Motion_Shobhit Nirwan, Summaries of Physics

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MOTION

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Shobhit Nirwan B e s t H a n d w r i t t e n N o t e s C L A S S 9 t h

Birds Flying Blades of fan^ Vehicles Moving We observe these kinds of phenomena in our everyday life? Isn’t it? All of these are examples of types of motions. #Kitaabi Definition: If a body changes its position with respect to time and surroundings, it is said to be in motion. We describe the location of an object by specifying a reference point. Some Important Terms: Relative: Beta tumhara school kitne dur hai? Baccha: Mere ghar se 2km dur. { To yaha par bacche ne “Ghar” ko as a reference point liya. } Therefore, to describe the position of an object we need to specify a reference point called the origin. Scalar and Vector Quantities Imagine you're playing a game of hide-and-seek with your friends. You want to tell your friends how far you are from the hiding spot and in which direction. Scalar Quantity: A scalar quantity is something that tells you only the amount or size or magnitude of something, without any information about the direction. For example, imagine you tell your friends, "I am 10 steps away from the hiding spot." Here, you're only giving them the distance, which is the scalar quantity. It doesn't matter if you went left or right or any other direction, just how far you are. L · ( - I X x

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L.P. 1-An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example Ans- - Yes. an object which has moved through a distance can have zero displacement if it comes back to its initial position. Example - If a person jogs in a circular park which is circular and completes one round. His Initial and Final Position are the same. Thus, his displacement is O. L.P. 2: Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object. Ans- (a) False (b) False [ NCERT Intext Pg100] Uniform Motion and Non-Uniform Motion Imagine you're riding your bicycle in a park. Uniform Motion: Uniform motion is when an object moves in a way that covers equal distances in equal intervals of time. It means the speed or rate at which the object is moving remains constant throughout its journey. For example, imagine you're riding your bicycle on a straight road, and you're maintaining a steady speed of 10 km/h. In this case, you're experiencing uniform motion because you're covering the same distance every hour (10 kilometers per hour). Non-Uniform Motion: Non-uniform motion is when an object moves in a way that covers unequal distances in equal intervals of time. It means the speed at which the object is moving keeps changing during its journey. For example, imagine you're riding your bicycle on a road with hills and valleys. As you go up a steep hill, your speed decreases, and when you go downhill, your speed increases. In this case, you're experiencing non-uniform motion because the distance covered in each hour will vary depending on the ups and downs of the road. -.

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In simple terms, uniform motion is when you're moving at a constant speed, covering the same distance in equal intervals of time. Non- uniform motion is when your speed keeps changing, and the distance covered in equal intervals of time is not the same. Rate of Motion Different objects may take different amounts of time to cover a distance. Some of them move fast and some move slowly. OR We can say the ‘Rate’ at which the objects move can be different. Also, different objects can move at the same rate. isi ‘Rate’ ko find karne ke liye we’ll study Speed, Velocity and acceleration. Speed Distance travelled by an object per unit time is called speed of an object. i.e. Average speed = total distance travelled total time taken If an object travels a distance s in time t then its speed v is, Speed is a scalar quantity, and its SI unit is metre per second (m/s) Speed with Direction: Velocity Direction bhi chahiye ?? To yaad karo direction kisme padha tha humne?? - Displacement me! So, Velocity of an object is defined as the displacement of an object per unit time. i.e. Velocity = total displacement Average Velocity = initial velocity + final velocity total time taken 2 Jab bhi koi object straight line me travel kar raha ho, aur uski speed variable ho tab hum average velocity nikalte hai.

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Time= 2 min 30 sec = (400x60)+30 = 150s Hence, velocity= displacement/time = 300/150 = 2m/sec Speed = distance/time = 300/150 = 2 m/sec (B) When he jogs from A to B and turns back to C Displacement= 300-100 = 200m Distance= 300+100 = 400m Time = 3 min 30 sec = (3x60) + 30 = 210 sec Velocity= Displacement / time = 400/210 = 1.90 m/sec Speed = distance / time = 200/210 = 0.95 m/sec L.P. 1- Distinguish between speed and velocity. L.P. 2- Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? Ans- Avg Speed is distance/time and Avg Velocity is displacement/ time. So for both of them to be equal, distance should be equal to the displacement. Which will only happen if the object moves only in a straight line. L.P. 3- What does the odometer of an automobile measure? Ans- The odometer of an automobile measures the distance travelled by a vehicle. [ NCERT Intext Pg102] Speed

• Speed is the distance travelled by an object in a given interval of time.
• Speed = Distance / Time
• Speed is a scalar quantity and has only magnitude. Velocity
• Velocity is the displacement shown by an object in a given interval of time.
• Velocity = Displacement / Time
• Velocity is a vector quantity and has both magnitude and direction. I I I! i! i H I X E .............. : ! !

