Lesson_1.2_Printable

Lesson_1.2_Printable - Velocity-time graph t=0 t=1 t=2 t=3...

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Velocity-time graph Consider an object moving velocity 2.0m 2.0 m 2.0 m 2.0 m 2.0 m Fig. 1The displacement is the same in every 1 s interval t=0 t=1 t=2 t=3 t=4 t=5 t =1 t =1 t =1 t =1 t =1 Time (s) 0 1 2 3 4 Velocity (ms -1 ) 2 2 2 2 2 with uniform velocity v ms -1 . v remains1 a constant and does not change with time. A graph of v against time t is a horizontal line time Displacement = uniform velocity × time v m/s t (s) x = v t 1
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time velocity v m/s t (s) x = v t Look at the area enclosed by the graph of v with the t axis . is a rectangle with length = and width = .s -1 It is a rectangle with length = t s and width = v m.s The area under the graph = v m.s -1 × t s = v t m = displacement (x) The area enclosed by the velocity graph with time axis is a measure of the displacement of the moving object. 2
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Area under the velocity-time graph The area enclosed by the graph of a function with the x-axis is the integral of the function. Since area enclosed by the graph of velocity against time is a measure of the displacement, we can conclude that: x vdt = velocity Displacement of a moving object can be obtained by integrating its velocity function. time v(t) x vdt = Displacement of a moving object from t = t 1 to t = t 2 is: 2 1 t t x vdt = 3
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Accelerated Motion Acceleration is caused by a change in velocity . Acceleration ( a ) is defined as the rate of change of velocity ( v ). v a t = change in velocity Acceleration = change in time 2 1 2 1 v v t t - = - Since acceleration is the rate of change of velocity, the instantaneous acceleration is the derivative of the velocity function with respect to time, and can be written as: 2 2 d x dt = -2 Acceleration is me asu m. r in s ed . 0 t dv dt v a Lim t ∆ → = = 4
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velocity v2 v time v1 t Fig. 4 Acceleration is the slope of the velocity-time graph. 5
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velocity v time t Fig. 5 The slope of velocity-time graph is negative. 6
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A car starts from rest and accelerates uniformly for 5 s until its velocity reaches 10 ms -1 . What is its acceleration? A car starts from rest means that v = 0 when t = 0 v = 10 ms -1 when t = 5 s v = 10 – 0 = 10 m.s -1 t = 5 – 0 = 5 s Acceleraiton = v t 1 10 . 5 m s s - = 2 2 . m s - = 7
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The definitions of displacement x , velocity v and acceleration a can be summarized as follows: x vdt = dv a t = dx v dt = v adt = We will now develop equations to describe uniformly accelerated motion. When time t = 0 , we will assume that: dt The object is at position x o and has a velocity v o It has a uniform acceleration a and at any time t After a time t , Its position is x and velocity is v f 8
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First we will obtain a relation between v f , v o , a and t . f
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This note was uploaded on 05/01/2011 for the course PHY 2048 taught by Professor George during the Fall '10 term at Edison State College.

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Lesson_1.2_Printable - Velocity-time graph t=0 t=1 t=2 t=3...

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