Kinematics Kinematics Describing Motion

Reference Frames Measurements of position, distance or speed must be with respect to a frame of reference. Measurements of position, distance or speed must be with respect to a frame of reference. What is the speed of a person with respect to the ground if she walks toward the back of the train at 5km/h while the train moves forward at 40 km/h? Coordinate axes are used to represent the frame of reference

Displacement Defined as change in position Defined as change in position What is the displacement of a person who walks 100 m East then 60 m West? Answer: 40 m East Displacement has both magnitude and direction – it is a vector Displacement has both magnitude and direction – it is a vector

Representing Displacement Let x 1 be position of object at time t 1 Let x 1 be position of object at time t 1 Let x 2 be position at time t 2 Let x 2 be position at time t 2 Then displacement  x = x 2 – x 1 Then displacement  x = x 2 – x 1  Greek letter delta) means change  Greek letter delta) means change If a person starts at x 1 = 40m and walks to the left until reaching x 2 = 10m what is the displacement? Answer:  x = x 2 – x 1 = 10m – 40m = -30m

Average Speed and Velocity Velocity is speed and direction Velocity is speed and direction Average speed = distance traveled ÷ time elapsed Average speed = distance traveled ÷ time elapsed Average velocity = displacement ÷ time elapsed Average velocity = displacement ÷ time elapsed Not always equal Not always equal Find average speed and velocity for a trip 60m North followed by 40m South in 10 seconds Answers: 10 m/s; 2 m/s N

Example During a four second interval a runner’s position changes from x 1 = 50m to x 2 = 10m. What was the average velocity? During a four second interval a runner’s position changes from x 1 = 50m to x 2 = 10m. What was the average velocity? v av =  x/  t = (x 2 – x 1 )/4s = -40m/4s = -10 m/s v av =  x/  t = (x 2 – x 1 )/4s = -40m/4s = -10 m/s What is the average speed? (always positive!) What is the average speed? (always positive!) Challenge: convert this velocity to kilometers per hour -10 m/s x 1 km/1000m x 3600 sec/ 1h = -36 km/h Pro tip: use conversion tables on inside front cover of your text

How Far? How far could a runner traveling at an average speed of 36 km/h go in 20. minutes? How far could a runner traveling at an average speed of 36 km/h go in 20. minutes? 20 min = 1/3 hour 20 min = 1/3 hour D x = v av  t = 36 km/h x 1/3h = 12 km = 1.2 x 10 4 m D x = v av  t = 36 km/h x 1/3h = 12 km = 1.2 x 10 4 m

Three Ways D = s x T D = s x T S = D/T S = D/T T = D/s T = D/s

Instantaneous Velocity Velocity at a particular instant of time Velocity at a particular instant of time Defined as average velocity over an infinitesimally short time interval Defined as average velocity over an infinitesimally short time interval v = lim(as  t --> 0)  x/  t v = lim(as  t --> 0)  x/  t Finite because both numerator and denominator approach zero; limit approaches a definite value Finite because both numerator and denominator approach zero; limit approaches a definite value

Acceleration Acceleration is how fast velocity changes Acceleration is how fast velocity changes Average acceleration = change of velocity ÷ time elapsed Average acceleration = change of velocity ÷ time elapsed a av = (v 2 – v 1 )/(t 2 –t 1 ) =  v/  t a av = (v 2 – v 1 )/(t 2 –t 1 ) =  v/  t

Example Find average acceleration of a car that accelerates along straight road from rest to 80 km/h in 5 seconds Find average acceleration of a car that accelerates along straight road from rest to 80 km/h in 5 seconds a av = (80 km/h – 0 km/h)/5s = 16 km/h/s a av = (80 km/h – 0 km/h)/5s = 16 km/h/s

Convert to m/s/s 80 km/h(1000m/1km)(1h/3600s) = 22.2 m/s 80 km/h(1000m/1km)(1h/3600s) = 22.2 m/s a av = (22.2 m/s – 0.0 m/s)/5.0 s = 4.4 m/s/s or 4. 4 m/s 2 a av = (22.2 m/s – 0.0 m/s)/5.0 s = 4.4 m/s/s or 4. 4 m/s 2 Pronounced “meters per second squared” Pronounced “meters per second squared” Challenge: Can an object with zero velocity have non zero acceleration? Challenge: Can an object with zero velocity have non zero acceleration?