LP- 4: What does the path of an object look like when it is in uniform motion? Ans- In uniform motion, the path of an object can be a straight line, curved line or a circle. It can have any shape. This is because in uniform motion, speed is constant, the direction of motion may change. LP- 5: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10^8 m/s. Ans- Speed= 3x108 m/s Time= 5 min = 5x60 sec = 300 seconds Distance= 3x10^8 m/s x 300 sec = 9 x 10^10 m Rate of Change of Velocity: Acceleration During a uniform motion, velocity remains constant through out, i.e. initial and final velocities are same. So change in velocity in uniform motion will be zero. But, In case of Non-Uniform motion, velocity varies with time, it has different values at different instants. Thus, the change in velocity of the object during any time interval is not zero. Hence, to measure the rate of change of velocity we use acceleration. Acceleration= Change in Velocity If the velocity of an object changes from an initial value ‘u’ to the final value ‘v’ in time ‘t’, the acceleration ‘a’, is a = v-u

• SI unit of acceleration is m/s
• Acceleration is +ve if it is in the direction of velocity.
• It is -ve when it is opposite to the direction of velocity.
• It is 0 when object moves with constant velocity. total time taken t I ! !! I I I " I L (^) - - ......... ....... - I

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Distance-Time Graphs Used to show Change in position of object with time. Distance: y-axis Time: x-axis The Slope of the distance-time graph is equal to speed of the object. (i) For object moving with Uniform Speed: In uniform speed- Distance travelled is directly proportional to time taken. So, the speed time graph is straight line. Let’s say we have to calculate speed of the object from a distance time graph, choose any two points let say ‘A’ and ‘B’ as marked. We can clearly see for A to B, Distance travelled = s2-s time taken = t2-t Hence, Speed= s2-s t2-t (ii) For object moving with Non-Uniform Speed: The nature of this graph shows non-linear variation of the distance travelled by the car with time.

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Velocity-Time Graphs Sometimes also called as speed time graph as velocity and speed are used interchangeably. Velocity: y-axis Time: x-axis The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of displacement. (i) For object moving with Uniform Velocity: If the object moves at uniform velocity, the height of its velocity-time graph will not change with time. It will be straight line parallel to the x- axis. We can find the distance moved by the car between car between time t1 and t by calculating the area under this. So distance ‘s’ moved by the car in the time(t2-t1) can be expressed as s= area of rectangle ABCD (shaded) = AC x CD = (40 km/hr) x (t2-t1)hr = 40 (t2-t1) km (ii) For object moving with Uniform Acceleration: Suppose a person sitting next to the driver records its velocity after every 5 seconds by noting the reading of the speedometer of the car. In this case, the velocity-time graph for the motion of the car is shown in the next page. The nature of the graph shows that velocity changes by equal amounts in equal intervals of time. Thus, for all uniformly accelerated motion, the velocity-time graph is a straight line.

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L.P. 3- What can you say about the motion of an object if its speedtime graph is a straight line parallel to the time axis? Since there is no change in the velocity of the object (Y-Axis value) at any point of time (X-axis value), the object is said to be in uniform motion. L.P. 4- What is the quantity which is measured by the area occupied below the velocity-time graph? Ans- Displacement Equations of Motion: There are three equations of motion: Here, u= initial velocity of the object a= uniform acceleration of the object v= final velocity of the object s= distance travelled by the object t= time taken 7

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L.P. 1- A bus starting from rest moves with a uniform acceleration of 0.1 m/s 2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled. Ans- (a) Given that the bus starts from rest, so u= Acceleration= 0.1 m/s^ Time= 2 min= 120 sec As we know, v= u + at v= 0 + (0.1) (120) v= 12 m/s (b) We have to find ‘s’ and we already have a, t, u and v with us. So we can use s= ut + 1/2 at^ = 0 + 1/2 (0.1) (120x120) = 720 m L.P. 2- A train is travelling at a speed of 90 km/hr. Brakes are applied so as to produce a uniform acceleration of -0.5 m/s^2. Find how far the train will go before it is brought to rest. Ans- Given, u= 90 km/hr = 25 m/s As brakes are applied so it will come to rest, so v= 0m/s Given, a= 0.5 m/s^ We have to find ‘s’ and we have u, v and s with us. So we can use, v^2=u^2 + 2as Filling all the values, we’ll get s=625m L.P. 3- A trolley, while going down an inclined plane, has an acceleration of 2 cm/s^2. What will be its velocity 3 sec after the start? Ans- As the trolley starts from rest, so u= a= 0.02 m/s^ t= 3 sec We have to find ‘v’ and we have a, t and u. So we can use, v=u+at [ NCERT Intext Pg109] -- 14 ............ ---

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