Chapter 2: 1-D Kinematics

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Presentation transcript:

Chapter 2: 1-D Kinematics Paul E. Tippens, Professor of Physics Southern Polytechnic State University Editing by Mr. Gehman © 2007

The Cheetah: A cat that is built for speed The Cheetah: A cat that is built for speed. Its strength and agility allow it to sustain a top speed of over 100 km/h. Such speeds can only be maintained for about ten seconds. Photo © Vol. 44 Photo Disk/Getty

Objectives: After completing this module, you should be able to: Define and apply concepts of average and instantaneous velocity and acceleration. Solve problems involving initial and final velocity, acceleration, displacement, and time. Demonstrate your understanding of directions and signs for velocity, displacement, and acceleration. Solve problems involving a free-falling body in a gravitational field.

Uniform Acceleration in One Dimension: Motion is along a straight line (horizontal, vertical or slanted). Changes in motion result from a CONSTANT force producing uniform acceleration. The cause of motion will be discussed later. Here we only treat the changes. The moving object is treated as though it were a point particle.

Reference Frames and Displacement Any measurement of position, distance, or speed must be made with respect to a reference frame. For example, if you are sitting on a train and someone walks down the aisle, their speed with respect to the train is a few miles per hour, at most. Their speed with respect to the ground is much higher.

Reference Frames and Displacement We make a distinction between distance and displacement. Displacement (blue line) is how far the object is from its starting point, regardless of how it got there. Distance traveled (dashed line) is measured along the actual path.

Distance and Displacement Distance is the length of the actual path taken by an object. Consider travel from point A to point B in diagram below: Distance is a scalar quantity (no direction): Contains magnitude only and consists of a number and a unit. (70 m, 40 mi/h, 10 gal)

Distance and Displacement Displacement is the straight-line separation of two points in a specified direction. A vector quantity: Contains magnitude AND direction, a number, unit & angle. (40 m east; 8 km/h, N)

Reference Frames and Displacement The displacement is written: Left: Displacement is positive. Right: Displacement is negative.

Distance and Displacement For motion along x or y axis, the displacement is determined by the x or y coordinate of its final position. Example: Consider a car that travels 8 m, E then 12 m, W. Net displacement D is from the origin to the final position: D 8 m,E x x = -4 x = +8 D = 4 m, W 12 m,W What is the distance traveled? 20 m !!

The Signs of Displacement Displacement is positive (+) or negative (-) based on LOCATION. The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION. Examples: 2 m -1 m -2 m The direction of motion does not matter!

Definition of Speed Speed is the distance traveled per unit of time (a scalar quantity). A B v = = s t 20 m 4 s s = 20 m v = 5 m/s Not direction dependent! Time t = 4 s

Examples of Speed Orbit 2 x 104 m/s Light = 3 x 108 m/s Car = 25 m/s Jets = 300 m/s Car = 25 m/s

Speed Examples (Cont.) Runner = 10 m/s Glacier = 1 x 10-5 m/s Snail = 0.001 m/s

Definition of Velocity Velocity is the displacement per unit of time. (A vector quantity.) A B d = 12 m v = 3 m/s, East Direction required! Time t = 4 s

Average Speed & Velocity Speed: how far an object travels in a given time interval Velocity includes directional information:

Avg. Speed vs. Avg. Velocity v = 25 m/s Not direction dependent! v = 10 m/s, East Direction required!

Speed Example Exercise FloJo, ’88 Olympics Converting to km/hr:

Total distance: s = 200 m + 300 m = 500 m Example 1. A runner runs 200 m, east, then changes direction and runs 300 m, west. If the entire trip takes 60 s, what is the average speed and what is the average velocity? Recall that average speed is a function only of total distance and total time: s2 = 300 m s1 = 200 m start Total distance: s = 200 m + 300 m = 500 m Avg. speed 8.33 m/s Direction does not matter!

Direction of final displacement is to the left as shown. Example 1 (Cont.) Now we find the average velocity, which is the net displacement divided by time. In this case, the direction matters. xo = 0 t = 60 s x1= +200 m xf = -100 m x0 = 0 m; xf = -100 m Direction of final displacement is to the left as shown. Average velocity: Note: Average velocity is directed to the west.

Total distance/ total time: Example 2. A sky diver jumps and falls for 600 m in 14 s. After chute opens, he falls another 400 m in 150 s. What is average speed for entire fall? 600 m 400 m 14 s 150 s A B Total distance/ total time: Average speed is a function only of total distance traveled and the total time required.

