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Chapter 2 Kinematics in One Dimension

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Mechanics: Study of motion in relation to force and energy, ie, the effects of force and energy on the motion of an object. Kinematics – Mechanics that describes how an object moves, no reference to the cause (force) or mass of the object. [Gk: kinema = movement]. Dynamics– Mechanics that deals with force: why objects move.

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To describe motion of an object, we need to specify its position, displacement, velocity and acceleration. Chap 2: Motion in one dimension only – along a straight line. Horizontal direction: position = x, displacement x, velocity v x, acceleration a x. Vertical direction: position = y, displacement y, velocity v y, acceleration a y. Chap 4 will discuss kinematics in 2-dimension [circular motion, projectile motion]. Introduction

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0 x + - + - y

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0 x + - + - y

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Consider a car driven at a perfectly steady 60 miles per hour on a long and straight section of the highway. In the first hour it will travel 60 mi. In the second hour it will cover another 60 mi. etc. We say the car is moving with uniform motion. Uniform Motion

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Uniform motion is a straight line motion where: Equal displacements are covered in any successive equal intervals of time. The velocity is constant. The graph of position versus time is a straight line.

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Uniform Motion t x xx xx 1 2 3 4 60 120 180 240 0 0 position velocity t v 1 2 3 4 20 40 60 80 0 0

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Non-uniform Motion An object in non-uniform motion – means velocity does not stay uniform (constant) during the motion. Either its magnitude or direction changes. Moving along a straight line with varying speed, or moving along a curved path. x(t) tt v(t)

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Average Velocity Velocity is the rate of change of position with time. The average velocity is displacement divided by the change in time. v ave = x/ t v ave = slope of position vs time graph v ave = (x f – x i )/(t f – t i ) x t xx tt

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Instantaneous velocity is the limit of average velocity as t gets small. It is the slope of the tangent line to the graph of x(t) versus t. It is the derivative of x(t) with respect to t. Instantaneous Velocity

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Graphs for Instantaneous Velocity x t t1t1 t3t3 v 3 >0 v 1 <0 v 2 =0 t2t2

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Determine the velocity of the car at times A, B, C and D. APositiveZeroNegative BPositiveZeroNegative CPositiveZeroNegative DPositiveZeroNegative x(t) t A BCD

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To find position from velocity x f = x i + Area under velocity-time curve between t i and t f Final position:

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The position of an object changes with time according to the expression x = t 2 + 3t - 9 (a) What is the velocity of the object at t = 3 s? (b) What is the position of the object when its velocity is 11 m/s? Example

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Area Under velocity-time Graph Area under the velocity-time graph = magnitude of the displacement over the time interval. time (s) velocity (m/s) uniform (constant) velocity t1t1 t2t2 Area = v(t 2 -t 1 ) = (x 2 – x 1 ) = x time (s) velocity (m/s)

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Example A sprinter starts from rest and runs with a constant acceleration for 6 s before reaching his maximum speed. He then runs at the maximum speed the rest of the way. If he ran the 200 m in 25.0 s, at what speed did he cross the finish line?

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Average acceleration = Motion with Constant Acceleration Constant acceleration – straight line motion where the speed changes at a constant rate. A car moving north. At time t=2sec, its speed is 4 m/s. At t=4 sec, its speed is 8 m/s. At t=6 sec, its speed is 12 m/s. Every 2 sec, its speed changes by 4 m/s. OR, every 1 sec, its speed changes by 2 m/s. It is moving with constant acceleration: a = 2 m/s/s = 2 m/s 2 north. Uniformly accelerated motion – when an object moves with constant acceleration.

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Uniform (Constant) Acceleration An object moving with uniform acceleration – means acceleration stays uniform (constant) throughout the motion. Its magnitude and direction stays the same. Its velocity changes uniformly. Moving along a straight line. a = slope t a(t) v(t) t vv tt

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Graphical Representation velocity, v (m/s) Time, t (s) Slope of the line = average acceleration O If the slope is zero, the object is moving with zero acceleration (constant velocity)

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Motion with Constant Acceleration t v a = slope of the velocity-time graph T/F? If the acceleration of an object is zero, it must be at rest. T/F? If an object is at rest, its acceleration must be zero. a<0 a>0

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Slope of the line = average acceleration = v/ t velocity, v (m/s) Time, t (s) O A B CD E 1. When is a = 0? 2. When is a < 0? 3. When is a = maximum? (A) OA (B) AB (C) BC (D) CD (E) DE

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Kinematic Equations for Const. Acceleration Consider an object moving with constant acceleration a. As the object moves, its velocity changes. a a Time = 0 Initial position = x i Initial velocity = v i Time = t Final position = x f Final velocity = v f vivi vfvf a

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Kinematic Equations for Const. Acceleration a ave = a inst Initial time = t i, final time = t f Time interval = t = t f - t i Initial position = x i, final position = x f Displacement x = x f - x i Initial velocity = v i, final velocity v f Change in velocity v = v f - v i Uniformly accelerated motion: a = const.

