Introduction to General Physics 1 - Motion II PDF

Introduction to General Physics 1 M 6.1.3 Physics of Motion II Dr Orlaith Brennan  Learning Outcomes On completion of this lecture, students will be able to… Interpret and sketch distance-time and velocity-time graphs for a range of real-life scenarios. Recall and apply...

Introduction to General Physics 1 M 6.1.3 Physics of Motion II Dr Orlaith Brennan  Learning Outcomes On completion of this lecture, students will be able to… Interpret and sketch distance-time and velocity-time graphs for a range of real-life scenarios. Recall and apply the equations of motion for an object moving in a straight line with constant acceleration. Define and discuss acceleration due to gravity. Demonstrate how the equations of motion relate to objects in free-fall. 2  Distance -Time Graphs A graph showing the magnitude of the distance of an object plotted against time is called a distance-time graph Straight line = The vertical (y) Constant, Steady or Uniform axis is the distance In this case, the travelled from straight line the start. represents a constant speed. The horizontal (x) axis is the time from the start. 3  Distance - Time Graph Slope of the graph = (y2-y1)/(x2-x1) = Distance/Time = Speed The slope of the graph is the speed of the object A distance-time graph for an object moving with constant 4 (uniform) speed is a straight line.  Distance - Time Graph Although distance is a scalar quantity, a distance-time graph typically represents a journey from a specific starting point of an object moving in a straight line. Therefore, distance is usually equivalent to displacement. If an object is stationary (not moving) the line is horizontal. If an object is moving at a steady speed, the line on the graph is straight and sloped (the steeper the line, the greater the speed of the object). 5  Distance -Time Graph What is the main information we can gain from a distance-time graph? 6  Problem (Past Exam) Sketch a graph of distance vs time for the skier’s journey. (0-1min): 40+100+40 = 180 m (1-2 min): 100 + 40 = 140 m (2-3 min): 100 m Total: 180 + 140 + 100 = 420 m 7  Velocity - Time Graph A graph showing the magnitude of the velocity of an object plotted against time is called a velocity-time graph The vertical axis of a velocity-time graph is the velocity of the object. The horizontal axis is the time from the start. 8  Velocity - Time Graph Acceleration is the rate of change of velocity with respect to time. Slope of the graph = (y2-y1)/(x2-x1) = Velocity/Time = Acceleration The slope is the acceleration A velocity-time graph with constant acceleration is a straight line What is the acceleration of the object represented in the graph? 9  Velocity - Time Graph If an object is moving with a constant velocity, the line on the graph is horizontal. If an object is moving with a constant acceleration, the line on the graph is straight, and sloped - the steeper the line, the greater the acceleration of the object. The red line sloping downwards represents an object that is slowing down with constant acceleration. 10  Velocity - Time Graph The area under the graph (between the x-axis and the line) is the distance travelled. What is the distance travelled by the object represented in the graph? What is the area under the graph? Blue rectangle and turquoise triangle. Rectangle Triangle X-axis: 6 units Area = ½ bh Y- axis: 8 units = ½ (4 s)(8 m/s) Area: 8 m/s x 6 s = 48 m = 16 m 11 Total area = 48 + 16 m = 64 m  Velocity - Time Graph Above the horizontal axis Below the horizontal axis Triangle Area = ½ bh Triangle = ½ (10)(60) Area = ½ bh = 300 m = ½ (10)(40) = 200 m Rectangle Area = 5(60) Triangle = 300 m Area = ½ bh = ½ (15)(40) Triangle = 300 m Area = ½ (60)(15) = 450 m Total distance: 300 + 300 + 450 + 200 + 300 = 1550 m 12  Velocity - Time Graph What is the main information we can gain from a velocity-time graph? 