Days 5 - 6 UNIT 1 Motion Graphs x t Lyzinski Physics

Day #5. Acceleration on x-t graphs. v-t graphs Day #5 * Acceleration on x-t graphs * v-t graphs * using v-t graphs to get a

x-t ‘s x t UNIFORM Velocity Speed increases as slope increases x t Object at REST x t Moving forward or backward x t x-t ‘s Object Positively Accelerating x t x t Changing Direction x t Object Speeding up Object Negatively Accelerating

POSITIVE OR NEGATIVE ACCEL? SPEEDING UP OR SLOWING DOWN? x t x t Slope of the tangent gives vinst Getting more sloped  speeding up Getting more + sloped  + Accel Getting less sloped  slowing down Slopes are getting less +  - Accel x t x t Getting less sloped  slowing down Slopes are getting less –  + Accel Getting more sloped  speeding up Slopes are getting more –  - Accel

An easy way to remember it  I’m Positive!!! I’m Negative!!!

Find the acceleration in each case. v1 = 10 m/s, v2 = 20 m/s, Dt = 5sec v1 = 10 m/s, v2 = -20 m/s, Dt = 10sec v1 = -9 km/h, v2 = -27 km/h, Dt = 3 h v1 = -9 km/h, v2 = 6 km/h, Dt = 3 h

v-t ‘s UNIFORM Positive (+) Acceleration Acceleration increases as slope increases v t v t Changing Direction v-t ‘s UNIFORM Velocity (no acceleration) v t v t Object at REST UNIFORM Negative (-) Acceleration

v-t graphs v(m/s) Constant + accel (slowing down) At rest t (s) 2 4 6 8 10 12 t (s) 8 6 4 2 -2 -4 At rest Constant + accel (slowing down) Constant - Vel Constant negative accel (speeding up) Constant negative accel (slowing down) Constant + Vel (constant speed) Constant + accel (speeding up)

How to get the velocity (v) at a certain time off a v-t graph v(m/s) 2 4 6 8 10 12 t (s) 8 6 4 2 -2 -4 Example: What is the velocity at t = 8 seconds? Go over to t = 8. Find the pt on the graph. -2 m/s Find the v value for this time.

Finding the average acceleration on a v-t graph v(m/s) 2 4 6 8 10 12 8 6 4 2 -2 -4 Example: What is the average acceleration between 0 & 2, 2 & 4, and 4 & 10 seconds? a0-2 = (v2 – v1) / Dt = rise / run = +4/2 = +2 m/s2 A2-4 = (v2 – v1) / Dt = rise / run = 0 m/s2 A4-10 = (v2 – v1) / Dt = rise / run = -7 / 6 = -1.17 m/s2

v-t graphs Slope of any segment is the AVERAGE acceleration v (m/s) The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration t (sec) t0 t1

Open to in your Unit 1 packet 3 30 20 10 -10 -20 -30 v = -30 m/s s = 30 m/s a = slope = (-30 m/s) / 16sec = -1.875 m/s2 a = slope = (+57 m/s) / 32sec = +1.78 m/s2 (approx)

Open to in your Unit 1 packet 3 30 20 10 -10 -20 -30 3) 4) 5) You can’t say. You know its speed at the start, but not where it is  Object is at rest whenever it crosses the t-axis  t = 0, 36, 80 sec Const – accel (object speeds up), const – vel, const + accel (slows down), const + accel (speeds up), const – accel (slows down)

Day #6. v-t graphs. slopes & areas of v-t graphs Day #6 * v-t graphs * slopes & areas of v-t graphs * instantaneous accelerations

x-t ‘s x t UNIFORM Velocity Speed increases as slope increases x t Object at REST x t Moving forward or backward x t x-t ‘s Object Positively Accelerating x t x t Changing Direction x t Object Speeding up Object Negatively Accelerating

v-t ‘s UNIFORM Positive (+) Acceleration Acceleration increases as slope increases v t v t Changing Direction v-t ‘s UNIFORM Velocity (no acceleration) v t v t Object at REST UNIFORM Negative (-) Acceleration

A Quick Review The slope between 2 points on an x-t graph gets you the _______________. The slope at a single point (the slope of the tangent to the curve) on an x-t graph gets you the ____________. The slope between 2 points on a v-t graph gets you the ____________. The slope at a single point (the slope of the tangent to the curve) on a v-t graph gets you the ____________. Average velocity Inst. velocity Avg. accel. Inst. accel.

