Rectilinear Motion in Mechanic & Dynamics

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Rectilinear Motion.

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An Overview of Mechanics Statics: T he study of bodies in equilibrium Dynamics: 1. Kinematics – concerned with the geometric aspects of motion 2. Kinetics - concerned with the forces causing the motion Mechanics: the study of how bodies react to forces acting on them

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POSITION AND DISPLACEMENT A particle travels along a straight-line path from origin O to P , thus covers some distance S which is it’s displacement .

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VELOCITY Velocity is a measure of the rate of change in the position of a particle. It is a vector quantity (it has both magnitude and direction). The magnitude of the velocity is called speed, with units of m/s or ft/s. The average velocity of a particle during a time interval  t is v avg =  r /  t The instantaneous velocity is the time-derivative of position. v = d r /dt Speed is the magnitude of velocity: v = ds/dt Average speed is the total distance traveled divided by elapsed time: (v sp ) avg = s T /  t

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ACCELERATION Acceleration is the rate of change in the velocity of a particle. It is a vector quantity. Typical unit is m/s 2 . The instantaneous acceleration is the time derivative of velocity. Vector form: a = d v / dt a = dv / dt = d 2 s/dt 2 Acceleration can be positive (speed increasing) or negative (speed decreasing). The derivative equations for velocity and acceleration can be manipulated to get a ds = v dv

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SUMMARY OF KINEMATIC RELATIONS: RECTILINEAR MOTION • Differentiate position to get velocity and acceleration. v = ds/dt ; a = dv/dt or a = v dv/ds • Integrate acceleration for velocity and position. • Note that s o and v o represent the initial position and velocity of the particle at t = 0. Velocity: ò ò = t o v v o dt a dv ò ò = s s v v o o ds a dv v or ò ò = t o s s o dt v ds Position:

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CONSTANT ACCELERATION The three kinematic equations can be integrated for the special case when acceleration is constant (a = a c ) to obtain very useful equations. A common example of constant acceleration is gravity; i.e., a body freely falling toward earth. In this case, a c = g = 9 . 81 m/s 2 downward. These equations are: t a v v c o + = 2 c o o t (1/2)a t v s s + + = ) s - (s 2a ) (v v o c 2 o 2 + =

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EXAMPLE Given: A motorcyclist travels along a straight road at a speed of 27 m/s. When the brakes are applied, the motorcycle decelerates at a rate of -6t m/s 2 . Find: The distance the motorcycle travels before it stops.

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EXAMPLE ( continued) Solution: 2) We can now determine the amount of time required for the motorcycle to stop (v = 0). Use v o = 27 m/s. 0 = -3t 2 + 27 => t = 3 s 1) Integrate acceleration to determine the velocity. a = dv / dt => dv = a dt => => v – v o = -3t 2 => v = -3t 2 + v o ò ò - = t o v v dt t dv o ) 6 ( 3) Now calculate the distance traveled in 3s by integrating the velocity using s o = 0: v = ds / dt => ds = v dt => => s – s o = -t 3 + v o t => s – 0 = -( 3) 3 + (27)(3) => s = 54 m ò ò + - = t o o s s dt v t ds o ) 3 ( 2

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If you know, then Answer… 1. The slope of a v-t graph at any instant represents instantaneous A) velocity. B) acceleration. C) position. D) jerk. 2. Displacement of a particle in a given time interval equals the area under the ___ graph during that time. A) a-t B) a-s C) v-t C) s-t

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S-T GRAPH Plots of position vs. time can be used to find velocity vs. time curves. Finding the slope of the line tangent to the motion curve at any point is the velocity at that point (or v = ds / dt ). Therefore, the v-t graph can be constructed by finding the slope at various points along the s-t graph.

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V-T GRAPH Plots of velocity vs. time can be used to find acceleration vs. time curves. Finding the slope of the line tangent to the velocity curve at any point is the acceleration at that point (or a = dv / dt ). Therefore, the a-t graph can be constructed by finding the slope at various points along the v-t graph. Also, the distance moved (displacement) of the particle is the area under the v-t graph during time  t.

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A-T GRAPH Given the a-t curve, the change in velocity (  v) during a time period is the area under the a-t curve. So we can construct a v-t graph from an a-t graph if we know the initial velocity of the particle.

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A-S GRAPH This equation can be solved for v 1 , allowing you to solve for the velocity at a point. By doing this repeatedly, you can create a plot of velocity versus distance. A more complex case is presented by the a-s graph. The area under the acceleration versus position curve represents the change in velocity (recall ò a ds = ò v dv ). a-s graph ½ (v 1 ² – v o ²) = = area under the ò s 2 s 1

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EXAMPLE Given: v-t graph for a train moving between two stations Find: a-t graph and s-t graph over this time interval Think about your plan of attack for the problem!

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EXAMPLE (continued) Solution: For the first 30 seconds the slope is constant and is equal to: a 0-30 = dv / dt = 12 /30 = 2/5 m/s 2 2 -2 5 5 a(m/s 2 ) t(s) Similarly, a 30-90 = 0 and a 90-120 = -2/5 m/s 2

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EXAMPLE (continued) The area under the v-t graph represents displacement. D s 0-30 = ½ (12)( 30) = 180m D s 30-90 = (60 )(12) = 720m D s 90-120 = ½ (12)( 30) = 180m 180 900 1080 30 90 120 t(s) s(m)

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CONCEPT QUIZ 1. If a particle starts from rest and accelerates according to the graph shown, the particle’s velocity at t = 20 s is A) 200 m/s B) 100 m/s C) 0 D) 20 m/s

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PROBLEM Given: The v-t graph shown Find: The a-t graph, average speed, and distance traveled for the 30 s interval

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PROBLEM For 0 ≤ t ≤ 10 a = dv / dt = 0 . 24 t m/s² For 10 ≤ t ≤ 30 a = dv / dt = 0.3 m/s² 0.3 2.4 a(m/s ²) 10 30 t(s) Solution:

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( continued) D s 0-10 = ò v dt = (1/3) (.12)( 10) 3 = 40 m D s 10-30 = ò v dt = ( 0.15 )(30) 2 + 9(30 ) – 0.15(10) 2 –9(10) + 40 = 850 m s 0-30 = 850 + 40 = 890 m v avg (0-30) = total distance / time = 890/30 = 29.6 m/s

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BRAIN STORMING 1. If a car has the velocity curve shown, determine the time t necessary for the car to travel 100 meters. A) 8 s B) 4 s C) 10 s D) 6 s t v 6 s 75 t v 2. Select the correct a-t graph for the velocity curve shown. A) B) C) D) a t a t a t a t

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