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PhysicsClass 9CBSE

Derive the third equation of motion (v² = u² + 2as) graphically.

PROMPT: A velocity-time graph showing uniform acceleration. The y-axis represents velocity (v) and the x-axis represents time (t). A straight line starts from initial velocity 'u' on the y-axis at t=0 and goes up to final velocity 'v' at time 't'. A horizontal dashed line extends from 'u' to the point (t,u) and a vertical dashed line extends from 'v' to the point (t,v). A point 'A' is at (0,u), 'B' is at (t,v), 'C' is at (t,u), and 'D' is at (0,0). The area under the graph OABD is calculated as the area of a trapezium.Consider an object moving with uniform acceleration 'a'. Let its initial velocity be 'u' at time t=0, and its final velocity be 'v' at time 't'.From the velocity-time graph, the displacement (s) is given by the area under the v-t graph, which is the area of trapezium OABD.Area of trapezium = (1/2) × (Sum of parallel sides) × heights = (1/2) × (OA + BD) × ODFrom the graph, OA = u, BD = v, and OD = t.So, s = (1/2) × (u + v) × t --- (Equation 1)From the first equation of motion, we know that v = u + at, which can be rearranged to find t:t = (v - u) / a --- (Equation 2)Substitute the value of 't' from Equation 2 into Equation 1:s = (1/2) × (u + v) × [(v - u) / a]s = (1/2a) × (v + u)(v - u)Using the algebraic identity (x + y)(x - y) = x² - y²:s = (1/2a) × (v² - u²)2as = v² - u²Therefore, v² = u² + 2as.

PhysicsClass 9CBSE

Derive the second equation of motion (s = ut + (1/2)at²) graphically.

PROMPT: A velocity-time graph showing uniform acceleration. The y-axis represents velocity (v) and the x-axis represents time (t). A straight line starts from initial velocity 'u' on the y-axis at t=0 and goes up to final velocity 'v' at time 't'. A horizontal dashed line extends from 'u' to the point (t,u) and a vertical dashed line extends from 'v' to the point (t,v). A point 'A' is at (0,u), 'B' is at (t,v), 'C' is at (t,u), and 'D' is at (0,0). The area under the graph OABD represents displacement. This area is divided into a rectangle OACD and a triangle ABC.Consider an object moving with uniform acceleration 'a'. Let its initial velocity be 'u' at time t=0, and its final velocity be 'v' at time 't'.From the velocity-time graph, the displacement (s) is given by the area under the v-t graph.Area of trapezium OABD = Area of rectangle OACD + Area of triangle ABCArea of rectangle OACD = length × breadth = OA × OD = u × tArea of triangle ABC = (1/2) × base × height = (1/2) × AC × BCFrom the graph, AC = OD = t.And BC = BD - CD = v - u.We know from the first equation of motion that v - u = at.So, Area of triangle ABC = (1/2) × t × (at) = (1/2)at².Therefore, Total Displacement (s) = ut + (1/2)at².

PhysicsClass 9CBSE

Derive the first equation of motion (v = u + at) graphically.

PROMPT: A velocity-time graph showing uniform acceleration. The y-axis represents velocity (v) and the x-axis represents time (t). A straight line starts from initial velocity 'u' on the y-axis at t=0 and goes up to final velocity 'v' at time 't'. A horizontal dashed line extends from 'u' to the point (t,u) and a vertical dashed line extends from 'v' to the point (t,v). A point 'A' is at (0,u), 'B' is at (t,v), 'C' is at (t,u), and 'D' is at (0,0). The slope of the line AB represents acceleration.Consider an object moving with uniform acceleration 'a'. Let its initial velocity be 'u' at time t=0, and its final velocity be 'v' at time 't'.From the velocity-time graph, acceleration is the slope of the velocity-time graph.Slope = (Change in velocity) / (Change in time)a = (v - u) / (t - 0)a = (v - u) / tat = v - uTherefore, v = u + at.

PhysicsClass 9CBSE

State the third equation of motion and explain the meaning of each term in it.

The third equation of motion is: v² = u² + 2asWhere:v = final velocity of the objectu = initial velocity of the objecta = acceleration of the objects = displacement of the object

PhysicsClass 9CBSE

State the second equation of motion and explain the meaning of each term in it.

The second equation of motion is: s = ut + (1/2)at²Where:s = displacement of the objectu = initial velocity of the objecta = acceleration of the objectt = time taken for the displacement 's'

PhysicsClass 9CBSE

State the first equation of motion and explain the meaning of each term in it.

The first equation of motion is: v = u + atWhere:v = final velocity of the objectu = initial velocity of the objecta = acceleration of the objectt = time taken for the velocity to change from u to v

PhysicsClass 9CBSE

Explain what is meant by positive acceleration and negative acceleration (retardation). Give one example for each.

Positive acceleration occurs when the velocity of an object increases in the direction of motion. For example, a car speeding up from a traffic light. Negative acceleration, also known as retardation or deceleration, occurs when the velocity of an object decreases, or its speed reduces, in the direction of motion. For example, a car applying brakes to slow down.

PhysicsClass 9CBSE

Define acceleration. What are its SI unit and common non-SI unit?

Its SI unit is meters per second squared (m/s²). A common non-SI unit is kilometers per hour squared (km/h²).

PhysicsClass 9CBSE

What is the key difference between speed and velocity? Give an example where two objects have the same speed but different velocities.

Speed is the rate at which an object covers distance. It is a scalar quantity, meaning it only has magnitude. Velocity is the rate at which an object changes its displacement. It is a vector quantity, meaning it has both magnitude and direction.Example: Two cars are traveling at 60 km/h. Car A is moving north, and Car B is moving south. Both cars have the same speed (60 km/h), but their velocities are different because they are moving in opposite directions.

PhysicsClass 9CBSE

A car travels 10 km in the first 15 minutes, 15 km in the next 15 minutes, and 20 km in the last 15 minutes. Is this an example of uniform or non-uniform motion? Justify your answer.

This is an example of non-uniform motion. Uniform motion is when an object covers equal distances in equal intervals of time. In this case, the car covers different distances (10 km, 15 km, 20 km) in equal intervals of time (15 minutes each). Therefore, its speed is changing, indicating non-uniform motion.

PhysicsClass 9CBSE

Explain the difference between distance and displacement. Provide a scenario where the distance covered is greater than the magnitude of the displacement.

Distance is the total path length covered by an object during its motion, irrespective of the direction. It is a scalar quantity. Displacement is the shortest straight-line distance between the initial and final positions of an object, along with its direction. It is a vector quantity.Scenario: Imagine a person walking from point A to point B, then from point B to point C, and finally from point C back to point A, forming a triangular path. If each side of the triangle is 5 meters long, the total distance covered would be 5m + 5m + 5m = 15 meters. However, since the person started at A and ended at A, their initial and final positions are the same, making the displacement zero.

PhysicsClass 9CBSE

Define motion and provide two examples of objects in motion and two examples of objects at rest.

Motion is defined as the change in position of an object with respect to its surroundings in a given interval of time. Examples of objects in motion include a moving car, a flying bird, a rolling ball, or a person walking. Examples of objects at rest include a book lying on a table, a parked car, a house, or a tree.

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