Object Slowing Down Called deceleration Called deceleration Find average acceleration of a car moving to the right(+x direction) 15.0 m/s when driver brakes to 5.0 m/s in 5.0 s? Find average acceleration of a car moving to the right(+x direction) 15.0 m/s when driver brakes to 5.0 m/s in 5.0 s? a av =  v/  t = (5.0 m/s – 15.0 m/s) ÷ 5.0 s = a av =  v/  t = (5.0 m/s – 15.0 m/s) ÷ 5.0 s = -2.0 m/s 2 (acceleration negative) -2.0 m/s 2 (acceleration negative) Would acceleration still be negative if car was moving to the left? Would acceleration still be negative if car was moving to the left? NO! Its acceleration vector would then point to the right and be positive. NO! Its acceleration vector would then point to the right and be positive.

Constant(Uniform) Acceleration Let t 1 = 0, t 2 = t = elapsed time Let t 1 = 0, t 2 = t = elapsed time Re-name x 1 = x 0 ; v 1 = v 0 Re-name x 1 = x 0 ; v 1 = v 0 Average velocity v av = (x – x 0 )/t Average velocity v av = (x – x 0 )/t a av = (v – v 0 )/t a av = (v – v 0 )/t From these it is possible to derive (next slide) From these it is possible to derive (next slide)

For Uniform Acceleration, a v = v 0 +at(a) v = v 0 +at(a) x = x 0 +v 0 t + 1/2at 2 (b) x = x 0 +v 0 t + 1/2at 2 (b) v 2 = v a(x-x 0 ) (c) v 2 = v a(x-x 0 ) (c) v av = (v 0 + v)/2(d) v av = (v 0 + v)/2(d) Each equation can be solved for any of the variables Each equation can be solved for any of the variables Problems can be solved more than one way Problems can be solved more than one way

v = v 0 +at (a) Describes change in velocity under uniform acceleration. tells how fast a particle will be going at time t if at time zero its velocity was v 0

x = x 0 +v 0 t + 1/2at 2 (b) Describes change in position under uniform acceleration Sometimes called "equation of motion." tells where a particle will be at time t, if at time zero it was at x 0 moving with velocity v 0

v 2 = v a(x-x 0 ) (c) Velocity equation Velocity equation Tells what velocity will be after a particle with initial velocity v 0 accelerates a distance x - x 0 with uniform acceleration a Tells what velocity will be after a particle with initial velocity v 0 accelerates a distance x - x 0 with uniform acceleration a Take square root to find v Take square root to find v Simplifies to v 2 = 2ad when v 0 = 0 and Simplifies to v 2 = 2ad when v 0 = 0 and d = displacement d = displacement

v av = (v 0 + v)/2 (d) Average (mean) velocity of particle for a trip under uniform acceleration Average (mean) velocity of particle for a trip under uniform acceleration Provides shortcut way to solve certain problems Provides shortcut way to solve certain problems

Special Case of Free Fall* v = v 0 –gt a = -g v = v 0 –gt a = -g y = y 0 +v 0 t - 1/2gt 2 y = y 0 +v 0 t - 1/2gt 2 v 2 = v g(y-y 0 ) v 2 = v g(y-y 0 ) With y up positive, g = 9.80 m/s/s With y up positive, g = 9.80 m/s/s *Free Fall is motion under the influence of gravity alone This form of equations assumes down is positive

Concept Check (1) The velocity and acceleration of an object The velocity and acceleration of an object –(a) must be in the same direction –(b) must be in opposite directions –(c) can be in the same or opposite directions –(d) must be in the same direction or zero (c) Can be in the same or opposite directions. Example: a rock thrown upward

