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PHYSICS 149: Lecture 10 • Chapter 4 – 4.1 Motion along a Line due to a Constant Net Force – 4.2 Visualizing Motion along a Line with Constant Acceleration Lecture 10 Purdue University, Physics 149 1 Exam Schedule • Midterm Exam 1 – – – – Wednesday, February 23, 18:30 – 19:30 Place: PHY 333 Chapters 1 - 4 Things to bring: • Purdue ID Card, Crib Sheet, #2 Pencil, Calculator • See syllabus for detailed information • Midterm Exam 2 – Wednesday, April 6, 18:30 – 19:30 – Place: PHYS 333 • Final Exam – Thursday, May 5, 15:20 – 17:20 – Place: PHY 333 Lecture 10 Purdue University, Physics 149 2 Midterm Exam 1 • The exam is closed book. • The exam is a multiple-choice test. • There will be 15 multiple-choice problems. – Each problem is worth 10 points. – Maximum possible score will be 150 points. • Note that total possible score for the course is 1,000 points (see the course syllabus) • The difficulty level is about the same as the level of textbook problems. • You may make a single crib sheet – You may write on both sides of an 8.5” × 11.0” sheet Lecture 10 Purdue University, Physics 149 3 Midterm Exam 1 • Do not forget to bring – – – – Purdue ID Card, Crib Sheet, #2 Pencil (with an eraser), Calculator • Adaptive learners should contact Prof. Neumeister ASAP. • Academic Dishonesty: – Do not cheat! Cheaters will be given an F grade in the course and will be reported to the Dean of Students. • Review session: – During next recitation session – Come and see your TA in the help center (private mini-reviews) Lecture 10 Purdue University, Physics 149 4 ILQ 1 A ball is thrown straight up in the air and returns to its initial position. During the time the ball is in the air, which of the following statements is true? A) B) C) D) Both average acceleration and average velocity are zero. Average acceleration is zero but average velocity is not zero. Average velocity is zero but average acceleration is not zero. Neither average acceleration nor average velocity are zero. C correct Vave = Δy/Δt = (yf – yi) / (tf – ti) = 0 aave = Δv/Δt = (vf – vi) / (tf – ti) Not 0 since Vf and Vi are not the same ! Lecture 10 Purdue University, Physics 149 5 ILQ 2 How could you determine acceleration from a graph of velocity versus time? A) From the slope of the line B) Impossible – you need a plot of velocity vs acceleration C) By reading it off the horizontal axis D) Impossible – you need a plot of acceleration versus time E) By reading it off the vertical axis Lecture 10 Purdue University, Physics 149 6 ILQ 3 A) B) C) Which x vs t plot shows positive acceleration? “The amount of distance traveled increases with each second that passes, so the slope should get steeper and steeper as long as the acceleration is positive.” Lecture 10 Purdue University, Physics 149 7 Review: Notions in Chapter 3 • Position Vector • Displacement vs. Distance • Average Velocity vs. Average Speed • Instantaneous Velocity (often called velocity) • Average Acceleration • Instantaneous Acceleration (often called acceleration) (They are vectors expect for distance and average speed) • Newton’s 2nd Law of Motion: Lecture 10 Purdue University, Physics 149 11 Determination of vx on Graph of x vs. t • The instantaneous velocity vx is the slope of the line tangent to the graph of x vs. t (or of x(t)) at the chosen time. Recall Lecture 10 Purdue University, Physics 149 12 Positive, Zero, and Negative Velocity vx: zero vx: positive vx: negative positive slope Î vx: positive & moving in +x-direction steeper slope Î faster moving Lecture 10 Purdue University, Physics 149 13 ILQ • The graph for position vs. time is given below for a car. What can you say about the velocity of the car over time? A) B) C) D) E) It speeds up all the time. It slows down all the time. It moves at constant velocity. Sometimes it speeds up and sometimes it slows. Not really sure. Lecture 10 Purdue University, Physics 149 14 Determination of ax on Graph of vx vs. t • The instantaneous acceleration ax is the slope of the line tangent to the graph of vx vs. t (or of vx(t)) at the chosen time. Recall lower positive acceleration higher positive acceleration Lecture 10 Purdue University, Physics 149 15 Determination of ∆x on Graph of vx vs. t • The displacement of ∆x during any time interval equals the area under the graph of vx(t). Recall: – Note: If vx is negative, the displacement is also negative. So, we count the area as negative. Lecture 10 Purdue University, Physics 149 16 ILQ • The graph at right represents the velocity of a car over time. The displacement of the car can be found by