![]() Using the value of t = 4s in P B = ½ (8)t ², we have the position of the bus at the time of overtaking is = 64m. For the bus to overtake the car, we must have: P B = P C Since the initial velocity of the bus, u = 0, hence we have P B = ½ (8)t²Īnd P C = velocity × Time = 16×t. Solution: A) Let the position of the bus be P B and the position of the car be P C. After how much time and at what distance, the bus overtake the car?Ī) t =4s, s = 64m B) t = 5s, s = 72m C) t = 8s, s = 58m D) None of These the same time, a car travelling with a constant velocity of 16 m/s overtakes and passes the bus. Hence the acceleration of the body is 2 m/s 2.Įxample 2: A bus starts from rest and moves with constant acceleration 8 m s − 2. In order to solve for m,we need to nd equations for motion in the x- and y-directions. Solution: Here, Final velocity v = 20 m/s and initial velocity u = 0 m/s (the body was at rest yo!). 3 Equations of motion: no air resistance We rst consider the situation of a projectile launched from a tower of height h onto some impact function, ignoring the eect of air resistance. \int _ vdvĮxample 1: A body starts from rest accelerate to a velocity of 20 m/s in a time of 10 s. The rate of change of velocity is acceleration, Now we know that:Īcceleration = (Final Velocity-Initial Velocity) / Time Taken Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”. Let us assume a body that has a mass “m” and initial velocity “u”. We know that the rate of change of velocity is the definition of body acceleration. You can download Motion Cheat Sheet by clicking on the download button belowĭerivation of the Equations of Motion Derivation of First Equation of Motion
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