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!

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.

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!!

Average & instantaneous accelerations are equal! 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!

Average velocity: vave = (x - x0)/t (1) vave = (½)(v + v0) (3) 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)

Constant Acceleration Equations Note Again: My preferred notation is slightly different than in our text!! Results (1-dimensional motion only!): 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 valid UNLESS a = Constant!!! Often, x0 = 0. Sometimes v0 = 0

Not valid UNLESS a = Constant!!! All we need for 1 dimensional constant-acceleration problems: Not valid UNLESS a = Constant!!!

Physics & Equations: Important!!! These equations & their applications are important, but Physics is not just 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.

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

Problem Solving Strategies Which equations relate 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.

Bottom Line: THINK! DO NOT BLINDLY APPLY FORMULAS!!!!

Example: Runway Design 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 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? Table of Knowns & Unknowns

Example: Runway Design 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? 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 this means take the square root! That obviously matters!

Example: Runway Design 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? 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 this means 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 m long!

Example: Acceleration of a Car 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? Obviously, it starts from rest!! Known: x0 = 0, x = 30 m, v0 = 0, a = 2.0 m/s2 Wanted: t.

Example: Acceleration of a Car 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? Obviously, it starts from rest!! Known: x0 = 0, x = 30 m, v0 = 0, a = 2.0 m/s2 Wanted: t. Use: x = x0 + v0t + (½)at2 = 0 + 0 + (½)at2 t = (2x/a)½ = 5.48 s NOTE! The square root obviously matters!

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. Estimate how fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver? Figure 2-21. Caption: Example 2-11. An air bag deploying on impact. Solution: Assume the acceleration is constant; the car goes from 100 km/h to zero in a distance of about 1 m (the crumple zone). This takes a time t = 0.07 s, so the air bag has to inflate faster than this. The seat belt keeps the driver in position, and also assures that the driver decelerates with the car, rather than by hitting the dashboard. (1) (2) (3) (4)

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. Estimate how 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. Figure 2-21. Caption: Example 2-11. An air bag deploying on impact. Solution: Assume the acceleration is constant; the car goes from 100 km/h to zero in a distance of about 1 m (the crumple zone). This takes a time t = 0.07 s, so the air bag has to inflate faster than this. The seat belt keeps the driver in position, and also assures that the driver decelerates with the car, rather than by hitting the dashboard. (1) (2) (3) (4)

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. Estimate how 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 acceleration a or distance x either! Estimate x = 1.0 m This has to be a 2 step problem! Figure 2-21. Caption: Example 2-11. An air bag deploying on impact. Solution: Assume the acceleration is constant; the car goes from 100 km/h to zero in a distance of about 1 m (the crumple zone). This takes a time t = 0.07 s, so the air bag has to inflate faster than this. The seat belt keeps the driver in position, and also assures that the driver decelerates with the car, rather than by hitting the dashboard. (1) (2) (3) (4)

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. Estimate how 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 acceleration a or distance x either! Estimate x = 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!! Figure 2-21. Caption: Example 2-11. An air bag deploying on impact. Solution: Assume the acceleration is constant; the car goes from 100 km/h to zero in a distance of about 1 m (the crumple zone). This takes a time t = 0.07 s, so the air bag has to inflate faster than this. The seat belt keeps the driver in position, and also assures that the driver decelerates with the car, rather than by hitting the dashboard. (1) (2) (3) (4)

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. Estimate how 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 acceleration a or distance x either! Estimate x = 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 !!! