MOTION Distance: ➢ Distance is a scalar quantity ➢ It means the length of the whole path from starting to ending. ➢ It is often indicated as the value of its vector siblings “displacement” Displacement: ➢ It is a vector quantity ➢ If we put a direction with the measurement of distance it will be a Displacement. ➢ It’s value is the direct length from the starting point to the finishing point. ➢ It’s direction towards the final position from initial position. Note: distance from 𝐴 𝑡𝑜 𝐵 𝑎𝑛𝑑 𝑓𝑟𝑜𝑚 𝐵 𝑡𝑜 𝐴 𝑖𝑠 𝑠𝑎𝑚𝑒; ℎ𝑒𝑛𝑐𝑒 𝐴𝐵 = 𝐵𝐴 But displacement from 𝐴 𝑡𝑜 𝐵 𝑎𝑛𝑑 𝑓𝑟𝑜𝑚 𝐵 𝑡𝑜 𝐴 𝑖𝑠 𝑛𝑜𝑡 𝑠𝑎𝑚𝑒 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒𝑖𝑟 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛. 𝑡ℎ𝑒𝑖𝑟 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑏𝑢𝑡 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑒𝑎𝑐ℎ 𝑜𝑡ℎ𝑒𝑟 Hence we can write 𝐴𝐵 = −𝐵𝐴 SPEED AND VELOCITY speed: ➢ Speed is related to distance. ➢ It is defined as the change of distance in per unit time. Suppose a particle travel a distance from the point 𝑥1 𝑡𝑜 𝑥2 within the time interval from 𝑡1 𝑡𝑜 𝑡2 . According to definition , speed= distance covered/time needed = ∆𝒙 𝒕𝟐 −𝒕𝟏 = ∆𝒙 ∆𝒕 Velocity: ➢ Velocity is related to the term displacement. ➢ It is the rate of change in displacement with time. ➢ For the term velocity to be measured we need to find out the straight length from the starting point to the ending point of the moving object and then divide the quantity with the time needed. ➢ Velocity is expressed with the letter “V” Mathematically, 𝑽 = 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 𝒕𝒊𝒎𝒆 = 𝒔 𝒕 VELOCITY AND SPEED ➢ Speed is often termed as the magnitude of velocity. ➢ Change of speed is occurred by change of its value. But change in velocity is occurred by either changing the direction or value or both of them. ➢ From this point, if an object travel with a fixed magnitude of velocity but continuously change its direction, we can not call it a uniform velocity. Because its direction is changing. In that case we can say “the object is moving with a non uniform velocity but it has a constant speed.” ➢ This type of arguments occur for object which are in circular motion. ➢ In circular motion , the linear velocity of the moving object is in the direction of the tangent drawn at its current position. As the object is moving , its position is changing instantaneously and so the direction of the velocity also changes. ➢ Most of the cases, the magnitude is constant but as for the direction is changing rapidly, The velocity is non uniform in circular motion. But as for speed , it is constant. Unit : unit for both speed and velocity is same. It is 𝑚𝑠 − . UNIFORM AND NON UNIFORM VELOCITY Uniform velocity: ➢ It a constant magnitude and its direction is not changing. ➢ Speed of light is an example of uniform velocity. The speed of sound is also of this category. Non uniform velocity: ➢ Either magnitude or direction or both of them are changing from time to time. ➢ The velocity of a falling body is non uniform velocity . ➢ With the definition of non uniform velocity, another term is under consideration which is the term known as “acceleration” or “deceleration/ retardation” ACCELERATION AND DECELERATION Acceleration: ➢ It is the rate of change (increase) in velocity with respect to time. ➢ It is vector quantity. Its direction is in