Concept Check(2) At the top of its path the velocity and acceleration of a brick thrown upward are At the top of its path the velocity and acceleration of a brick thrown upward are (a) both non zero (a) both non zero (b) both zero (c) velocity is zero acceleration is non zero (c) velocity is zero acceleration is non zero (d) acceleration is zero, velocity is non zero (d) acceleration is zero, velocity is non zero (c) is correct; v = 0, a = 9.80 m/s/s downward

Problem Solving Tips Read and re-read the problem Read and re-read the problem Make a diagram with all given info Make a diagram with all given info Ask yourself “what is problem asking?” Ask yourself “what is problem asking?” Ask which physics principles apply Ask which physics principles apply Look for most applicable equations Look for most applicable equations Be sure problem lies within their range of validity Be sure problem lies within their range of validity

Special Tricks Break problem up into parts, Break problem up into parts, like up and down part of path of an object thrown up like up and down part of path of an object thrown up Use symmetry Use symmetry Take situation to an extreme and look for a constraint Take situation to an extreme and look for a constraint Choose reference frame that makes problem easiest Choose reference frame that makes problem easiest

Problem solving… Do algebraic calculations Do algebraic calculations Be aware you may have to solve equations simultaneously Be aware you may have to solve equations simultaneously Do arithmetic at end Do arithmetic at end Check units and significant figures Check units and significant figures Ask, “is answer reasonable?” Ask, “is answer reasonable?”

Your Choice You may choose up positive or negative; same with left or right You may choose up positive or negative; same with left or right You may put the zero of coordinates anywhere you choose You may put the zero of coordinates anywhere you choose Generally make choices that minimize number of negative quantities Generally make choices that minimize number of negative quantities

Examples What time is required for a car to travel 30.0 m while accelerating from rest at a uniform 2.00 m/s 2 What time is required for a car to travel 30.0 m while accelerating from rest at a uniform 2.00 m/s 2 x = ½ a t 2 x = ½ a t 2 t 2 = 2x/a t 2 = 2x/a t = (2x/a) 1/2 t = (2x/a) 1/2 t = (2(30m)/2.00 m/s 2 ) = 5.48s t = (2(30m)/2.00 m/s 2 ) = 5.48s a = 2.00m/s 2 x 0 = 0 v 0 = 0 x = 30m

Distance to Brake A car traveling 28 m/s brakes at -6.0 m/s 2. What distance is required to stop. A car traveling 28 m/s brakes at -6.0 m/s 2. What distance is required to stop. Method 1. Use v 2 = v a(x-x 0 ) Method 1. Use v 2 = v a(x-x 0 ) Solve for x = x 0 + (v 2 – v 0 2 ) ÷ 2a Solve for x = x 0 + (v 2 – v 0 2 ) ÷ 2a x = (0 – 28 m/s 2 ) ÷ 2(-6.0 m/s 2 ) = 65m x = (0 – 28 m/s 2 ) ÷ 2(-6.0 m/s 2 ) = 65m Method 2. Use v av = (v 0 + v)/2 = 14 m/s Method 2. Use v av = (v 0 + v)/2 = 14 m/s v = v 0 +at ; t = (v – v 0 )/a = -28m/s ÷ -6.0 m/s 2 = 4.67 s v = v 0 +at ; t = (v – v 0 )/a = -28m/s ÷ -6.0 m/s 2 = 4.67 s x = v av t = 14 m/s (4.67s) = 65 m x = v av t = 14 m/s (4.67s) = 65 m

Moral of Story There is always more than one way to solve a (kinematics) problem! There is always more than one way to solve a (kinematics) problem! You do not have to use the teacher’s way or the textbook way so long as the method and answer are correct. You do not have to use the teacher’s way or the textbook way so long as the method and answer are correct. Warning: Sometimes it is possible to get a correct answer using an incorrect method – no credit Warning: Sometimes it is possible to get a correct answer using an incorrect method – no credit

Free Fall Which falls faster, an elephant or a mouse? Which falls faster, an elephant or a mouse?

