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Motion with Constant Acceleration

Motion with Constant Acceleration. Reminder!. I think that, logically, the topics in Chs. 2 & 3 should be covered in a slightly different order than your book does them. So, I will not cover them in the same order as the book. Further, I will treat Chs. 2 & 3 as if they were only one chapter!.

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Motion with Constant Acceleration

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  1. Motion with Constant Acceleration

  2. Reminder! • I think that, logically, the topics in Chs. 2 & 3 should be covered in a slightly different order than your book does them. • So, I will not cover them in the same order as the book. • Further, I will treat Chs. 2 & 3 as if they were only one chapter!

  3. Constant Acceleration • In many practical situations: • The magnitude of the acceleration is uniform(constant) • The motion is in a straight line • It is useful to derive some equations which apply in this special case ONLY!!! • The kinematic equations for constant (uniform) acceleration in one dimension.

  4. Constant Acceleration • The derivation is in the text. It is also partially on the next few slides! So, read it on your own! • In the derivation, its useful to change notation slightly. Note:My preferred notation is slightly different than our text!!

  5. Note:My preferred notation is slightly different than that in our text!! • t1 0 =time when the problem begins • x1 x0 =initial position (at t1= 0, often x0 = 0) • v1 v0 =initial velocity (at t1= 0) • t2 t =time when we wish to know other quantities • x2 x =position at time t • v2 v =velocity at time t • a acceleration = constant Average & instantaneous accelerations are equal!

  6. By definition we have: Average velocity: vave = (x - x0)/t (1) Acceleration(average = instantaneous): a = (v - v0)/t (2) Average velocity(another form): vave = (½)(v + v0) (3)

  7. Constant Acceleration Equations Note Again:My preferred notation is slightly different than in our text!! • Results (1-dimensional motiononly!): v = v0 + at (1) x = x0 + v0 t + (½)a t2 (2) v2 = (v0)2 + 2a (x - x0) (3) vave = (½) (v + v0) (4) • Not validUNLESS a = Constant!!! Often, x0 = 0. Sometimes v0 = 0

  8. All we need for 1 dimensional constant-acceleration problems: Not validUNLESS a = Constant!!!

  9. Physics & Equations:Important!!! • These equations & their applications are important, but Physics is notjust a collection of formulas to memorize & blindly apply! • Physics is a set of PHYSICAL PRINCIPLES. • Blindly searching for the “equation which will work for this problem” can be DANGEROUS!!!! • On exams, you get to have an 8.5´´ 11´´ sheet with anything written on it (both sides) you wish. On quizzes, I will give you relevant formulas.

  10. Problem Solving Strategies • Readthe whole problem. Make sure you understand it. • Decide on the objects of study & what the time interval is. • Sketcha diagram & choose coordinate axes. • Write down the known quantities, & the unknown ones needed. • What physics applies? Plan an approach to a solution.

  11. Problem Solving Strategies • Which equationsrelate known & unknown quantities? Are they valid in this situation? Solve algebraically for the unknown quantities, & check that your result is sensible (correct dimensions). • Calculate the solution, round it to appropriate number of significant figures. • Look at the result - is it reasonable? Does it agree with a rough estimate? • Check the units again.

  12. Bottom Line: THINK! DO NOTBLINDLY APPLY FORMULAS!!!!

  13. You’re designing an airport. A plane that will use this airport must reach a speed of vmin = 100 km/h (27.8 m/s). It can accelerate at a = 2 m/s2. (a) If the runway is x = 150 m long, can this plane reach the speed of before it runs off the end of the runway? (b) If not, what is the minimum length required for the runway? Example: Runway Design

  14. You’re designing an airport. A plane that will use this airport must reach a speed of vmin = 100 km/h (27.8 m/s). It can accelerate at a = 2 m/s2. (a) If the runway is x = 150 m long, can this plane reach the speed of before it runs off the end of the runway? (b) If not, what is the minimum length required for the runway? Example: Runway Design Table of Knowns & Unknowns

  15. You’re designing an airport. A plane that will use this airport must reach a speed of vmin = 100 km/h (27.8 m/s). It can accelerate at a = 2 m/s2. (a) If the runway is x = 150 m long, can this plane reach the speed of before it runs off the end of the runway? (b) If not, what is the minimum length required for the runway? Example: Runway Design Table of Knowns & Unknowns Solutions (a)v2 = (v0)2 + 2a(x – x0) v2= 0 + 2(2.0)(150 – 0) = 600 m/s2 So v = (600)½ = 24.5 m/s Note that thismeans take the square root! That obviously matters!

  16. You’re designing an airport. A plane that will use this airport must reach a speed of vmin = 100 km/h (27.8 m/s). It can accelerate at a = 2 m/s2. (a) If the runway is x = 150 m long, can this plane reach the speed of before it runs off the end of the runway? (b) If not, what is the minimum length required for the runway? Example: Runway Design Table of Knowns & Unknowns Solutions (a)v2 = (v0)2 + 2a(x – x0) v2= 0 + 2(2.0)(150 – 0) = 600 m/s2 So v = (600)½ = 24.5 m/s Note that thismeans take the square root! That obviously matters! (b) Use Eq. (3) again with v = vmin = 27.8 m/s. Solve for x – x0 = [v2 – (v0)2]/(2a) x = [(27.8)2 – 0]/[2(2.0)] So x = 193 m. To be safe, make the runway 200 mlong!

