Lecture 3 Purdue University, Physics 220 * Lecture 03 Motion with Constant Acceleration Textbook Sections 3.1 PHYSICS 220 Purdue University, Physics 220 Graphing calculator are not allowed during exams. Because they it can store nearly everything in memory. Lecture 3 Purdue University, Physics 220 * Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * Where is velocity zero? Where is velocity positive? Where is velocity negative? Where is speed largest? Where is acceleration zero? Where is acceleration positive? position vs. time velocity vs. time Exercise If speed is increasing, v and a are in same direction. If speed is decreasing, v and a are in opposite direction. Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * A skydiver is falling straight down, along the negative y direction. During the initial part of the fall, her speed increases from 16 to 28 m/s in 1.5 s. Which of the following is correct? A) v>0, a>0 B) v>0, a<0 C) v<0, a>0 D) v<0, a<0 Question During a later part of the fall, after the parachute has opened, her speed decreases from 48 to 26 m/s in 11 s. Which of the following is correct? A) v>0, a>0 B) v>0, a<0 C) v<0, a>0 D) v<0, a<0 Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * A skydiver is falling straight down, along the negative y direction. During the initial part of the fall, her speed increases from 16 to 28 m/s in 1.5 s. Which of the following is correct? A) v>0, a>0 B) v>0, a<0 C) v<0, a>0 D) v<0, a<0 Question During a later part of the fall, after the parachute has opened, her speed decreases from 48 to 26 m/s in 11 s. Which of the following is correct? A) v>0, a>0 B) v>0, a<0 C) v<0, a>0 D) v<0, a<0 correct v a correct v a Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * iClicker Draw v vs t A ball is tossed from the ground up a height of two meters above the ground and falls back down. v v v -2 -2 t 4 3 -2 t 4 3 -2 t 4 3 -2 v t 4 3 v t 4 3 A B C D E Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * Tossed Ball Draw v vs t Draw y vs t Draw a vs t v t y t a t A ball is tossed from the ground up a height of two meters above the ground and falls back down. Purdue University, Physics 220 Apply Newton’s law to calculate how things will move Lecture 3 Purdue University, Physics 220 * Spacecraft with engine on a=F/m (Newton’s second law) v = v0 + at (equation for straight line) How to find position versus time relation? x = x0 + v0t + 1/2 at2 v2 = v02 + 2a(x-x0) (Demonstrate at home) Can we relate velocity only to position change? Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * Question 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 Correct v=at Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * 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 Dx Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * Newton’s Second Law SF = ma Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * A force F acting on a mass m1 results in an acceleration a1. The same force acting on a different mass m2 results in an acceleration a2 = 2a1. What is the mass m2? (1) 2m1 (2) m1 (3) 1/2 m1 F a1 m1 F a2 = 2a1 m2 F=ma F= m1a1 = m2a2 = m2(2a1) Therefore, m2 = m1/2 Or in words…twice the acceleration means half the mass Example Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * Question Compare the average horizontal force on a car accelerating from 0 to 60 mph in 6 seconds to that when it reaches 60 mph in 12 seconds. A) F6 = F12 B) F6 = 2 F12 C) F6 = 4 F12 F = m a F = m (vf-vi)/t m(vf-vi)=F/t ½ the time requires twice the force Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * Problem Solving Recipe Identify the object(s) of interest. Draw the free-body diagram for each object, showing all the forces acting on it. Choose a coordinate system. Find the net force along each axis. Use Newton’s second law to find the acceleration along each axis. Find the velocity and displacement along each axis from the acceleration. Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * Free Body Diagrams Isolate the object of interest Identify all forces acting on object and represent them as vectors Choose a coordinate system (e.g., x,y,z) Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * 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). Neglect friction. T N W x–direction y x Example Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * y x X direction: Tractor SF = ma Fw – T = mtractora Fw = T + mtractora T W N X direction: Trailer SF = ma T = mtrailera T Fw W N Combine: Fw = mtrailera + mtractora Fw = (mtrailer + mtractor ) a 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 meters to the right. Calculate the driving force on the tractor. Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * Example y x T W N T Fw W N Combine: Fw = mtrailera + mtractora Fw = (mtrailer + mtractor ) a Acceleration: Dx = v0t +1/2 a t2 a = 2 Dx / t2 = 0.6 m/s2 FW=900kg0.6m/s2 FW = 540 Newtons 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 meters to the right. Calculate the driving force on the tractor. Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * If an object is acted on by two finite constant forces, is it possible for the object to move at constant velocity? A) No, it will accelerate B) Yes, the forces must be perpendicular C) No, it will follow a curved path D) Yes, the forces must be equal and opposite E) Yes, the forces must be in the same direction iClicker Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * Frame of Reference Motion is relative. Motion can be described only after a frame of reference is chosen. A frame of reference may be in motion with respect to other frames of reference. The description of motion in one frame of reference may be mathematically transformed into that in another frame of reference. In a frame of reference, different coordinate systems may be used to describe motion. Purdue University, Physics 220 Lecture 3 Purdue University, Physics 220 * If no net force acts on a body (FNET = 0) there is no acceleration acting on the body, i.e. it moves with constant velocity M F1 F2 F3 F4 Any reference frame in which Newton’s laws are valid is called an inertial frame of reference Inertial Frame of Reference Purdue University, Physics 220