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Kinematic Equations for Constant Acceleration v f = v i + a t x f = x i +v i t + ½a t 2 v f 2 = v i 2 + 2a x Average velocity v ave = (v i + v f )/2 Uniformly accelerated motion: a = constant. Time: Initial = t i, final = t f Positions: Initial = x i, final = x f Velocity: Initial = v i, final = v f

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Example A train traveling at a constant speed of 22 m/s, comes to an incline with a constant slope. While going up the incline the train slows down with a constant acceleration of magnitude 1.4 m/s 2. Draw a graph of v x versus t. What is the speed of the train after 8.0s on the incline? How far has the train traveled up the incline after 8.0 s?

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Free Fall Free fall: Only force of gravity acting on an object making it fall. Effect of air resistance is negligible. Motion in vacuum. Motion of objects in air where air resistance is negligible (not feather). Force of gravity acting on an object near the surface of the earth is F = W = mg. Acceleration of any object in free fall: a = 9.8 m/s 2 down (a y = -9.8 m/s 2 ). g = 9.8 m/s 2 hence a y = -g

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Free Fall +y +x a y = -9.8 m/s 2 a x = 0

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Free Fall contd… 1. a y = -g, regardless of mass of object.

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2. a y = -g, regardless of initial velocity +y +x a y = -9.8 m/s 2, a x = 0 v 0 = 0v 0 = -15 m/s v 0 = +15 m/s

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3. Free Fall: Motion is symmetric. +y +x a y = -9.8 m/s 2, a x = 0 v 0 = +5 m/s At the maximum height: v y = 0 Speed at equal heights will be equal. Equal time going up and down.

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Example: A stone is launched straight up by a slingshot. Its initial speed is 19.6 m/s and the stone is 1.5 m above the ground when launched. (a) How high above the ground does the stone rise? (b) How much time elapses before the stone hits the ground?

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In the figure above showing a graph of position (x) versus time (t), when is the object (A) at rest? (B) moving to the left? (C) moving fastest? (D) moving slowest?

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A car moves at a constant velocity of magnitude 20 m/s. At time t = 0, its position is 50 m from a reference point. What is its position at (i) t = 2s? (ii) t = 4s (iii) t = 10s? Example

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A stray dog ran 3 km due north and then 4 km due east. What is the magnitude of its average velocity if it took 45 min? Example

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Which position-versus-time graph represents the motion shown in the motion diagram?

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Which position-versus-time graph goes with the velocity-versus-time graph at the top? The particle’s position at t i = 0 s is x i = –10 m.

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The average acceleration is the change in velocity divided by the change in time. SI unit = m/s 2 Slope of velocity-time graph. t v(t) tt vv A C E B D Acceleration (a)

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The instantaneous acceleration a s at a specific instant of time t is given by the derivative of the velocity Instantaneous Acceleration

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To find velocity from acceleration v f = v i + Area under acceleration-time curve between t i and t f OR graphically: If we know the initial velocity, v i, and the instantaneous acceleration, a inst, as a function of time, t, then the final velocity can be obtained:

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Finding velocity from acceleration

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Example The velocity of an object varies as v(t) = 2t 2 - 10t + 18 (a)Find the acceleration of the object at t = 3 s (b) At what time is the velocity of the object 9 m/s (c) What is the displacement of the object from t = 0s to t = 2s?

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A car moving south slows down with at a constant acceleration of 3.0 m/s 2. At t = 0, its velocity is 26 m/s. What is its velocity at t = 3 s? A.35 m/s south B.17 m/s south C.23 m/s south D.29 m/s south E.17 m/s north 1234567891011121314151617181920 2122232425262728293031323334353637383940 4142434445464748495051525354555657585960 6162636465

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A car initially traveling at 18.6 m/s begins to slow down with a uniform acceleration of 3.00 m/s 2. How long will it take to come to a stop? A.55.8 s B.15.6 s C.6.20 s D.221.6 s E.None of these 1234567891011121314151617181920 2122232425262728293031323334353637383940 4142434445464748495051525354555657585960 6162636465

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A ball is kicked straight up from ground level with initial velocity of 22.6 m/s. How high above the ground will the ball rise? A.9.8 m B.3.00 m C.1.15 m D.26.1 m E.19.6 m

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A ball is kicked straight up from ground level with initial velocity of 22.6 m/s. How high above the ground will the ball rise? A.9.8 m B.3.00 m C.1.15 m D.26.1 m E.19.6 m

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1. A car initially traveling at a velocity v o begins to slow down with a uniform deceleration of 1.20 m/s 2 and comes to a stop in 26.0 seconds. Determine the value of v o. A.31.2 m/s B.21.7 m/s C.27.2 m/s D.24.8 m/s E.None of these

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The minimum stopping distance for a car traveling at a speed of 30 m/s is 60 m, including the distance traveled during the driver’s reaction time of 0.5 s. What is the minimum stopping distance for the same car traveling at a speed of 40 m/s? Example

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