  17. How long does it take a car to cross a 30 m wide intersection after the light turns green if it accelerates at a constant 2.0 m/s2? Example: Acceleration of a Car Known:x0 = 0, x = 30 m, v0 = 0, a = 2.0 m/s2 Wanted:t. Obviously, it starts from rest!!

  18. How long does it take a car to cross a 30 m wide intersection after the light turns green if it accelerates at a constant 2.0 m/s2? Example: Acceleration of a Car Known:x0 = 0, x = 30 m, v0 = 0, a = 2.0 m/s2 Wanted:t. Obviously, it starts from rest!! • Use:x = x0 + v0t + (½)at2 = 0 + 0 + (½)at2 • t = (2x/a)½ = 5.48 s • NOTE! The square root obviously matters!

  19. Example: Air Bags You need to design an air bag system that can protect the driver at a speed of 100 km/h = 28 m/s (60 mph) if the car hits a brick wall. Estimatehow fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver? (1) (2) (3) (4)

  20. Example: Air Bags You need to design an air bag system that can protect the driver at a speed of 100 km/h = 28 m/s (60 mph) if the car hits a brick wall. Estimatehow fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver? Known: x0 = v0 = 28 m/s v = 0 Car obviously stops when crash ends!  Wanted unknown: t. (1) (2) (3) (4)

  21. Example: Air Bags You need to design an air bag system that can protect the driver at a speed of 100 km/h = 28 m/s (60 mph) if the car hits a brick wall. Estimatehow fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver? Known: x0 = v0 = 28 m/s, v = 0 Car obviously stops when crash ends! Wanted unknown: t.But we don’t know accelerationa or distance x either! Estimatex = 1.0 m This has to be a 2 step problem! (1) (2) (3) (4)

  22. Example: Air Bags You need to design an air bag system that can protect the driver at a speed of 100 km/h = 28 m/s (60 mph) if the car hits a brick wall. Estimatehow fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver? Known: x0 = v0 = 28 m/s v = 0 Car obviously stops when crash ends!  Wanted unknown: t.But we don’t know accelerationa or distance x either! Estimatex = 1.0 m. This has to be a 2 step problem! First, use (2) to solve for a: 0 = (v0)2 + 2a(x – 0) so a = - (v0)2∕(2x) = - (28)2∕(2) = - 390 m/s2 This is a HUGE acceleration!! (1) (2) (3) (4)

  23. Example: Air Bags You need to design an air bag system that can protect the driver at a speed of 100 km/h = 28 m/s (60 mph) if the car hits a brick wall. Estimatehow fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver? Known: x0 = v0 = 28 m/s v = 0 Car obviously stops when crash ends!  Wanted unknown: t. But we don’t know accelerationa or distance x either! Estimatex = 1.0 m This has to be a 2 step problem! First, use (2) to solve for a: 0 = (v0)2 + 2a(x – 0) so a = - (v0)2∕(2x) = - (28)2∕(2) = - 390 m/s2 This is a HUGE acceleration!! Now, use (1) to solve for t: 0 = v0 + at so t = - (v0) ∕a = 0.07 s !!! (1) (2) (3) (4)

  24. Example: Air Bags You need to design an air bag system that can protect the driver at a speed of 100 km/h = 28 m/s (60 mph) if the car hits a brick wall. Estimatehow fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver? Known: x0 = v0 = 28 m/s v = 0 Car obviously stops when crash ends!  Wanted unknown: t. But we don’t know accelerationa or distance x either! Estimatex = 1.0 m This has to be a 2 step problem! First, use (2) to solve for a: 0 = (v0)2 + 2a(x – 0) so a = - (v0)2∕(2x) = - (28)2∕(2) = - 390 m/s2 This is a HUGE acceleration!! Now, use (1) to solve for t: 0 = v0 + at so t = - (v0) ∕a = 0.07 s !!! (1) (2) (3) (4)

  25. Example: Estimate Breaking Distances   v = v0 = constant = 14 m/s t = 0.50 s, a = 0, x = v0t = 7 m a = - 6.0 m/s2, x0 = 7 m v decreases from 14 m/s to zero

  26. Example: Estimate Breaking Distances   v = v0 = constant = 14 m/s t = 0.50 s, a = 0, x = v0t = 7 m a = - 6.0 m/s2, x0 = 7 m v decreases from 14 m/s to zero Note:The 2nd time interval is the actual braking period when the car slows down & comes to a stop.Stopping distancedepends on 1) the driver’s reaction time, 2)the car’s initial speed, 3) the car’s acceleration.

  27. Example: Estimate Breaking Distances   v = v0 = constant = 14 m/s t = 0.50 s, a = 0, x = v0t = 7 m a = - 6.0 m/s2, x0 = 7 m v decreases from 14 m/s to zero v0 = 14 m/s, v = 0 v2 = (v0)2 + 2a(x – x0) x = x0 + [v2 - (v0)2]/(2a) x = 7 m + 16 m = 23 m Note:The 2nd time interval is the actual braking period when the car slows down & comes to a stop.Stopping distancedepends on 1) the driver’s reaction time, 2)the car’s initial speed, 3) the car’s acceleration.