Galileo’s Experiment He asked, which would reach the ground first, a marble or a cannonball? He asked, which would reach the ground first, a marble or a cannonball? Courtesy Dan Heller Photography

Galileo’s Discovery All objects accelerate to earth equally (regardless of mass) All objects accelerate to earth equally (regardless of mass) Air resistance must be neglected for this to be true Air resistance must be neglected for this to be true Acceleration due to gravity a = Acceleration due to gravity a = g = 9.80 m/s 2 g = 9.80 m/s 2

Velocity Reached In free fall an object’s speed increases by about 10 m/s in each second. In free fall an object’s speed increases by about 10 m/s in each second. v = gt ~ 10t v = gt ~ 10t g ~ 10 m/s 2 g ~ 10 m/s 2 Time to fall(s) Speed (m/s)

Distance Fallen From Rest Distance fallen in t seconds Distance fallen in t seconds d = 1/2gt 2 d = 1/2gt 2 timedistance

Example: Ball Thrown Upward Person throws ball upward with v = 15.0 m/s. (a) How high does it go? (b) How long is it in the air? Person throws ball upward with v = 15.0 m/s. (a) How high does it go? (b) How long is it in the air? Choose y positive up, negative down Choose y positive up, negative down Choose y = 0 at throw height Choose y = 0 at throw height Use v 2 = v ay Use v 2 = v ay y = (v 2 – v 0 2 )/2a = (0 – (15.0m/s) 2 )/2(- 9.80m/s 2) = 11.5 m y = (v 2 – v 0 2 )/2a = (0 – (15.0m/s) 2 )/2(- 9.80m/s 2) = 11.5 m

Time In The Air Use y = v 0 t + ½ a t 2 Use y = v 0 t + ½ a t 2 0 = (15.0 m/s)t + ½(-9.80 m/s 2 )t 2 0 = (15.0 m/s)t + ½(-9.80 m/s 2 )t 2 Factor: (15.0 m/s – 4.90 m/s 2 t)t = 0 Factor: (15.0 m/s – 4.90 m/s 2 t)t = 0 Two solutions: t =0 corresponds to instant ball thrown Two solutions: t =0 corresponds to instant ball thrown T = 15.0 m/s ÷ 4.90 m/s 2 = 3.06 s corresponds to instant ball returns to ground T = 15.0 m/s ÷ 4.90 m/s 2 = 3.06 s corresponds to instant ball returns to ground

Another Way? Can you think of another way? Can you think of another way? Find time to rise from v = v 0 +at Find time to rise from v = v 0 +at t = (v – v 0 )/a = (15m/s – 0)/9.80 m/s = 1.53s t = (v – v 0 )/a = (15m/s – 0)/9.80 m/s = 1.53s Then use average velocity 7.5 m/s x 1.53s = 11.5 m to find height. Then use average velocity 7.5 m/s x 1.53s = 11.5 m to find height. Total time in air is double 1.53s by symmetry. Gravity gives back speed on the way down at the same rate it took it away on the way up.

Puzzlers What is the velocity of a ball thrown upward at its maximum height? What is the velocity of a ball thrown upward at its maximum height? What is its acceleration at that height? What is its acceleration at that height? Are velocity and acceleration always in the same direction Are velocity and acceleration always in the same direction

Useful Form of Kinematics Equations for Free Fall v = v 0 - gt(a) v = v 0 - gt(a) y = y 0 +v 0 t - 1/2gt 2 (b) y = y 0 +v 0 t - 1/2gt 2 (b) v 2 = v g(y-y 0 ) (c) v 2 = v g(y-y 0 ) (c) v av = (v 0 + v)/2(d) v av = (v 0 + v)/2(d) Here assuming up is positive Here assuming up is positive g = 9.80 m/s 2 g = 9.80 m/s 2

Graphing Motion Position Graph (x or y vs. t) Position Graph (x or y vs. t) –Slope is velocity –If curved slope defined as slope of tangent to the curve at that point Velocity Graph Velocity Graph –Slope is acceleration –Area under graph is displacement Acceleration Graph Acceleration Graph –Constant and non zero for uniform acceleration

Concept Check (3) Which of these could be the velocity graph of a rock thrown upward, then falling downward? (Assume up is positive) Which of these could be the velocity graph of a rock thrown upward, then falling downward? (Assume up is positive) (a)(b) (c) (a)(b) (c) (d)(e)