Distance(s)- simply pertains to length (scalar)e.g. 5m, 5km Displacement (s)- change in position specified by magnitude + direction (vector) e.g. 5m, north , 5m, 20˚ S of E. Speed (v) – distance per unit time (scalar) where v= speed s= distance t= time
Velocity ( v )- displacement per unit time where: v = velocity s = displacement t = time
Acceleration – time rate of change of velocity
if Where: a= acceleration t= time v0= Initial velocity vi= final velocity Uniformly Accelerated Motion: the motion in a straight line in which the rate changes uniformly
Sample problems: 1.An airplane taking off from a landing field has a run of 1200ft. If it starts from rest, moves with a constant acceleration and makes he run in 30s, with what velocity in m/s did it take off?
Illustration: 1ft=0.3048m (1) From we can derive
3. a runner “A” can run the mile race in 4.25min. Another runner “B” requires 4.55mins to finish this distance. If they start out together and maintain their normal speed, how far apart will they be at the end of the race?
A stone is thrown from a window with an initial horizontal velocity of 10m/s. if the window is 20m high, and the ground is level, a. in how many seconds will the stone reach the ground? b. how far in the ground will it reach?
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. No movement External forces A=B A B Unbalance force rotate A=B
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. a=? Note: N is always ┴ to the plane N FBD- free body diagram- shows all the forces acting on the object 30° F W
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