A) B) C) D) adding the slopes of each section of the graph. adding the area 1 to area 2. subtracting area 2 from area 1. subtracting area 1 from area 2. Lecture 10 Purdue University, Physics 149 17 Summary for Interpreting Graphs • On a graph of x(t), the slope at any point is vx • On a graph of vx(t), the slope at any point is ax • On a graph of vx(t), the area under the graph during any time interval is the displacement ∆x during that time interval. – Note: If vx is negative, the displacement is also negative. So, we count the area as negative. • On a graph of ax(t), the area under the curve is ∆vx(t), the change in vx during that time interval. Lecture 10 Purdue University, Physics 149 18 Problem Solving Strategy • Decide what objects will have Newton’s second law applied to them. • Identify all the external forces acting on that object. • Draw a free-body diagram to show all the forces acting on the object. • Choose a coordinate system. If the direction of the net force is known, choose axes so that the net force is along one of the axes. • Find the net force by adding the forces as vectors. • Use Newton’s second law to relate the net force to the acceleration. • Relate the acceleration to the change in the velocity vector during a time interval. Lecture 10 Purdue University, Physics 149 19 Newton’s Second Law: F = ma A tractor T is pulling a trailer M with a constant acceleration. If the forward acceleration is 1.5 m/s2, calculate the force on the trailer (m=400 kg) due to the tractor (m=500 kg). x–direction y N T x W Lecture 10 Purdue University, Physics 149 20 1-D Motion with Constant Acceleration • If the net force on an object is constant, the acceleration of the object is also constant, both in magnitude and direction (recall Newton’s 2nd Law). • If the acceleration a is constant, xf = xi + vi⋅(tf–ti) + ½⋅a⋅(tf–ti)2 (xf – xi) = ½⋅(vf+vi)⋅(tf–ti) vf = vi + a⋅(tf–ti) vf2 – vi2 = 2⋅a⋅(xf–xi) where xi and vi are the initial position and velocity at an initial time ti. And, xf and vf are the final position and velocity at a final time tf. Lecture 10 Purdue University, Physics 149 21 1-D Motion with Constant Acceleration • vf = vi + a⋅(tf–ti) (if a is constant) – This is simply the definition of average acceleration. In this case, aav = a (= const). • (xf – xi) = ½⋅(vf+vi)⋅(tf–ti) (if a is constant) – vav = ½⋅(vf+vi) (if a is constant) – From the definition of average velocity, (xf – xi) = vav⋅(tf–ti) = ½⋅(vf+vi)⋅(tf–ti) Lecture 10 Purdue University, Physics 149 22 1-D Motion with Constant Acceleration • xf = xi + vi⋅(tf–ti) + ½⋅a⋅(tf–ti)2 (if a is constant) – (xf – xi) = ½⋅(vf+vi)⋅(tf–ti) = ½ ⋅ {[vi+a⋅(tf–ti)]+vi} ⋅ (tf–ti) = vi⋅(tf–ti) + ½⋅a⋅(tf–ti)2 • vf2 – vi2 = 2⋅a⋅(xf–xi) (if a is constant) – (xf – xi) = ½⋅(vf+vi)⋅(tf–ti) = ½⋅(vf+vi) ⋅ (vf–vi)/a = (vf2–vi2) / (2⋅a) Lecture 10 Purdue University, Physics 149 23 Constant Acceleration x • x = x0 + v0t + 1/2 at2 • v = v0 + at • v2 = v02 + 2a(x-x0) t v at2 Δx = v0t + 1/2 Δv = at v2 = v02 + 2a Δx t a t Lecture 10 Purdue University, Physics 149 tt 24 Example A tractor T (m=500 kg) is pulling a trailer M (m=400 kg). It starts from rest and pulls with constant force such that after 10 seconds it has moved 30 y meters to the right. Calculate the horizontal force on the tractor. x-direction: Tractor ΣF = ma Fw – T = mtractora Fw = T + mtractora Lecture 10 N T W x-direction: Trailer ΣF = ma T = mtrailera x T N Fw W Combine: Fw = mtrailera + mtractora Fw = (mtrailer + mtractor ) a Purdue University, Physics 149 25 Example A tractor T (m=500 kg) is pulling a trailer M (m=400 kg). It starts from rest and pulls with constant force such that after 10 seconds it has moved 30 y meters to the right. Calculate the horizontal force on the tractor. x Combine: Fw = mtrailera + mtractora Fw = (mtrailer + mtractor ) a Acceleration: Δx = v0t +0.5 a t2 a = 2 Δx / t2 = 0.6 m/s2 Lecture 10 N W T T N Fw W FW = 900kg×0.6m/s2 FW = 540 Newtons Purdue University, Physics 149 26 Kinematics Example A car is traveling 30 m/s and applies its breaks to stop after a distance of 150 m. How fast is the car going after it has traveled ½ the distance (75 meters) ? A) v < 15 m/s B) v = 15 m/s C) v > 15 m/s This tells us v2 proportional to Δx Lecture 10 Purdue University, Physics 149 27 ILQ: Acceleration A car accelerates uniformly from rest. If it travels a distance D in time t then how far will it travel in a time 2t? A) D/4 B) D/2 C) D D) 2D E) 4D Correct x=1/2 at2 Follow up question: If the car has speed v at time t then what is the speed at time 2t? A) v/4 B) v/2 C) v D) 2v E) 4v Lecture 10 Correct v=at Purdue University, Physics 149 28