the direction of velocity. ➢ It denoted by "𝑎“ . Deceleration: ➢ It is the negative deceleration. ➢ Denoted by "𝑎“ with a negative sign preceding . Mathematically, 𝑎 = ∆𝑣 ∆𝑡 = (𝑣 − 𝑢)/𝑡 Unit: unit of acceleration and deceleration is 𝒎𝒔𝟐 UNIFORM AND NON UNIFORM ACCELERATION Uniform acceleration: ➢ The velocity will increase or decrease with a constant value in each second. Example: the increase in velocity of a free falling body is an example of uniform acceleration. The velocity of a free falling object increases by 9.8 𝑚𝑠1 in each second from the beginning of its motion. That means , it the first second it will get the velocity 9.8𝑚𝑠1 ; after two seconds, its velocity will increase by another 9.8𝑚𝑠1 and it will become 19.6 𝑚𝑠1 Note: all of these are in ideal condition that means we only consider The gravitational force on the object and ignore all other force like drag, Fluid friction etc on the object. But in practical experiment, all other force are acting on the object and For this reason we get some error between the experimental value and mathematical value. UNIFORM AND NON UNIFORM ACCELERATION ✓ In our level we only deal with uniform acceleration or deceleration. ✓ All our equation of motion is designed considering uniform acceleration. ✓ To deal with non uniform acceleration, we need to know integration and differentiation. Note: we should always keep in mind that every acceleration is associated with a force acting on the object. If we apply force on an object it will accelerate and when the force is withdrawn , the acceleration is also gone. EQUATION OF MOTIONS To find out the value of several variables in linear motion, we need some equations. These equations are named as equations of motion. So far, we are introduced to four (three are mostly used) such equations for our mathematical problem solving and other derivations. 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 1 𝑣 = 𝑢 ± 𝑎𝑡 Here, 𝑣 = 𝑓𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑜𝑐𝑖𝑡𝑦 𝑢 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎 = 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡 = 𝑡𝑖𝑚𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 1 𝑣 = 𝑢 ± 𝑎𝑡 𝑖𝑓 𝑢 = 0𝑚𝑠 −1 ; 𝑎 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ∴ 𝑣 = 𝑎𝑡 ∴𝑣∝𝑡 From that relation we can say that, “if any object starting from resting condition with uniform acceleration, its velocity is proportional to time” 𝑣1 𝑣2 = 𝑡1 𝑡2 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 − 2 1 2 𝑠 = 𝑢𝑡 ± 𝑎𝑡 2 Here, 𝑠 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 − 2 Now, if 𝑢 = 0 𝑚𝑠 −1 ; 𝑎 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 1 2 𝑠 = 𝑎𝑡 2 ∴ 𝑠 ∝ 𝑡2 So, an object starting from resting condition with uniform acceleration, its displacement is proportional to the square of time required. 𝑠1 𝑠2 = 𝑡12 𝑡22 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 − 3 𝑣 2 = 𝑢2 ± 2𝑎𝑠 Now if, 𝑢 = 0 𝑚𝑠 −1 ; 𝑎 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑣 2 = 2𝑎𝑠 ∴ 𝑣2 ∝ 𝑠 𝑠 ∝ 𝑣2 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 − 3 So we can say “any object starting from rest, with uniform acceleration, its displacement is proportional to the square of its velocity” 𝑠1 𝑠2 = 𝑣12 𝑣22 Form the relation we can state that “any object starting from resting condition with uniform acceleration, its displacement is proportional to the square of its velocity” 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 − 4 𝑠= 𝑢+𝑣 2 𝑡 N:B: this formula is not used most of the case. We use other three equations for solving problems. 