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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter Goal: To learn how to solve problems about motion in a straight line. Chapter 2. Kinematics in One Dimension

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Student Learning Objectives To differentiate clearly between the concepts of position, velocity, and acceleration. To translate kinematic information between verbal, pictorial, graphical, and algebraic representations. To utilize basic ideas of differentiation and integration both symbolically and graphically in kinematics problems. To understand free-fall motion and motion along an inclined plane. To begin the development of a robust problem-solving strategy for solving quantitative kinematics problems.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook problem # 1b Sketch a position vs. time graph for the following on the white board: Trucker Bob starts the day 120 miles west of Denver. He drives east for 3 hours at a steady 60 mph before stoping for his coffee break. Let Denver be located at x = 0 mi. and assume the x-axis (which is the position axis) points east (i.e. positive direction on axis represents east). Also assume he started driving at t = 0 hrs.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Solution to Workbook problem # 1b

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Non-uniform motion A volunteer will go to the far right hand side of the room. At t=0, start moving left very slowly and gradually speed up to top speed, passing the established origin on the way. Draw a motion diagram (without coordinate axis), a position vs. time graph, velocity vs. time graph and an acceleration vs time for this motion on the white board. Assume our volunteer is moving with constant acceleration.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook Problem #5 The figure shows the position vs. time graphs for two objects, A and B, that are moving along the same axis. a.At the instant t=1s, is the speed of A greater than, less than, or equal to the speed of B? Explain: b.Do A and B ever have the same speed? If so when?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook Problem #5 - Answer a.At the instant t=1s, the speed of A is greater than the speed of B. The slope of the line A is steeper than the slope of the line tangent to the point at t = 1s for B. b.A and B have the same speed when the slope of the line tangent to B is the same as that of A, at about 2.5 s.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook Problem #6

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook Problem #6 a,b,c,d answers

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook #6 e, f

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Which velocity-versus-time graph goes with the position-versus-time graph on the left?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Which velocity-versus-time graph goes with the position-versus-time graph on the left?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Using the velocity graph to determine displacement Redraw and put results in a table next to graph.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Using the velocity graph to determine displacement x f = x i + the area under the velocity curve v x between t 0 and t f. t=1sx 1 = 5 m t=2sx 2 = 20 m t=3sx 3 = 45 m t= 4sx 4 = 80 m t=5sx 5 = 80 m (a tad unrealistic – nobody’s brakes are that good!)

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Using the velocity graph to determine displacement x f = x i + the area under the velocity curve v x between t 0 and t f. t=1sx 1 = 5 m But the velocity at t= 1s is equal to 10 m/s, according the graph. Why did the object only move 5m?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Using the velocity graph to determine position When acceleration is constant: where (v 0 + v 1 )/2 = v avg For the first second v avg = 5m/s

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Motion with Constant Acceleration A jet plane has an acceleration of 2m/s 2, starting from rest. Make a velocity vs. time graph for t = 0, 1, 2, 3, and 4s on the white board without using calculators or formulas. Make a position vs. time graph for the same time period. without resorting to a formula. Assume x 0 = 0.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Motion with Constant Acceleration Establish velocity values by taking the area under the acceleration curve at each second and adding v 0 : Establish position values by taking the area under the velocity curve at each second and adding x 0.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Motion with Constant Acceleration Establish velocity values by taking the area under the acceleration curve at each second and adding v 0 : Establish position values by taking the area under the velocity curve at each second and adding x 0.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. x f = x i + v x ∆t for constant velocity for constant acceleration Is time important? If not, default to eq #3 above.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

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Example Problem #1 Bob throws the ball straight up at 20 m/s releasing the ball 1.5 m above the ground. a.What is the maximum height of the ball above the ground? b.What is the ball’s impact speed when it hits the ground? Step 1: pictorial representation, using particle model.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Bob throws the ball straight up at 20 m/s releasing the ball 1.5 m above the ground. Draw a coordinate axis showing 0 and positive direction. Show important points in the motion, i.e. starting and ending points, where acceleration changes, points indicated by the problem. Use the particle model and select appropriate symbols for the kinematic quantities at those points. Acceleration symbols go between points It’s not necessary to draw Bob!

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. List knowns and quantities you need to find. Use a motion diagram and vector subtraction if you need to confirm the direction of acceleration. KnownsFind y 0 = 1.5my 1, v 2 v 0 = 20 m/s t 0 = 0s a 0 = -9.8m/s 2 v 1 = 0 m/s x 2 = 0 m

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. KnownsFind y 0 = 1.5my 1, v 2 v 0 = 20 m/s t 0 = 0s a 0 = -9.8m/s 2 v 1 = 0 m/s x 2 = 0 m Bob throws the ball straight up at 20.0 m/s releasing the ball 1.5 m above the ground. a.What is the maximum height of the ball above the ground? b.What is the ball’s impact speed when it hits the ground? For part a, is time important? The answer is no, so use equation #3 with appropriate symbols: v 1 2 = v a 0 ∆y 1,0 where ∆y 1,0 = y 1 - y 0 Solve for ∆y 1,0 = - v 0 2 /2a ∆y 1,0 = 20.4 m, so y 1 = 21.9 m. Assess: If I could throw a ball straight up at about 40 mph, 22 m (70-ish feet) seems like a reasonable number

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. KnownsFind y 0 = 1.5mv 2 v 0 = 20 m/s t 0 = 0s a 0 = -9.8m/s 2 v 1 = 0 m/s x 2 = 0 m y 1 = 21.9 m Bob throws the ball straight up at 20.0 m/s releasing the ball 1.5 m above the ground. a.What is the maximum height of the ball above the ground? b.What is the ball’s impact speed when it hits the ground? For part b, is time important? The answer is no, so use equation #3 with appropriate symbols: v 2 2 = v a 0 ∆y 2,1 where ∆y 2.1 = - y 1 Solve for v 2 = +/- (- 2a 0 y 1 ) 1/2 v 2 = m/s Assess: reasonable, since it fell just a tad more than when it was launched.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example – 3 equations in 3 unknowns A sprinter can accelerate with constant acceleration for 4.0 s before reaching top speed. He can run the 100 m dash in 10 s. what is his speed as he crosses the finish line? Draw a pictorial representation, list knowns and “finds”, and decide whether time is important.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example – 3 equations in 3 unknowns A sprinter can accelerate with constant acceleration for 4.0 s before reaching top speed. He can run the 100 m dash in 10 s. what is his speed as he crosses the finish line? Answer: 12.5 m/s

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Instantaneous Acceleration The instantaneous acceleration a s at a specific instant of time t is given by the derivative of the velocity:

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook exercise #23

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook exercise #23 -Answer

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Non-constant acceleration The graph shows an acceleration vs time graph of a particle moving along the x axis. Its initial velocity is v 0x = 8.0 m/s at t=0 s. a.Is the particle speeding up, or slowing down? b.What is the velocity at t = 4.0s.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Non-constant acceleration The graph shows an acceleration vs time graph of a particle moving along the x axis. Its initial velocity is v 0x = 8.0 m/s at t=0 s. a.The particle is speeding up b.What is the velocity at t = 4.0s? v(4s) = 16 m/s

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC # 19 – inclined plane A skier is gliding along at 3.0 m/s on horizontal frictionless snow. He starts down a 10° incline. His speed at the bottom is 15 m/s. a.What is the length of the incline? b.How long does it take to reach the bottom? Draw a pictorial representation, list knowns and unknowns, and decide whether time is important, in either part a or b.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC # 19 – inclined plane A skier is gliding along at 3.0 m/s on horizontal frictionless snow. He starts down a 10° incline. His speed at the bottom is 15 m/s. Positive direction to be in the direction of motion of the skier Story starts when he hits the incline.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC # 19 – inclined plane A skier is gliding along at 3.0 m/s on horizontal frictionless snow. He starts down a 10° incline. His speed at the bottom is 15 m/s. a.What is the length of the incline? 64 m b.How long does it take to reach the bottom? 7.1 s

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 2 objects at different acceleration Ball A rolls along a frictionless, 2 meter long horizontal surface at a speed of 1.0 m/s. Ball B is released from rest at the top of a 2.0 m long ramp at an angle of 10 degrees at the exact instant that Ball A passes by. Will B overtake A before reaching the bottom of the ramp? If so, at what position?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 2 objects at different accelerations Ball A rolls along a frictionless, 2 meter long horizontal surface at a speed of 1.0 m/s. Ball B is released from rest at the top of a 2.0 m long ramp at an angle of 10 degrees at the exact instant that Ball A passes by. Will B overtake A before reaching the bottom of the ramp? If so, at what position? KnownFind x 0A = x 0B =0 m x 1A = x 1B t 0A = t 0B = 0 swhere v 0A = v 1B = 1m/s t 1A = t 1B v 0B = 0 m/s a A = 0 m/s/s a B = g sin 10° m/s/s x 1A = x 1B t 1A = t 1B Note: is x<2 m?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Problem - Answer Ball A rolls along a frictionless horizontal surface at a speed of 1.0 m/s. Ball B is released from rest at the top of a 2.0 m long ramp at an angle of 10 degrees at the exact instant that Ball A passes by. Will B overtake A before reaching the bottom of the ramp? If so, at what position? Yes, at 1.18 m.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Free Fall Problem for 2 objects A pellet gun is fired straight downward from the edge of a cliff that is 15 m above the ground. The pellet strikes the ground with a speed of 27 m/s. How far above the cliff would the pellet have gone if the gun had been fired straight upward (Note: the gun has the same muzzle speed whether fired up or down). For 2 object problems, there is always at least one quantity that is the same for both objects. Use your pictorial representation and list of variables to figure out the important quantity or quantities.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Free Fall Problem (#46)-Answer v 0D = -v 0U = m/s. How far the bullet goes is y 1U = 22.3m

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC # 42 t (s)v (mph) Measured velocity data for a Porsche 944 Turbo is as follows: a. Is the acceleration constant? Graph the data and find out. b. How would you estimate the car’s acceleration at 2s and at 8s? Do so and give your answer in SI units. c. How would you estimate the distance traveled in the 10s? Do so. 1mph = m/s

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC # 42 a(2s) = 11.5 mph/s - average between t=4s and t=0s a(8s) = 5.4 mph/s – average between t=10s and t = 6s to estimate distance use rectangles with a width of 2s and a height of v avg for the interval ∆x = approx 217m

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What was that Quadratic formula again?? A cave explorer drops a stone from rest into a hole. The speed of sound is a constant 343 m/s in air. The sound of the stone hitting the bottom is heard 1.50 s after the stone is dropped. How deep is the hole?

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What was that Quadratic formula again?? - Answer A cave explorer drops a stone from rest into a hole. The speed of sound is a constant 343 m/s in air. The sound of the stone hitting the bottom is heard 1.50 s after the stone is dropped. How deep is the hole? The quadratic formula yields both a negative and positive result for t 1stone. The former is not physically possible. t 1stone = 1.47 s The hole is 10.6 m deep.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Track Problems Position s is measured as distance from the origin, (s = 0) along the track. Right is positive, left negative, regardless of up/down x, v, and a vs. t graphs should be stacked as shown with the same time (horizontal) scale for all three. Graphs should show qualitatively correct slopes.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Track Problem #36 (end of chapter) The ball goes down and then up to the left, turns and rolls all the way to the bottom. Draw position, velocity and acceleration graphs for the ball, all with the same time scale. Assume the ball has enough speed to reach the top. Recall position is measured as distance from the arbitrary origin (far left) therefore two right portions of track will show as the same position on the position graph.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Track Problem #36 - Answer The ball first speeds up in a negative direction, starting from rest. Then it slows down, still in the negative direction, and has a turning point at s = 0. Then it speeds up in the positive direction.

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