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King Fahd University of Petroleum & Minerals. Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 2. Objective. To introduce the concepts of position, displacement, velocity, and acceleration. To study particle motion along a straight line.
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King Fahd University of Petroleum & Minerals Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 2
Objective • To introduce the concepts of position, displacement, velocity, and acceleration. • To study particle motion along a straight line.
Rectilinear KinematicsSection 12.2 • Rectilinear : Straight line motion • Kinematics : Study the geometry of the motion dealing with s, v, a. • Rectilinear Kinematics : To identify at any given instant, the particle’s position, velocity, and acceleration. (All objects such as rockets, projectiles, or vehicles will be considered as particles “has negligible size and shape” particles : has mass but negligible size and shape
Position • Position : Location of a particle at any given instant with respect to the origin r : Displacement ( Vector ) s : Distance ( Scalar )
Distance & Displacement • Displacement : defined as the change in position. • r : Displacement ( 3 km ) • s : Distance ( 8 km ) Total length • For straight-line Distance = Displacement s = r D s = D r Vector is direction oriented Dr positive (left ) Dr negative (right)
Velocity & Speed • Velocity : Displacement per unit time • Average velocity : • V = Dr / Dt • Speed : Distance per unit time • Average speed : • usp = sT / Dt (Always positive scalar ) • Speed refers to the magnitude of velocity • Average velocity : uavg = Ds / Dt
Velocity (con.) • Instantaneous velocity : • For straight-line Dr = Ds
Problem • A particle moves along a straight line such that its position is defined by s = (t3 – 3 t2 + 2 ) m. Determine the velocity of the particle when t = 4 s. At t = 4 s, the velocity = 3 (4)2 – 6(4) = 24 m/s
Acceleration • Acceleration : The rate of change in velocity {(m/s)/s} • Average acceleration : • Instantaneous acceleration : • If v ‘ > v “ Acceleration “ • If v ‘ < v “ Deceleration”
Problem • A particle moves along a straight line such that its position is defined by s = (t3 – 3 t2 + 2 ) m. Determine the acceleration of the particle when t = 4 s. • At t = 4 a(4) = 6(4) - 6 = 18 m/s2
Problem • A particle moves along a straight line such that its position is defined by s = (t3 – 12 t2 + 36 t -20 ) cm. Describe the motion of P during the time interval [0,9] Total time = 9 seconds Total distance = (32+32+81)= 145 meter Displacement = form -20 to 61 = 81 meter Average Velocity = 81/9= 9 m/s to the right Speed = 9 m/s Average speed = 145/9 = 16.1 m/s Average acceleration = 27/9= 3 m/s2 to the right
Relation involving s, v, and aNo time t Position s Velocity Acceleration
Problem 12.18 • A car starts from rest and moves along a straight line with an acceleration of a = ( 3 s -1/3 ) m/s2. where s is in meters. Determine the car’s acceleration when t = 4 s. Rest t = 0 , v = 0
Free Fall • Ali and Omar are standing at the top of a cliff of heightH. Both throw a ball with initial speedv0, Ali straightdownand Omar straightup. The speed of the balls when they hit the ground arevAandvOrespectively.Which of the following is true: (a)vA < vO(b) vA = vO (c) vA > vO Ali v0 Omar v0 H vA vO
Free fall… • Since the motion up and back down is symmetric, intuition should tell you that v = v0 • We can prove that your intuition is correct: Equation: This looks just like Omar threw the ball down with speed v0, sothe speed at the bottom shouldbe the same as Ali’s ball. • Omar v0 v= v0 H y = 0
Free fall… • We can also just use the equation directly: Ali : same !! Omar: Ali v0 Omar y = H v0 y = 0
Time dependent acceleration Constant acceleration Summary This applies to a freely falling object:
