What will be the value of its retardation?

Solution

v = 0 km/hr

u = 20 km/hr

t = 0.5 hrs

Retardation, a’ = (v-u)/t = (0-20)/0.5

= -200/5 = -40 km hr-2

In this  video numericals based on acceleration has been taken for better understanding of the student..

More numericals  on acceleration has been discussed in this lecture . 

Basic concepts of distance time graph are taught to student in this video.

The distance Time graph is a line graph that denotes the distance versus time findings on the graph. Drawing a distance-time graph is simple. For this, we first take a sheet of graph paper and draw two perpendicular lines on it conjoining at O. The horizontal line is the X-axis, while the verticle line is the Y-axis. On these axes, we write the quantities or readings of our observation. Now let’s take a few observations of a traveling car:

S.no.          Time                Distance

1                    0                          0                                                                                                                                                                                2                 5min                     10km                                                                                                                                                                           3                 10min                    20km                                                                                                                                                                                              4                  15min                  30km                                                                                                                                                                                      5                    20min               40km

Using the above information our graph shall be as follows:

The distance Time graph is a line graph that denotes the distance versus time findings on the graph. Drawing a distance-time graph is simple. For this, we first take a sheet of graph paper and draw two perpendicular lines on it conjoining at O. The horizontal line is the X-axis, while the verticle line is the Y-axis. On these axes, we write the quantities or readings of our observation. Now let’s take a few observations of a traveling car:

Now when we draw a graph, we take our observations on any axis, here we have taken distance on the x-axis while time on the y-axis. Point O is a zero (o) i.e our point of departure or beginning. Nowhere we notice that the line when joined with our different observations, is a slant line. This shows that the speed during the course of traveling is constant.

The concepts of distance time graph have been taken in more depth in this video.

In this video student will learn how to interpret and solve numericals based on distance time graph.

In this lecture student will understand the concepts of speed time graph including reading and interpretation of speed time graph and the concept of slope of distance time graph.

First Equation: v = u + at

Final velocity = Initial velocity + Acceleration × Time

Graphical Derivation

Suppose a body has initial velocity ‘u’ (i.e., velocity at time t = 0 sec.) at point ‘A’ and this velocity changes to ‘v’ at point ‘B’ in ‘t’ secs. i.e., final velocity will be ‘v’.

For such a body there will be an acceleration.

a = Change in velocity/Change in Time

⇒ a = (OB - OA)/(OC-0) = (v-u)/(t-0)

⇒ a = (v-u)/t

⇒ v = u + at

Second Equation: s = ut + ½ at2

Distance travelled by object = Area of OABC (trapezium)

= Area of OADC (rectangle) + Area of ∆ABD

= OA × AD + ½ × AD × BD

= u × t + ½ × t × (v – u)

= ut + ½ × t × at

⇒ s = ut + ½ at2 (∵a = (v-u)/t)

Uniform Circular Motion

Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. At all instances, the object is moving tangent to the circle. Since the direction of the velocity vector is the same as the direction of the object's motion, the velocity vector is directed tangent to the circle as well. The animation at the right depicts this by means of a vector arrow.

An object moving in a circle is accelerating. Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction. The direction of the acceleration is inwards. The animation at the right depicts this by means of a vector arrow.

The final motion characteristic for an object undergoing uniform circular motion is the net force. The net force acting upon such an object is directed towards the center of the circle. The net force is said to be an inward or centripetal force. Without such an inward force, an object would continue in a straight line, never deviating from its direction. Yet, with the inward net force directed perpendicular to the velocity vector, the object is always changing its direction and undergoing an inward acceleration.

Joseph jogs from one end A to the other end B of a straight 300m road in 2 minutes 50 seconds and then turns around and jogs 100m back to a point C in another one minute . What are Joseph average speeds and velocities in jogging ?