NEW CONCEPT When you find the area “under the curve” on a v-t graph, this gets you the displacement during the given time interval. v t v t The “area under the curve” is really the area between the graph and the t-axis. This is NOT the area under the curve 

Find the area under the curve from …. 0-4 seconds. 4-6 6-10 0-10 The displacement during the first 4 seconds is -20m A = ½ (4)(-10) = -20m A = ½ (2)(-10) = -10m The displacement during the next 2 seconds is -10m A = ½ (4)(15) = 30m The displacement during the next 4 seconds is 30m v t 15 -10 A = -20 + (-10) + 30 = 0m 4 6 10 The OVERALL displacement from 0 to 10 seconds is zero (its back to where it started)

How to find the displacement from one time to another from a v-t graph v(m/s) 2 4 6 t (s) 8 6 4 2 -2 -4 Example: What is the displacement from t = 2 to t = 10? Find the positive area bounded by the “curve” 12 m + 4 m = 16 m 8 10 12 Find the negative area bounded by the “curve” (-2.25 m) + (-4.5 m) = - 6.75 m Add the positive and negative areas together  Dx = 16 m + (-6.75 m) = 9.25 m

How to find the distance traveled from one time to another from a v-t graph v(m/s) t (s) 8 6 4 2 -2 -4 Example: What is the distance traveled from t = 2 to t = 10? Find the positive area bounded by the “curve” 12 m + 4 m = 16 m 8 10 12 Find the negative area bounded by the “curve” (-2.25 m) + (-4.5 m) = - 6.75 m 2 4 6 Add the MAGNITUDES of these two areas together  distance = 16 m + 6.75 m = 22.75 m

How to find the average velocity during a time interval on a v-t graph v(m/s) t (s) 8 6 4 2 -2 -4 Example: What is the average velocity from t = 2 to t = 10? The DISPLACEMENT is simply the area “under” the curve. Dx = 16 m + (-6.75 m) = 9.25 m 12 m + 4 m = 16 m (-2.25 m) + (-4.5 m) = - 6.75 8 10 12 2 4 6 The AVG velocity = Dx / Dt = 9.25 m / 8 s = 1.22 m/s

How to find the final position of an object using a v-t graph (and being given the initial position) v(m/s) t (s) 8 6 4 2 -2 -4 Example: What is the final position after t = 10 seconds if xi = 40 m? The DISPLACEMENT during the 1st 10 sec is simply the area “under” the curve. Dx = 20 m + (-6.75 m) = 13.25 m 4 m 12 m + 4 m = 20 m (-2.25 m) + (-4.5 m) = - 6.75 8 10 12 2 4 6 Dx = x2 – x1  x2 = Dx + x1 = 13.25 m + 40 m = 53.25 m

v-t graphs Slope of any segment is the AVERAGE acceleration v (m/s) The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration The area under the curve between any two times is the CHANGE in position (the displacement) during that time period. t (sec) t0 t1

Open to in your Unit 1 packet 3 30 20 10 -10 -20 -30 6) 7) Distance travelled = |area| = | ½ (16)(-30) | = 240m s = d/t = 240 m / 16 sec = 15 m/s Displacement = |area| = ½ (16)(-30) + 8 (-30) = -480m v = Dx/t = -480 m / 24 sec = -20 m/s

Open to in your Unit 1 packet 3 30 20 10 -10 -20 -30 8) Find all the areas “under the curve” from 0 to 44 sec Area = ½ (16)(-30) + 12(-30) + ½ (8)(-30) + ½ (8)(30) = - 600 m Area = Dx = - 600 m  Dx = x2 – x1  -600m = x2 – (-16m)  x2 = - 616m

Open to in your Unit 1 packet 4 +3.3 m/s2 +10 m/s 0 m/s +75 m

10 -10 5 -2 m/s2 0 m 50 m 30 m

14 & 34 sec +.35 m/s2 + 8 m/s + 2 m/s2 approx 0.8 m/s2 9) 10) 11) 12) 13) 14 & 34 sec +.35 m/s2 + 8 m/s + 2 m/s2 approx 0.8 m/s2