13  Problem (Past Exam) An object takes 10 s to come to a stop under a constant acceleration. During this time, it travels a distance of 10 m. Use a graph to determine what was the object’s initial velocity. Unknown initial velocity but travels 10m in a time of 10s. Area under the graph is the distance travelled. Base is time and Height is velocity. Area = ½ bh 10 = ½ (10)(h) h = 2 = velocity = 2m/s 14  Constant Acceleration Acceleration is the rate of change of velocity with respect to time. If the object is gaining velocity by the same amount each second, it has a constant acceleration. 15  Equations of Motion Object Moving in a Straight Line with Constant Acceleration Velocity as a function of time: v = u + at Displacement as a function of time: s = ut + ½at2 Velocity as a function of displacement: v2 = u2 + 2as v: final velocity t: time s: distance 16 u: initial velocity a: acceleration  Equations of Motion Velocity as a function of time: v = u + at From acceleration we have a = (v – u)/ t at = v – u at + u= v or v = u + at 17  Equations of Motion Displacement as a function of time: s = ut + ½at2 Average velocity vave= (u + v )/2 We know vave = s/t Therefore s = t vave or s = t (u + v)/2 Substitute v = u + at s = t (u + u + at)/2 = t (2u + at)/2 18 s = ut + ½ at2  Equations of Motion Velocity as a function of displacement: v2 = u2 + 2as v = u + at Square both sides v2 = (u + at)2 = (u + at)(u + at) = u 2 + 2uat + (at)2 = u 2 + 2a (ut + ½at2 ) But s = ut + ½at2 v2 = u 2 + 2as 19  Equations of Motion s = ut + ½at2 v = u + at a is constant 20  Problem A motorbike has a constant acceleration in a straight line of 7 m/s2 in a given direction. Calculate the velocity of the bike after 20 seconds if the initial velocity of the bike is 2 m/s. What information do I have and Write out the equations to see what do I need to find? how your information fits. a = 7 m/s2 v = u + at v=? s = ut + ½at2 t = 20 s v2 = u2 + 2as u = 2 m/s v = u + at = 2 + 7(20) 21 = 142 m/s  Problem A van starting from rest accelerates at a constant rate of 5 m/s2, calculate the velocity of the van after it travels 40 m. u= 0 v = u + at a = 5 m/s2 s = ut + ½at2 v=? v2 = u2 + 2as s = 40 m v2 = 02 + 2(5)(40) v2 = 400 v = 20 m/s 22  Problem (Past Exam) An object starting from rest, travelling in a straight line, has a constant acceleration before reaching a velocity of 5 m/s in a time of 6 seconds. How far did it travel during this time? Distance = Speed x Time = 5x6 = 30 m m/s m/s 23  Constant Acceleration What is a common example of constant acceleration? 24  Acceleration Due to Gravity If an object is dropped, the force of gravity or weight of the object, causes it to fall downwards with a constant acceleration. The acceleration is ≃ 10 m/s2 or every 1 second the object gains in velocity by 10 m/s. Ball falling for ½ second at 22 25 flashes per second. 26 https://www.youtube.com/watch?v=E43-CfukEgs  Acceleration due to Gravity In the absence of air resistance (in a vacuum), all objects if released near the earth’s surface, will fall downwards with the same acceleration. This acceleration is called acceleration due to gravity (g). The value of g varies slightly depending where you are on earth. It increases slightly as you move away from the equator to the poles. Average value on earth of g = 9.81 m s-2 (correct to two decimal places) 27  Problem (Past Exam) During a typical skydive, the first half minute is free-fall followed by a period when your parachute is open. Estimate how far you would fall during this free-fall phase. a = 9.8 m/s2 u=0 t = 30 s s = ut + ½ at2 = ½ (9.8)(30)2 = 4410 m 28  Problem (Past Exam) The graph below shows a typical speed-time graph for an actual parachute jump. Use this graph to estimate the parachutist’s approximate acceleration during the first 10 seconds. Acceleration is slope of the graph during the first 10 seconds. (35-0)/(10-0) = 3.5 m/s2 29 Why is the acceleration not closer to 10 m/s2?