Chapter 2 Kinematics in One Dimension

Chapter 2 Kinematics in One Dimension

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

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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.

Describing Motion Mechanics is the study of motion of objects and the related concepts of force and energy. Kinematics is the description of how objects move. Dynamics deals with force and why objects move. Objects that move without rotating are in translational motion.

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.

Distance and Displacement Distance is the length of the actual path taken by an object. Any measurement of distance, position or speed must be made with respect to a frame of reference. Consider travel from point A to point B in diagram below: A s = 20 m B Distance s is a scalar quantity (no direction): Contains magnitude only and consists of a number and a unit. (20 m, 40 mi/h, 10 gal)

Distance and Displacement Displacement (Δx) is the straight-line separation of two points in a specified direction. Δx = x x 0 D = 12 m, 20 o A θ B A vector quantity: Contains magnitude AND direction, a number, unit & angle. (12 m, 30 0 ; 8 km/h, N)

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 = 4 m, W What is the distance traveled? 20 m!! D x = -4 8 m,e x x = +8 12 m,w

The Signs of Displacement Displacement is positive (+) or negative (-) based on LOCATION. Examples: The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION. 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 s = 20 m B s v = = t 20 m v = 5 m/s 4 s Time t = 4 s Not direction dependent!

Definition of Velocity Velocity is the displacement per unit of time. (A vector quantity.) s = 20 m A D=12 m 20 o Time t = 4 s B v D = = t 12 m 4 s v = 3 m/s at 20 0 N of E Direction required!

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: s 2 = 300 m start s 1 = 200 m Total distance: s = 200 m + 300 m = 500 m total path 500 m Average speed = = Avg. speed time 60 s 8.33 m/s Direction does not matter!

Example 1 (Cont.) Now we find the average velocity, which is the net displacement divided by time. In this case, the direction matters. v = x f t x x 0 = 0 m; x f = -100 m v 0 x f = -100 m 100 m 0 = = 1.67 m/s 60 s Average velocity: x o = 0 t = 60 s x 1 = +200 m Direction of final displacement is to the left as shown. v = 1.67 m/s, West Note: Average velocity is directed to the west.

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? Total distance/ total time: v v = x + x ta + tb 1000 m 164 s A B = = 600 m + 400 m 14 s + 150 s v = 6.10 m/s A B 14 s 625 m Average speed is a function only of total distance traveled and the total time required. 142 s 356 m

Examples of Speed Orbit Light = 3 x 10 8 m/s 2 x 10 4 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

End Section 2.1

Average Speed and Instantaneous Velocity The average speed depends ONLY on the distance traveled and the time required. s = 20 m C A Time t = 4 s B The instantaneous velocity is the magnitude and direction of the speed at a particular instant. (v at point 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.

Average and Instantaneous v Average Velocity: v avg x x x = = t t t 2 1 2 1 Instantaneous Velocity: v inst x = ( t 0) t x 2 x 1 t x Displacement, x slope t x t 1 t 2 Time

End Sections 2.2 and 2.3

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 acceleration is same as direction of force. The acceleration is proportional to the magnitude of the force.

Example of Acceleration + t = 3 s Force v 0 = +2 m/s v f = +8 m/s 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.

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

Average and Instantaneous a a avg v v v = = t t t 2 1 2 1 a inst v = ( t 0) t slope v 2 v 1 t v t v t 1 t 2 time

Example 3 (No change in direction): A constant force changes the speed of a car from 8 m/s to 20 m/s in 4 s. What is average acceleration? + Force t = 4 s v 1 = +8 m/s v 2 = +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? + v 1 = +8 m/s t = 4 s Force v 2 = +20 m/s Step 5. Recall definition of average acceleration. a avg v v v = = t t t 2 1 2 1 20 m/s - 8 m/s a = = + 3 m/s 4 s a = +3 m/s, rightward

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 v f = -5 m/s v o = +20 m/s E Step 1. Draw a rough sketch. Step 2. Choose the eastward direction as positive. Step 3. Label given info with + and - signs.

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, v o = +20 m/s, east (+) Final velocity, v f = -5 m/s, west (-) The change in velocity, v = v f - v 0 v = (-5 m/s) - (+20 m/s) = -25 m/s

Example 4: (Continued) + Force v f = -5 m/s v o = +20 m/s E v = (-5 m/s) - (+20 m/s) = -25 m/s v a avg = = t v f - v o t f - t o a = -25 m/s 5 s a = - 5 m/s 2 Acceleration is directed to left, west (same as F).

Signs for Displacement + C Force D E A B v f = -5 m/s v o = +20 m/s a = - 5 m/s 2 Time t = 0 at point A. What are the signs (+ or -) of displacement at B, C, and D? At B, x is positive, right of origin At C, x is positive, right of origin At D, x is negative, left of origin

Signs for Velocity + x = 0 C Force D E A B v f = -5 m/s v o = +20 m/s a = - 5 m/s 2 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 + C Force D E A B v f = -5 m/s v o = +20 m/s a = - 5 m/s 2 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.

Definitions Average velocity: v avg x x x = = t t t 2 1 2 1 Average acceleration: a avg v v v = = t t t 2 1 2 1

Velocity for constant a Average velocity: Average velocity: v avg x xf x = = t t t f 0 0 v avg = v 0 + 2 v f Setting t o = 0 and combining we have: v + v x= x + 0 f t 0 2

Constant Acceleration Acceleration: a avg v vf v = = t t t f 0 0 Setting t o = 0 and solving for v, we have: v = v + at f 0 Final velocity = initial velocity + change in velocity

Formulas based on definitions: v + v x x 0 f t 0 = + v = v + at f 0 2 Derived formulas: 1 0 0 2 x = x + v t + at 2 1 0 f 2 x = x + v t at 2 2 2 0 0 2 ax ( x) = v v f For constant acceleration only

Use of Initial Position x 0 in Problems. 0 v + v x= x + 0 f t 0 2 1 0 0 2 x = x + v t + at 1 0 f 2 x = x + v t at 2 2 2 2 0 0 2 ax ( x) = v v f v = v + at f 0 0 0 0 If you choose the origin of your x,y axes at the point of the initial position, you can set x 0 = 0, simplifying these equations. The x o term is very useful for studying problems involving motion of two bodies.

