Graphical Analysis Part III. Motion Graphs. Basic Equations. Velocity is Constant. acceleration is zero. and. becomes

Transcription

1 Graphical Analysis Part III Motion Graphs Basic Equations d = vt+ 0 1 at v = v 0 + at Velocity is Constant acceleration is zero and becomes 1 d = v 0 t+ at d = vt 1

2 Velocity is Constant the slope of d vs t is VELOCITY the slope of v vs t is ACCELERATION (i.e. 0) the area under the curve of v vs t is DISTANCE Acceleration is constant Assume v 0 =0, then 1 d + at = v 0 t becomes v = v 0 + at becomes d = 1 at v=at Acceleration is constant

3 The slope of the d vs t graph is velocity The slope of the tangent line at a specific point is the velocity at that point The slope of the d vs t graph is velocity The slope of the tangent line at a specific point is the velocity at that point The slope of the d vs t graph is velocity The slope of the secant line is the average velocity between the two points 3

4 Telling acceleration from a d vs t graph object is accelerating; (velocity is increasing) object is decelerating; (velocity is decreasing) acceleration is zero; (velocity is constant) Slope of v vs t graph is acceleration The area under the v vs t curve is distance Area = 1/ b h = 1/ t v = 1/ t at = 1/ a t = d 4

5 The area under the v vs t curve is distance The area equal the distance traveled in the first 5 seconds Area = 1/ b h = 1/ 5 s 10 m/s = 5 m The area under the v vs t curve is distance The shaded are is the distance traveled from the 5th to 7th second Area = 1/ h (b + b ) = 1/ s 4 m/s = 4 m The area under the v vs t curve is distance Note that: A 7 = 1/ b h = 1/ 7 s 14 m/s = 49 m A = A 7 - A 5 = 49 m - 5 m = 4 m 5

6 Area Formulas to Know Rectangle: A = b h Triangle: A = 1/ b h Trapezoid: A = 1/ h (b + b ) For a complex graph: break into parts to find total area The area under the a vs t curve is velocity Area = b h = t a = v 6

7 Summary AB accelerating BC constant velocity CD decelerating DE not moving EF accelerating FG constant velocity GH decelerating 7

8 velocity is + acceleration is + object is speeding up velocity is + acceleration is - object is slowing down velocity is - acceleration is + object is slowing down 8

9 velocity is - acceleration is - object is speeding up consider a car moving with a constant, rightward (+) velocity - say of +10 m/s. If the position-time data for such a car were graphed, then the resulting graph would look like the graph at the right. Note that a motion described as a constant, positive velocity results in a line of constant and positive slope when plotted as a position-time graph. consider a car moving with a rightward (+), changing velocity - that is, a car that is moving rightward but speeding up or accelerating. a motion described as a changing, positive velocity results in a line of changing and positive slope when plotted as a position-time graph. 9

10 Constant Velocity Positive Velocity Changing Velocity Positive Velocity Whatever characteristics the velocity has, the slope will exhibit the same (and vice versa). If the velocity is constant, then the slope is constant (i.e., a straight line). If the velocity is changing, then the slope is changing (i.e., a curved line). If the velocity is positive, then the slope is positive (i.e., moving upwards and to the right). Describe the motion of the objects depicted by the two plots below. Include such information as the direction of the velocity vector (i.e., positive or negative), whether there is a constant velocity or an acceleration, and whether the object is moving slow, fast, from slow to fast or from fast to slow. 10

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12 Describe the motion depicted by the following velocitytime graph. Tell the direction of motion (+ or - direction), the velocity and acceleration and any changes in speed (speeding up or slowing down) during the various time intervals (e.g., intervals A, B, and C). 1

13 Describe the motion depicted by the following velocitytime graph. Tell the direction of motion (+ or - direction), the velocity and acceleration and any changes in speed (speeding up or slowing down) during the various time intervals (e.g., intervals A, B, and C). Consider the graph below. Note that the slope is not positive but rather negative; that is, the line slopes in the downward direction. Note also that the line on the graph does not pass through the origin. Slope calculations are relatively easy when the line passes through the origin since one of the points is (0,0). But that is not the case here. Determine the slope of the line. Determine the velocity (i.e., slope) of the object as portrayed by the graph below. 13

14 1. Consider the graph at the right. The object whose motion is represented by this graph is... (include all that are true): moving in the positive direction. moving with a constant velocity. moving with a negative velocity. slowing down. changing directions. speeding up. moving with a positive acceleration. moving with a constant acceleration. The velocity-time graph for a two-stage rocket is shown below. Use the graph and your understanding of slope calculations to determine the acceleration of the rocket during the listed time intervals. t = 0-1 second t = 1-4 second t = 4-1 second Consider the velocity-time graph below. Determine the acceleration (i.e., slope) of the object as portrayed by the graph 14

15 Find the shaded area for each below. Make sure you include units. Find the shaded area for each below. Make sure you include units. Find the shaded area for each below. Make sure you include units. 15