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Motion Distance(d)- simply pertains to length (scalar) e.g. 5m, 5km Displacement (d)- change in position specified by magnitude + direction (vector) e.g. 5m, north , 5m, 20˚ S of E. Speed (v) – distance per unit time (scalar) Velocity (v) – displacement per unit time (vector) -ve velocity is in the opposite direction v= speed/velocity 𝑣= 𝑑 𝑡 d= distance/displacement t= time • Acceleration – time rate of change of velocity 𝑣𝑓 − 𝑣𝑖 𝑎= 𝑡 if Where: ∆𝑣 = 𝑣𝑓 − 𝑣𝑖 a= acceleration t= time vi= Initial velocity vf= final velocity Uniformly Accelerated Motion: the motion in a straight line in which the rate changes uniformly Equations 1. 𝑑 = 1 2 𝑣𝑖 + 𝑣𝑓 𝑡 2. 𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡 3. 𝑑 = 𝑣_𝑖𝑡 + 1 𝑎𝑡 2 2 4. 𝑣𝑓 2 = 𝑣𝑖 2 + 2𝑎𝑑 NEWTON’S LAWS OF MOTION First law: law of Inertia a body at rest remains at rest and a body in motion continues to move in straight line at constant speed, unless an external unbalanced force acts on it. A=B No movement A Unbalance force B A=B rotate Second law: Law of Acceleration an external unbalance force acting on an object produces an acceleration in the direction of the net force, an acceleration that is directly proportional to the unbalanced force and inversely proportional to the mass of the body Where: F=force m=mass a= acceleration Where: w = weight m = mass g = gravity Third law: Interaction for every force that a first body exerts upon second body, there is a force equal in magnitude but opposite in direction that a second body exerts upon the first body. A 100N box is sliding down a frictionless plane inclined at an angle of 30° from the horizontal. Find the acceleration of the box. Note: N is always ┴ to the plane N FBD- free body diagram- shows all the forces acting on the object F W Solution: UNIFORM CIRCULAR MOTION - motion of an object in a curved or circular motion 10 rev. in 5s Frequency, F = the no. of revolution per unit time Period, T = the time required for one complete revolution. Where: v= linear velocity r= radius Radian- angle subtended by the arc of a circle whose length is equal to the radius of the same circle. x r Where: ω = angular velocity = angle turned through t = time elapse Where: v= linear velocity r= radius ω =angular velocity Using linear velocity if Centripetal force Using angular velocity SAMPLE PROBLEM If the radius of the circular path of the stone is 0.5m and its period is 0.5s. What is its constant speed? Given: r= 0.5m T= 0.5s Required : v=? Solution: What is the angular velocity of a stone which makes 10 rev in 5 seconds? The radius of the circular path is 0.5m Given: no. of revolution = 10 T= 5s r=0.5m Required: ω=? Solution: A mass of 0.5kg is whirled in a horizontal circle of radius 2m. If it makes 5 rev in 5s Find: a. speed b. acceleration c. centripetal force Given: m=0.5 kg r=2m t=5s rev=5 T=1s Solution: End of presentation