Average Speed and Instantaneous Velocity The average speed depends ONLY on the distance traveled and the time required. The instantaneous velocity is how fast and in what direction an object is moving in a particular instant. (v at point C) A B s = 20 m Time t = 4 s C

- - The Signs of Velocity + Velocity is positive (+) or negative (-) based on direction of motion. + - First choose + direction; then v is positive if motion is with that direction, and negative if it is against that direction. + - +

Velocity in Position-Time Graphs This is a graph of p vs. t for an object moving with const. velocity. If we take the slope of this line (“rise over run”) we get Notice that the slope of the p-t curve can be (+) or (-).

Definition of Acceleration An acceleration is the change in velocity per unit of time. (A vector quantity.) A change in velocity requires the application of a push or pull (force). A formal treatment of force and acceleration will be given later. For now, you should know that: The direction of accel- eration is same as direction of force. The acceleration is proportional to the magnitude of the force.

Example of Acceleration + vf = +8 m/s v0 = +2 m/s t = 3 s Force The wind changes the speed of a boat from 2 m/s to 8 m/s in 3 s. Each second the speed changes by 2 m/s. Wind force is constant, thus acceleration is constant.

Acceleration You are driving in a car. What can you do to cause the car to accelerate? Speed it up (step on gas) Slow it down (step on brake) Change the direction (turn the wheel)

The Signs of Acceleration Acceleration is positive (+) or negative (-) based on the direction of force. + Choose + direction first. Then acceleration a will have the same sign as that of the force F —regardless of the direction of velocity. F a (+) a(-) F

Negative. Acceleration vs. Deceleration There is a difference between negative acceleration and deceleration: Negative acceleration is acceleration in the negative direction as defined by the coordinate system. Deceleration occurs when the acceleration is opposite in direction to the velocity.

Acceleration To calculate acceleration, we use the following formula: This can also be written as

Example Exercises from Notes Accelerating Plane An acceleration of +4.35 m/s2 can also be written as +4.35 m/s/s or +4.35 m s-2. This means that every second, the velocity changes by +4.35 m/s

Acceleration Which one of these accelerations is the largest? 60 mi/hr-s 60 mi/hr-min 60 mi/hr-hr

Ex. Exercises from Notes, cont. Car Slowing Down Notice that in computing Δv, you always subtract final from initial: v-v0 Question: An object moving at -25 m/s with an acceleration of +5 m/s/s. What is its velocity after 7 s?

+ Force t = 4 s v1 = +8 m/s v2 = +20 m/s Example 3 (No change in direction): A constant force changes the speed of a car from 8 m/s to 20 m/s in 4s. What is the average acceleration? + Force t = 4 s v1 = +8 m/s v2 = +20 m/s Step 1. Draw a rough sketch. Step 2. Choose a positive direction (right). Step 3. Label given info with + and - signs. Step 4. Indicate direction of force F.

Example 3 (Continued): What is average acceleration of car? + v1 = +8 m/s t = 4 s v2 = +20 m/s Force Step 5. Recall definition of average acceleration.

+ E Force vf = -5 m/s vo = +20 m/s Example 4: A wagon moving east at 20 m/s encounters a very strong head-wind, causing it to change directions. After 5 s, it is traveling west at 5 m/s. What is the average acceleration? (Be careful of signs.) + Force E vf = -5 m/s vo = +20 m/s Step 1. Draw a rough sketch. Step 2. Choose the eastward direction as positive. Step 3. Label given info with + and - signs.

Choose the eastward direction as positive. Example 4 (Cont.): Wagon moving east at 20 m/s encounters a head-wind, causing it to change directions. Five seconds later, it is traveling west at 5 m/s. What is the average acceleration? Choose the eastward direction as positive. Initial velocity, vo = +20 m/s, east (+) Final velocity, vf = -5 m/s, west (-) The change in velocity, Dv = vf - v0 Dv = (-5 m/s) - (+20 m/s) = -25 m/s

+ Example 4: (Continued) E Force vf = -5 m/s aavg = = Dv Dt vf - vo vo = +20 m/s vf = -5 m/s E Dv = (-5 m/s) - (+20 m/s) = -25 m/s aavg = = Dv Dt vf - vo tf - to a = -25 m/s 5 s Acceleration is directed to left, west (same as F). a = - 5 m/s2

+ Signs for Displacement E Force C D B A vf = -25 m/s vo = +20 m/s a = - 5 m/s2 A B C D Time t = 0 at point A. What are the signs (+ or -) of displacement moving from? A to B B to C C to D

+ Signs for Velocity E Force vo = +20 m/s vf = -25 m/s a = - 5 m/s2 A B C D x = 0 What are the signs (+ or -) of velocity at points B, C, and D? At B, v is zero - no sign needed. At C, v is positive on way out and negative on the way back. At D, v is negative, moving to left.

Signs for Acceleration + Force vo = +20 m/s vf = -25 m/s a = - 5 m/s2 A B C D What are the signs (+ or -) of acceleration at points B, C, and D? At B, C, and D, a = -5 m/s, negative at all points. The force is constant and always directed to left, so acceleration does not change.