𝒇𝒓𝒆𝒆 𝒇𝒂𝒍𝒍𝒊𝒏𝒈 𝒃𝒐𝒅𝒚 there are three rules for describing the motion of free falling object from above. The falling body need to fulfill some condition to be appropriate for the formulae: 1. All the body must start from resting condition i.e. 𝑢 = 0𝑚𝑠 −1 2. The object must travel only with the force of gravity on it. We ignore all other forces like drag, buyounce, viscous force etc on the falling body. 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 − 1 After fulfilling the two condition, The time needed for multiple falling object with different mass for reaching the ground is same. The falling time is not depending on the mass of the object. (law of mass) 𝒇𝒓𝒆𝒆 𝒇𝒂𝒍𝒍𝒊𝒏𝒈 𝒃𝒐𝒅𝒚 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 − 2 Starting from rest , if the object fall freely, its velocity is proportional to time. Mathematically, 𝑣∝𝑡 ∴ 𝑣1 𝑣2 = 𝑡1 𝑡2 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 − 3 Starting from rest , if the object fall freely, its displacement is proportional to the square to the time. Mathematically, 𝑠 ∝ 𝑡2 ∴ 𝑠1 𝑠2 = 𝑡12 𝑡22 𝑶𝒃𝒋𝒆𝒄𝒕 𝒎𝒐𝒗𝒊𝒏𝒈 𝒗𝒆𝒓𝒕𝒊𝒄𝒂𝒍𝒍𝒚 𝒖𝒑𝒘𝒂𝒓𝒅 The objects that are moving vertically upward, must have the following parameters and directions: 1. It must have an initial velocity and its direction is always upward. 2. We must keep in mind that the object is moving in vacuum . So its acceleration is equal to 9.8𝑚𝑠 −2 and its direction is always vertically downward. 3. The object comes to halt for a moment when it reaches the highest point of its trajectory. 4. From the picture, it is obvious that the vertically upward moving body crosses any point In its trajectory for two different time. One is when the object moving upward and the other Is when the object is falling back to the ground. 5. At those points, velocity of the object is equal but in opposite direction. (from the image) 𝑺𝒐𝒎𝒆 𝒊𝒎𝒑𝒐𝒓𝒕𝒂𝒏𝒕 𝒒𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝟏. 𝒉𝒊𝒈𝒉𝒆𝒔𝒕 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒕𝒓𝒆𝒋𝒆𝒄𝒕𝒐𝒓𝒚 At the highest point of trajectory, 𝑣 = 0𝑚𝑠 −1 Now from equation, 𝑣 2 = 𝑢2 − 2𝑔ℎ At highest point, ℎ = 𝐻𝑚𝑎𝑥 , 𝑣 = 0𝑚𝑠 −1 02 = 𝑢2 − 2𝑔𝐻𝑚𝑎𝑥 ∴ 𝑯𝒎𝒂𝒙 = 𝒖𝟐 𝟐𝒈 𝟐. 𝒕𝒊𝒎𝒆 𝒏𝒆𝒆𝒅𝒆𝒅 𝒇𝒐𝒓 𝒓𝒆𝒂𝒄𝒉𝒊𝒏𝒈 𝒕𝒉𝒆 𝒉𝒊𝒈𝒉𝒆𝒔𝒕 𝒑𝒐𝒊𝒏𝒕 From the equation, 𝑣 = 𝑢 − 𝑔𝑡 At highest point, 𝑣 = 0𝑚𝑠 −1 𝑢 = 𝑔𝑡 ∴𝒕= 𝒖 𝒈 𝑺𝒐𝒎𝒆 𝒊𝒎𝒑𝒐𝒓𝒕𝒂𝒏𝒕 𝒒𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝟑. 𝑻𝒊𝒎𝒆 𝒏𝒆𝒆𝒅𝒆𝒅 𝒇𝒐𝒓 𝒇𝒂𝒍𝒍𝒊𝒏𝒈 𝒃𝒂𝒄𝒌 𝒕𝒐 𝒈𝒓𝒐𝒖𝒏𝒅 the time needed for reaching the highest point and falling back is equal ∴ 𝒕𝒊𝒎𝒆 𝒏𝒆𝒆𝒅𝒆𝒅 𝒇𝒐𝒓 𝒇𝒂𝒍𝒍𝒊𝒏𝒈 𝒃𝒂𝒄𝒌, 𝒕𝟐 = 𝟒. 𝑻𝒐𝒕𝒂𝒍 𝒕𝒊𝒎𝒆 𝒐𝒇 𝒇𝒍𝒊𝒈𝒉𝒕 Time of flight = 𝒕𝟏 + 𝒕𝟐 = 𝟐𝒖 𝒈 𝒖 𝒈 𝑽𝒂𝒓𝒊𝒐𝒖𝒔 𝒕𝒚𝒑𝒆𝒔 𝒐𝒇 𝒈𝒓𝒂𝒑𝒉𝒔 1. 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑉𝑠 𝑡𝑖𝑚𝑒 𝑔𝑟𝑎𝑝ℎ: ✓ In this graph, the magnitude of velocity remains constant with time. The graph looks like, ✓ It has a slope zero (0). 𝑽𝒂𝒓𝒊𝒐𝒖𝒔 𝒕𝒚𝒑𝒆𝒔 𝒐𝒇 𝒈𝒓𝒂𝒑𝒉𝒔 2. 𝑛𝑜𝑛 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑉𝑠 𝑡 ✓ in this case , the magnitude of the velocity changes with time. But rate of change is constant. ✓ For this specific case, we have two sub cases: a) Starting from rest i.e. 