Explain Rest, Motion, ……etc.

Explain Rest, Motion, Distance, Displacement, Speed, Velocity, Acceleration, average speed, and average velocity. also, write the equation of motion with proof. write 10 numerical of class 9 standard points of view of chapter rest and motion. write in detail with proper examples.

Certainly! Here’s an advanced explanation of the concepts of rest, motion, distance, displacement, speed, velocity, acceleration, average speed, and average velocity, along with the equations of motion and some detailed numerical examples suitable for a class 9 level:

Rest: When an object does not change its position with respect to its surroundings, it is said to be at rest. For example, a parked car is at rest because it is not moving.

Motion: Motion is the change in position of an object with respect to a frame of reference. This change is observed over time. For instance, a moving car changes its position concerning the road as time progresses.

Distance: Distance is the total length of the path traveled by an object during motion. It is a scalar quantity, meaning it has magnitude but no direction. The distance traveled by a car from point A to point B is the actual length of the path it takes.

Displacement: Displacement is the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. For example, if a car moves from point A to point B, its displacement is the straight-line distance between A and B, along with the direction from A to B.

Speed: Speed is the rate at which an object covers distance. It is the magnitude of the velocity vector and is always positive. The average speed of a car over a certain distance is calculated by dividing the total distance by the total time taken.

Velocity: Velocity is the rate of change of displacement. It is a vector quantity, meaning it has both magnitude and direction. If a car moves from point A to point B in a certain time, its velocity is the displacement from A to B divided by the time taken, along with the direction.

Acceleration: Acceleration is the rate of change of velocity. It is a vector quantity. An object is said to be accelerating if its speed, direction, or both are changing. Acceleration is calculated as the change in velocity divided by the time taken for the change.

Average Speed: Average speed is the total distance traveled divided by the total time taken. It gives an overall idea of how fast an object is moving, without considering the direction.

Average Velocity: Average velocity is the total displacement divided by the total time taken. It gives an overall idea of the rate at which an object’s position is changing, considering both magnitude and direction.

Equations of Motion:

The equations of motion describe the relationship between an object’s initial velocity, final velocity, acceleration, displacement, and time taken. They are derived from the basic definitions of motion and are valid for objects moving with uniform acceleration.

The three equations of motion are:

1. 𝑣=𝑢+𝑎𝑡v=u+at
2. 𝑠=𝑢𝑡+12𝑎𝑡2s=ut+21​at2
3. 𝑣2=𝑢2+2𝑎𝑠v2=u2+2as

Let’s prove the equations of motion graphically using velocity-time graphs. We’ll use the following conventions:

• Initial velocity 𝑢u is the velocity at time 𝑡=0t=0.
• Final velocity 𝑣v is the velocity at time 𝑡t.
• Acceleration 𝑎a is constant.

1. First Equation of Motion 𝑣=𝑢+𝑎𝑡v=u+at: Consider a velocity-time graph where the initial velocity 𝑢u is a point on the velocity axis and the acceleration 𝑎a is represented by a constant slope.The area under the acceleration-time graph represents the change in velocity, which is 𝑎𝑡at. Therefore, the final velocity 𝑣v at time 𝑡t is given by 𝑢+𝑎𝑡u+at.
2. Second Equation of Motion 𝑠=𝑢𝑡+12𝑎𝑡2s=ut+21​at2: Consider a velocity-time graph where the initial velocity 𝑢u is a point on the velocity axis and the acceleration 𝑎a is represented by a constant slope.The area under the acceleration-time graph represents the change in velocity, which is 𝑎𝑡at. The total area under the velocity-time graph represents the displacement 𝑠s. This area can be divided into two parts: the area of the trapezoid representing the displacement due to initial velocity (𝑢𝑡ut) and the area of the triangle representing the displacement due to acceleration (12𝑎𝑡221​at2). Therefore, the total displacement 𝑠s is given by 𝑢𝑡+12𝑎𝑡2ut+21​at2.
3. Third Equation of Motion 𝑣2=𝑢2+2𝑎𝑠v2=u2+2as: Consider a velocity-time graph where the initial velocity 𝑢u is a point on the velocity axis and the acceleration 𝑎a is represented by a constant slope.The area under the acceleration-time graph represents the change in velocity, which is 𝑎𝑡at. The final velocity 𝑣v can be calculated as 𝑢+𝑎𝑡u+at. Using the formula 𝑣2=𝑢2+2𝑎𝑠v2=u2+2as, we substitute 𝑣=𝑢+𝑎𝑡v=u+at to get (𝑢+𝑎𝑡)2=𝑢2+2𝑎𝑠(u+at)2=u2+2as, which simplifies to 𝑢2+2𝑢𝑎𝑡+𝑎2𝑡2=𝑢2+2𝑎𝑠u2+2uat+a2t2=u2+2as. Canceling out 𝑢2u2 on both sides and rearranging, we get 𝑣2=𝑢2+2𝑎𝑠v2=u2+2as.

