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Can anyone share or write the formulas for motion chapter ?

Answer» LOGINJOIN NOWPhysics\xa0>\xa0Motion\xa0>\xa0Equations of MotionEquations of MotionCricket fan? Hockey fan? Soccer fan? What is the first thing that is taught when you first start training for these or any other sports? It is understanding the correct motion, speed acceleration or the Equations of Motion. Once you master the Equations of Motion you will be able to predict and understand every motion in the world.Equations of Motion For Uniform AccelerationAs we have already discussed earlier,\xa0motion is the state of change in position of an object over time. It is described in terms of\xa0displacement,\xa0distance,\xa0velocity,\xa0acceleration,\xa0time and speed. Jogging, driving a car, and even simply taking a walk are all everyday examples of\xa0motion. The relations between these quantities are known as the equations of motion.In case of\xa0uniform acceleration, there are three equations of motion which are also known as the laws of constant acceleration. Hence, these equations are used to derive the components like displacement(s), velocity (initial and final), time(t) and acceleration(a). Therefore they can only be applied when acceleration is constant and motion is a straight line. The three equations are,v = u + atv² = u² + 2ass = ut + ½at²where, s = displacement; u = initial velocity; v = final velocity; a = acceleration; t = time of motion. These equations are referred as SUVAT equations where SUVAT stands for displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (T)Browse more Topics under MotionIntroduction to Motion and its ParametersGraphical Representation of MotionUniform Circular MotionYou can download Motion Cheat Sheet by clicking on the download button belowDerivation of the Equations of Motionv = u + atLet us begin with the first equation, v=u+at. This equation only talks about the acceleration, time,\xa0the initial and the final velocity. Let us assume a body that has a mass “m” and initial velocity “u”. Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”. Now we know that:Acceleration = Change in velocity/Time TakenTherefore, Acceleration = (Final Velocity-Initial Velocity) / Time TakenHence, a = v-u /t or at = v-uTherefore, we have: v = u + atv² = u² + 2asWe have, v = u + at. Hence, we can write t = (v-u)/aAlso, we know that, Distance = average velocity × TimeTherefore, for constant acceleration we can write:\xa0Average velocity\xa0= (final velocity + initial velocty)/2 = (v+u)/2Hence, Distance (s) = [(v+u)/2] × [(v-u)/a]or s = (v² – u²)/2aor 2as = v² – u²or v² = u² + 2ass = ut + ½at²Let the distance be “s”. We know thatDistance = Average velocity\xa0× Time. Also, Average velocity = (u+v)/2Therefore, Distance (s) = (u+v)/2\xa0× tAlso, from v = u + at, we have:s = (u+u+at)/2\xa0× t = (2u+at)/2\xa0× ts = (2ut+at²)/2 = 2ut/2 + at²/2or s = ut +½ at²\xa0Learn more:Newtons Law of MotionUniform Circular MotionMore Solved Examples For YouExample 1: A body starts from rest accelerate to a velocity of 20 m/s in a time of 10 s. Determine the acceleration of the boy.Solution: Here, Final velocity v = 20 m/s and initial velocity u = 0 m/s (the body was at rest yo!). Therefore, Time taken t = 10 s. Hence, using the equation v = u +at.a = (v-u )/t= (20 – 0 ) /10= 2 m/s2Hence the acceleration of the body is 2 m/s2.Example 2: A bus starts from rest and moves with constant acceleration\xa08ms−2. At. the same time, a car travelling with a constant velocity 16 m/s overtakes and passes the bus. After how much time and at what distance, the bus overtakes the car?A) t =4s, s = 64m B) t = 5s, s = 72m C) t = 8s, s = 58m D) None of TheseSolution: A) Let the position of the bus be PB\xa0and the position of the car be PC. From s = ut +½ at², we haveSince the initial velocity of the bus, u = 0, hence we have PB\xa0= ½ (8)t²And PC\xa0= velocity × Time = 16×t. For the bus to overtake the car, we must have: PB\xa0= PCHence, ½ (8)t² = 16×t. Therefore, t = 4s.Using the value of t = 4s in PB\xa0= ½ (8)t ², we have the position of the bus at the time of overtaking is = 64m.CLASSESBOARDSEXAMSAbout Us\xa0Press\xa0Customer Stories\xa0Jobs\xa0Educators\xa0Blog\xa0Bytes\xa0Contact Us\xa0FAQsTerms Of ServicePrivacy Policy
to identity distance and displacement by short trick in examination is if the speed is given then it is distance and if velocity is given then it is displacement..hope this will help you in your examination.
And also pls tell that how to identify distance and displacement
formula of motion..speed = st average speed= total distance/ total time taken.velocity =st.average velocity =( u+v) /2 displacement = average velocity*time takenAcceleration =(v-u)/tkinematical equation..(i)v=u+at.(ii)s=ut+1/2 ut square.(iii)2as=v sq.- u sq.s- displacementv- final velocity/ speedt- timeu- initial velocitya- acceleration.hope this will helpful for you.


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