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Kinematics Distance versus Displacement distance = 7 blocksblocks east displacement= 3 blocks distance = 7 distance 10 displacement = 3=blocks south A 7 blocks B 23 3 blocks 7.6 blocks C displacement = 7.6 blocks, 23 south of east Definition The displacement is a vector that points from an object’s initial position toward its final position and has a magnitude that equals the shortest distance between the two positions. SI Unit - meter (m) Speed versus Velocity A 7 blocks B leave: arrive: 9:20 AM 9:31 AM displacement average velocity average speed distance elapsedelapsed time time 7 7 blocks blocks east 11 minutes 11 minutes 0.64 0.64 blocks/minute, east blocks/min ute 0.64 blocks/min Definition The average velocity is a vector. Its magnitude is the displacement divided by the elapsed time. x x0 v t t0 SI Unit - meter per second (m/s) Example: You are driving your car at 27 m/s. You look away from the road for 2.0 s to tune in a decent radio station. How far does your car go during that time? x v x vt (27 m/s)(2.0 s) = 54 m t Example: You are driving your car along a straight, level road for 8.4 km at 70.0 km/h, when your car stops, having run out of gas. Over the next 30 minutes, you walk another 2.0 km to a gas station. What is your average velocity from the beginning of your drive to the time when you arrive at the gas station? Reasoning: Considering the definition of average velocity, we’ll need to know the total displacement and the elapsed time. There are two segments in which you are moving at two different speeds. We’ll need to consider each segment separately and add the two together. Displacement: Car: This is given as 8.4 km Walk: We are given that you walk 2.0 km Since both of these are in the same direction, we can just add them together… x = 8.4 + 2.0 = +10.4 km Elapsed Time: Car: The car is traveling at 70.0 km/h for a distance of 8.4 km. We can use this to figure out how long you were driving. x v t x 8.4 km t 0.12 h v 70.0 km/h Walk: We are told that you walk for 30 minutes or 0.5 h. So the total elapsed time is 0.62 h. x 10.4 km v 17 km/h t 0.62 h Instantaneous Velocity This refers to how fast something is moving at a particular instant in time. Definition The average acceleration is a vector. Its magnitude is the change in velocity divided by the elapsed time. v v 0 a t t0 SI Unit - meter per second squared (m/s2) At time t = 0 s, the plane starts from rest. At time t = 29 s, the plane’s velocity is 260 km/h and it takes off. v v o 260 km h 0 km h km h a 9.0 t to 29 s 0 s s Acceleration and Decreasing Velocity v v o 13 m s 28 m s a 5.0 m s 2 t to 12 s 9 s Equations of Kinematics Five Variables x - displacement v0 - initial velocity at time t0 = 0 s v - final velocity at time t a - acceleration (constant) t - elapsed time since t0 = 0 s Reasoning Strategy 1. Make a drawing. 2. Decide which directions are to be called positive (+) and negative (-). 3. Write down the values that are given for any of the five kinematic variables. 4. Verify that the information contains values for at least three of the five kinematic variables. Select the appropriate equation. 5. When the motion is divided into segments, remember that the final velocity of one segment is the initial velocity for the next. 6. Keep in mind that there may be two possible answers to a kinematics problem. Speed Trap The state police placed speed sensors at two locations, 201 m apart, on an interstate highway. A motorcycle passes the first sensor at time t0 = 0 s and passes the second sensor 7.25 s later. Unaware of the sensors, the motorcyclist increases his speed by maintaining a constant acceleration of 1.88 m/s2 between them. (a) Discuss the direction of each of the following vectors: displacement, velocity, and acceleration. (b) What is the speed of the motorcycle as it passes each sensor? In which direction does each of the following vectors for the motorcycle point: displacement, velocity, and acceleration? Answer: Let’s assume that the direction the motorcycle is initially moving is the positive x direction. The motorcycle continues moving in this direction; therefore, the velocity and displacement are in the positive x direction. Finally, since the speed is increasing, the acceleration must also point in the positive x direction. Answer: Three of the five kinematic variables are given in the statement of the problem: motorcycle data x a v v0 t 201 m +1.88 m/s2 ? ? 7.25 s By knowing three of the variables, the other two can be found. 1 2 1 2 x v 0t at x at 2 v0 2 t 1 (201 m) (1.88 m/s2 )(7.25 s) 2 2 7.25 s 20.9 m/s Once the initial speed is known, we can find the final speed. 2 2 v v 0 2ax 2 v v0 2ax 2 2 (20.9 m/s) 2(1.88 m/s )(201 m) 34.5 m/s A car will appear at the bottom that travels at a constant velocity. The car is leaking oil at a regular time interval. At the instant the police car will accelerate to catch up. By a freakish quirk of fate, the police car also leaks oil at the same rate. When the cars are side by side, are the cars traveling at the same velocity? A car, starting from rest, accelerates in a straight line path at a constant rate of 2.5 m/s2. How far will the car travel in 10 seconds? x=? x vot at vo = 0 m/s 1 2 v=? 2 a = 2.5 m/s2 t = 10 s x at 1 2 2 1 2 2.5 m/s 10 s 2 2 125 m A car starts from rest and accelerates at a constant rate in a straight line. In the first second, the car moves 2.0 m. How fast will the car be moving at the end of the second second? Part 1 x = 2.0 m x vot at 1 2 2 vo = 0 m/s v=? x at 1 2 2 a=? 2 x 2(2.0 m) t = 1.0 s a 2 4.0 m/s 2 1.0 s 2 t Part 2 x=? vo = 0 m/s v=? a = 4.0 m/s2 t = 2.0 s v v o at 0 m/s 4.0 m/s2 (2.0 s) = 8.0 m/s Motion in Two Dimensions vx vox a xt x 1 2 vox vx t x voxt a x t 1 2 2 v v 2a x x 2 x 2 ox v y voy a y t y voyt a yt 1 2 2 y 1 2 v oy vy t v v 2ay y 2 y 2 oy Galileo’s Experiments Galileo’s Experiments The x part of the motion occurs exactly as it would if the y part did not occur at all, and vice versa. Drop the Ball

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