  28. Example: Braking distances continued v = const.  Plots for this case: Velocity vs time v(t) Position vs time x(t) v = v0 + at  v(t) x = x0 + v0t + (½)at2 x(t) x = v0t

  29. Example: Fastball Known: x0 = 0, x = 3.5 m, v0 = 0, v = 44 m/s Wanted: a

  30. Example: Fastball Known: x0 = 0, x = 3.5 m, v0 = 0, v = 44 m/s Wanted: a Use: v2 = (v0)2 + 2a (x - x0)  a = (½)[v2 - (v0)2]/(x - x0) = 280 m/s2 !

  31. Example: 2 Moving Objects: Police & Speeder A car, speeding at v0S = 150 km/h (42 m/s) passes a still police car(v0P = 0)which immediately takes off (accelerates!) in hot pursuit. Using simple assumptions, such as that the speeder continues at constant speedv0S = 42 m/s (& also that the accelerationaPof the police car is constant!), ESTIMATEhow long it takes the police car to overtake the speeder. Then ESTIMATEthe police car’s speed at that moment & decide if the assumptions were reasonable.

  32. Example: 2 Moving Objects: Police & Speeder A car, speeding at v0S = 150 km/h (42 m/s) passes a still police car(v0P = 0)which immediately takes off (accelerates!) in hot pursuit. Using simple assumptions, such as that the speeder continues at constant speedv0S = 42 m/s (& also that the accelerationaPof the police car is constant!), ESTIMATEhow long it takes the police car to overtake the speeder. Then ESTIMATEthe police car’s speed at that moment & decide if the assumptions were reasonable. Note!Before working this problem, we need to work another problem, which will give us an ESTIMATEof the acceleration aPof the police car. In order to do this, we take numbers from ads for the type of car the police drive. These claim that this car can accelerate from rest to 100 km/h (28 m/s) in 5.0 s. Using v = v0 + aPtwith these numbers gives 28 = 0 + aP(5)or aP = 5.6 m/s2. So, to solve this problem of the police car catching up to the speeder, we use this ESTIMATEfor the acceleration aP

  33. Problem now restated is: A car, speeding at v0S = 150 km/h (42 m/s) passes a still police car (v0P = 0)which immediately takes off (accelerates!) in hot pursuit.Assume that the speeder continues at constant speedv0S = 42 m/s& that aP = 5.6 m/s2. ESTIMATEhow long it takes the police car to overtake the speeder. Then ESTIMATEthe police car’s speed at that time & decide if the assumptions were reasonable. Solution:The speeder moves at constant speed v0S = 42 m/sso, at some time t later it has moved a distance xS = v0St. In that same time t the police car has moved a distance xP = (½)aPt2 When the police car catches the speeder, the two distances must be the same. So, we equate them and solve for t: xS = v0St = xP = (½)aPt2. This is a quadratic equation for t, which has 2 solutions; t = 0 & t = 15 s.

  34. The Problem also asks: ESTIMATEthe police car’s speed at that time (t = 15 s) & decide if the assumptions were reasonable. Use : vP = v0P + aPt Gives: vP = 84 m/s (300 km/h ≈ 190 mph!) Not only unreasonable, but also very dangerous! For the assumptions we’ve made, the xversus t & v versus tcurves are shown here: More reasonablev versus tcurves are:

  35. Example: Carrier Landing A jet lands on an aircraft carrier at velocity v0 = 140 km/h (63 m/s). a) Calculate the acceleration (assumed constant) if it stops in t = 2.0 s due to the arresting cable that snags the airplane & stops it. b) If it touches down at position x0 = 0, calculate it’s final position.

  36. Example: Carrier Landing A jet lands on an aircraft carrier at velocity v0 = 140 km/h (63 m/s). a) Calculate the acceleration (assumed constant) if it stops in t = 2.0 s due to the arresting cable that snags the airplane & stops it. b) If it touches down at position x0 = 0, calculate it’s final position. Solutions a) v0 = 63 m/s, t = 2.0 s = time to stop. When it is stopped, v = 0. So, use v = v0 + at = 0, which gives a = - (v0/t) = - (63/2) = -31.5 m/s2

  37. Example: Carrier Landing A jet lands on an aircraft carrier at velocity v0 = 140 km/h (63 m/s). a) Calculate the acceleration (assumed constant) if it stops in t = 2.0 s due to the arresting cable that snags the airplane & stops it. b) If it touches down at position x0 = 0, calculate it’s final position. Solutions a) v0 = 63 m/s, t = 2.0 s = time to stop. When it is stopped, v = 0. So, use v = v0 + at = 0, which gives a = - (v0/t) = - (63/2) = -31.5 m/s2 b) Usex = x0 + v0t + (½)at2, which gives x = x0 + v0t + (½)at2 = 0 + (63)(2) + (½)(-31.5)(2)2 = 63 m

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