These graphical proofs provide a visual understanding of how the equations of motion are derived from velocity-time graphs.

1. First Equation of Motion 𝑣=𝑢+𝑎𝑡v=u+at: Consider a velocity-time graph with initial velocity 𝑢u and constant acceleration 𝑎a. The graph is a straight line with slope 𝑎a passing through the point (0,𝑢)(0,u).The area under the acceleration-time graph represents the change in velocity, which is 𝑎𝑡at. Therefore, the final velocity 𝑣v at time 𝑡t is given by 𝑢+𝑎𝑡u+at.
2. Second Equation of Motion 𝑠=𝑢𝑡+12𝑎𝑡2s=ut+21​at2: Consider the same velocity-time graph as before. To find the displacement 𝑠s, we need to find the area under the velocity-time graph.The total area can be divided into two parts: the area of the rectangle representing 𝑢𝑡ut and the area of the triangle representing 12𝑎𝑡221​at2. The total displacement 𝑠s is the sum of these two areas, given by 𝑢𝑡+12𝑎𝑡2ut+21​at2.
3. Third Equation of Motion 𝑣2=𝑢2+2𝑎𝑠v2=u2+2as: Consider the same velocity-time graph as before. To derive this equation, we need to relate the final velocity 𝑣v to the initial velocity 𝑢u and the displacement 𝑠s.Using the first equation of motion, 𝑣=𝑢+𝑎𝑡v=u+at, we can express 𝑡t in terms of 𝑢u, 𝑣v, and 𝑎a: 𝑡=𝑣−𝑢𝑎t=av−u​.Substituting this into the second equation of motion, 𝑠=𝑢𝑡+12𝑎𝑡2s=ut+21​at2, we get: 𝑠=𝑢(𝑣−𝑢𝑎)+12𝑎(𝑣−𝑢𝑎)2s=u(av−u​)+21​a(av−u​)2Simplifying, we get: 𝑠=𝑢(𝑣−𝑢)𝑎+(𝑣−𝑢)22𝑎s=au(v−u)​+2a(v−u)2​ 𝑠=𝑢𝑣−𝑢2𝑎+𝑣2−2𝑢𝑣+𝑢22𝑎s=auv−u2​+2av2−2uv+u2​ 𝑠=2𝑢𝑣−2𝑢2+𝑣2−2𝑢𝑣+𝑢22𝑎s=2a2uv−2u2+v2−2uv+u2​ 𝑠=𝑣2−𝑢22𝑎s=2av2−u2​Rearranging, we get: 2𝑎𝑠=𝑣2−𝑢22as=v2−u2 𝑣2=𝑢2+2𝑎𝑠v2=u2+2as

These calculations demonstrate the graphical proofs of the equations of motion. The graphical representation helps visualize how the equations are derived from the areas under the velocity-time graphs.

10 problems related to rest and motion along with their solutions:

1. Problem: An object travels 50 meters in 10 seconds. Calculate its average speed. Solution: Average speed = Total distance ÷ Total time = 50 m ÷ 10 s = 5 m/s.
2. Problem: A car accelerates uniformly from rest to a speed of 20 m/s in 5 seconds. Calculate its acceleration. Solution: Using the equation 𝑎=𝑣−𝑢𝑡a=tv−u​, where 𝑣=20v=20 m/s (final velocity), 𝑢=0u=0 m/s (initial velocity), and 𝑡=5t=5 s (time taken), we get 𝑎=20−05=205=4a=520−0​=520​=4 m/s².
3. Problem: A train travels a distance of 300 km at an average speed of 60 km/h. Calculate the time taken for the journey. Solution: Time = Distance ÷ Speed = 300 km ÷ 60 km/h = 5 hours.
4. Problem: A ball is thrown vertically upwards with a speed of 20 m/s. Calculate the maximum height it reaches. (Assume acceleration due to gravity 𝑔=10g=10 m/s²) Solution: Using the equation 𝑣2=𝑢2+2𝑎𝑠v2=u2+2as, where 𝑣=0v=0 m/s (final velocity at highest point), 𝑢=20u=20 m/s (initial velocity), 𝑎=−10a=−10 m/s² (acceleration due to gravity, negative as it opposes the motion), we get 0=202+2×(−10)×𝑠0=202+2×(−10)×s. Solving for 𝑠s, we get 𝑠=40020=20s=20400​=20 m.
5. Problem: A car travels a distance of 40 meters at a speed of 10 m/s and then comes to rest with a uniform deceleration of 2 m/s². Calculate the total time taken for the motion. Solution: Time taken to travel 40 meters at 10 m/s is 4010=41040​=4 seconds. Using the equation 𝑣=𝑢+𝑎𝑡v=u+at, where 𝑣=0v=0 m/s (final velocity), 𝑢=10u=10 m/s (initial velocity), and 𝑎=−2a=−2 m/s² (deceleration), we get 0=10−2𝑡0=10−2t. Solving for 𝑡t, we get 𝑡=102=5t=210​=5 seconds. Therefore, the total time taken is 4+5=94+5=9 seconds.
6. Problem: A motorcycle accelerates from rest at 2 m/s² for 5 seconds. Calculate its final velocity. Solution: Using the equation 𝑣=𝑢+𝑎𝑡v=u+at, where 𝑢=0u=0 m/s (initial velocity), 𝑎=2a=2 m/s² (acceleration), and 𝑡=5t=5 seconds (time), we get 𝑣=0+2×5=10v=0+2×5=10 m/s.
7. Problem: A runner runs 100 meters in 10 seconds and then walks back to the starting point in 20 seconds. Calculate the average speed of the runner for the entire trip. Solution: Total distance covered = 100 meters (to + fro), Total time taken = 10 seconds (running) + 20 seconds (walking) = 30 seconds. Average speed = Total distance ÷ Total time = 100 ÷ 30 = 3.33 m/s.
8. Problem: A ball is thrown horizontally from the top of a building with a speed of 10 m/s. Calculate the time it takes for the ball to reach the ground. (Assume the building is 20 meters high and neglect air resistance) Solution: The time taken for the ball to reach the ground can be calculated using the equation 𝑠=𝑢𝑡+12𝑔𝑡2s=ut+21​gt2, where 𝑠=20s=20 m (height of the building), 𝑢=10u=10 m/s (horizontal speed of the ball), and 𝑔=10g=10 m/s² (acceleration due to gravity). Plugging in the values, we get 20=10𝑡+12×10×𝑡220=10t+21​×10×t2. Simplifying, we get 𝑡2+2𝑡−4=0t2+2t−4=0. Solving for 𝑡t, we get 𝑡=2t=2 seconds (ignoring the negative value).
9. Problem: A car accelerates uniformly from rest and reaches a speed of 30 m/s in 6 seconds. Calculate the distance traveled by the car during this time. Solution: Using the equation 𝑠=𝑢𝑡+12𝑎𝑡2s=ut+21​at2, where 𝑢=0u=0 m/s (initial velocity), 𝑎=306=5a=630​=5 m/s² (acceleration), and 𝑡=6t=6 seconds (time), we get 𝑠=0×6+12×5×62=0+12×5×36=90s=0×6+21​×5×62=0+21​×5×36=90 meters.
10. Problem: A stone is thrown vertically upwards with a speed of 15 m/s. Calculate the time it takes for the stone to reach the highest point. (Assume acceleration due to gravity 𝑔=10g=10 m/s² and neglect air resistance) Solution: Using the equation 𝑣=𝑢+𝑔𝑡v=u+gt, where 𝑣=0v=0 m/s (final velocity at the highest point), 𝑢=15u=15 m/s (initial velocity), and 𝑔=−10g=−10 m/s² (acceleration due to gravity, negative as it opposes the motion), we get 0=15−10𝑡0=15−10t. Solving for 𝑡t, we get 𝑡=1510=1.5t=1015​=1.5 seconds.