𝑢 = 0𝑚𝑠 −1 For this condition, 𝑣 = 0 + 𝑎𝑡 ∴ 𝑣 = 𝑎𝑡 This graph is originated from the origin point of the reference axis and the graph looks like: * slope of this graph indicates the magnitude of the acceleration 𝑠𝑙𝑜𝑝𝑒, 𝑚 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑥𝑖𝑠 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑥𝑖𝑠 = 𝑣−𝑢 ∆𝑡 =𝑎 𝑽𝒂𝒓𝒊𝒐𝒖𝒔 𝒕𝒚𝒑𝒆𝒔 𝒐𝒇 𝒈𝒓𝒂𝒑𝒉𝒔 b) The object starting with a finite initial velocity i.e 𝑢 ≠ 0𝑚𝑠 −1 For this condition, 𝑣 = 𝑢 + 𝑎𝑡 (𝑓𝑜𝑟 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛) The graph looks like, * the intersect of the graph from vertical axis is the initial velocity , 𝑢 * slope of the graph also indicates the acceleration of the object. And it is constant. 𝑽𝒂𝒓𝒊𝒐𝒖𝒔 𝒕𝒚𝒑𝒆𝒔 𝒐𝒇 𝒈𝒓𝒂𝒑𝒉𝒔 𝑂𝑡ℎ𝑒𝑟 𝑔𝑟𝑎𝑝ℎ𝑠 𝑓𝑟𝑜𝑚 𝑛𝑜𝑛 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛) 𝑉𝑠 𝑡 ✓ For the graph, acceleration is constant. So the graph of accleleration Vs time will be like, ✓ As we know, every acceleration is associated with a force acting on the object. From newton’s law, we know 𝐹 = 𝑚𝑎 As acceleration 𝑎 is constant, the force is also constant with time. So force Vs time graph looks like, 𝑽𝒂𝒓𝒊𝒐𝒖𝒔 𝒕𝒚𝒑𝒆𝒔 𝒐𝒇 𝒈𝒓𝒂𝒑𝒉𝒔 3. 𝑛𝑜𝑛 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑑𝑒𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑏𝑗𝑒𝑐𝑡) 𝑉𝑠 𝑡𝑖𝑚𝑒 𝑔𝑟𝑎𝑝ℎ: ✓ The graph has a negative slope as its velocity is decreasing with time. ✓ It can be uniform or non uniform. The graph looks like, 𝑉 −−−−→ 𝑡 Note: calculations are similar to previous graphs. 𝑽𝒂𝒓𝒊𝒐𝒖𝒔 𝒕𝒚𝒑𝒆𝒔 𝒐𝒇 𝒈𝒓𝒂𝒑𝒉𝒔 4. 𝑛𝑜𝑛 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑛𝑜𝑛 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑉𝑠 𝑡 ✓ For this graph, the rate of change of velocity is not constant with time. ✓ It may change in a zigzag fashion. May be sometimes its slope is moving upward (acceleration), may be sometimes its slope is moving downward (deceleration). The graph looks like, 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 𝒇𝒓𝒐𝒎 𝒈𝒓𝒂𝒑𝒉 ✓ The area of velocity Vs time graph indicates the displacement in the timeframe. ✓ So, if we have any graph of velocity Vs time and asked to find out the displacement in a definite time limit, we can find out the result in two ways: 1. Either by finding out acceleration in different parts of graph and then use the equation of motion. 2. Or we can find out the area of the graph and it is our expected result. Proof: from the graph below, suppose we are asked to find out the displacement in time 𝑡 from the graph. starting from the point 𝐷 where its velocity is 𝑢. At time 𝑡 , the object is at point 𝐵 and at that point its velocity is 𝑣. We need to find out the displacement in time 𝑡. So from our previous claim, we have to find out the total area of the graph between 𝑡 = 0𝑠 𝑡𝑜 𝑡 = 𝑡𝑠 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 𝒇𝒓𝒐𝒎 𝒈𝒓𝒂𝒑𝒉 We can divide the graph into two geometric figure, 1. OEAD rectangle 2. BAD triangle ∴ 𝑡𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 𝑂𝐸𝐴𝐷 + 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐵𝐴𝐷 ∴ ∎𝑂𝐸𝐴𝐷 = 𝑢𝑡 1 2 1 2 ∴ ∆𝐵𝐴𝐷 = × 𝑡 × 𝑣 − 𝑢 = 𝑎𝑡 2 1 2 Total area = 𝑢𝑡 + 𝑎𝑡 2 = S So, the area is equal to the displacement for time 𝑡 . Note: we use this method for solving the objectives. But in CQ we have to do all calculation manually from the graph to find out the displacement. 𝑷𝒓𝒂𝒄𝒕𝒊𝒄𝒆 𝒑𝒓𝒐𝒃𝒍𝒆𝒎 𝒓𝒆𝒍𝒂𝒕𝒆𝒅 𝒕𝒐 𝒈𝒓𝒂𝒑𝒉 1. 3. 2. 4. 𝑷𝒓𝒂𝒄𝒕𝒊𝒄𝒆 𝒑𝒓𝒐𝒃𝒍𝒆𝒎 𝒓𝒆𝒍𝒂𝒕𝒆𝒅 𝒕𝒐 𝒈𝒓𝒂𝒑𝒉 5. 6. 7. 8. 𝑷𝒓𝒂𝒄𝒕𝒊𝒄𝒆 𝒑𝒓𝒐𝒃𝒍𝒆𝒎 𝒓𝒆𝒍𝒂𝒕𝒆𝒅 𝒕𝒐 𝒈𝒓𝒂𝒑𝒉 9. 10.