Review of Symbols and Units Displacement (x, x o ); meters (m) Velocity (v, v o ); meters per second (m/s) Acceleration (a); meters per s 2 (m/s 2 ) Time (t); seconds (s) Review sign convention for each symbol

The Signs of Displacement Displacement is positive (+) or negative (-) based on LOCATION. 2 m -1 m -2 m The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION.

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

Acceleration Produced by Force Acceleration is (+) or (-) based on direction of force (NOT based on v). F a(-) A push or pull (force) is necessary to change velocity, thus the sign of a is same as sign of F. F a(+) More will be said later on the relationship between F and a.

Example 6: A airplane flying initially at 400 ft/s lands on a carrier deck and stops in a distance of 300 ft. What is the acceleration? v = 0 300 ft + F +400 ft/s v o X 0 = 0 Step 1. Draw and label sketch. Step 2. Indicate + direction and F direction.

Example: (Cont.) +400 ft/s v = 0 300 ft v o + F X 0 = 0 Step 3. List given; find information with signs. List t =?, even though time was not asked for. Given: v o = +400 ft/s v = 0 x = +300 ft Find: a =?; t =?

Continued... v = 0 x 300 ft + F Step 4. Select equation that contains a and not t. -v 2 o a = = 2x -(400 ft/s) 2 2(300 ft) +400 ft/s v o X 0 = 0 0 0 2a(x -x o ) = v 2 - v o 2 Initial position and final velocity are zero. a = - 267 ft/s 2 Because Why is Force the acceleration is in a negative negative? direction!

End of Sections 2.4, 2.5, and 2.6

Acceleration Due to Gravity Every object on the earth experiences a common force: the force due to gravity. This force is always directed toward the center of the earth (downward). g W The acceleration due to gravity is relatively constant near the Earth s surface. Earth

Gravitational Acceleration In a vacuum, all objects fall with same acceleration. Equations for constant acceleration apply as usual. Near the Earth s surface: a = g = 9.80 m/s 2 or 32 ft/s 2 Directed downward (usually negative).

Experimental Determination of Gravitational Acceleration. t The apparatus consists of a device which measures the time required for a ball to fall a given distance. y Suppose the height is 1.20 m and the drop time is recorded as 0.650 s. What is the acceleration due to gravity?

Experimental Determination of Gravity (y 0 = 0; y = -1.20 m) y = -1.20 m; t = 0.495 s 1 2 y = v0t + 2 at ; v0 = 0 a 2y 2( 1.20 m) = = 2 2 t (0.495 s) Acceleration of Gravity: 2 a = 9.79 m/s Acceleration a is negative because force W is negative. y W t +

a = - y v = + 0 Sign Convention: A Ball Thrown Vertically Upward a = - v y = + UP = + Release Point a = - y v = + - a = - v y = - 0 y v= Negative a = - Tippens Displacement is positive (+) or negative (-) based on LOCATION. Velocity is positive (+) or negative (-) based on direction of motion. Acceleration is (+) or (-) based on direction of force (weight).

Example 7: A ball is thrown vertically upward with an initial velocity of 30 m/s. What are its position and velocity after 2 s, 4 s, and 7 s? Step 1. Draw and label a sketch. Step 2. Indicate + direction and force direction. Step 3. Given/find info. a = g + a = -9.8 ft/s 2 t = 2, 4, 7 s v o = + 30 m/s y =? v =? v o = +30 m/s

Finding Displacement: Step 4. Select equation that contains y and not v. 0 1 0 0 2 y = y + v t + at y = (30 m/s)t + ½(-9.8 m/s 2 )t 2 Substitution of t = 2, 4, and 7 s will give the following values: 2 a = g v o = 30 m/s + y = 40.4 m; y = 41.6 m; y = -30.1 m

Finding Velocity: Step 5. Find v from equation that contains v and not x: v = v + at f 0 a = g + v f = 30 m/s + ( 9.8 m/s ) t Substitute t = 2, 4, and 7 s: 2 v o = 30 m/s v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s

Example 7: (Cont.) Now find the maximum height attained: Displacement is a maximum when the velocity v f is zero. v f = + = 2 30 m/s ( 9.8 m/s ) t 0 a = g + t = 30 m/s ; t = 3.06 s 2 9.8 m/s To find y max we substitute t = 3.06 s into the general equation for displacement. v o = +96 ft/s y = (30 m/s)t + ½(-9.8 m/s 2 )t 2

Example 7: (Cont.) Finding the maximum height: y = (30 m/s)t + ½(-9.8 m/s 2 )t 2 t = 3.06 s a = g + Omitting units, we obtain: y = (30)(3.06) + ( 9.8)(3.06) 1 2 y = 91.8 m - 45.9 m 2 v o =+30 m/s y max = 45.9 m

Summary of Formulas v + v x x 0 f t 0 = + v = v + at f 0 2 Derived Formulas: 1 0 0 2 x = x + v t + at 2 1 0 f 2 x = x + v t at 2 2 2 0 0 2 ax ( x) = v v f For Constant Acceleration Only

End of Sections 2.7 and 2.8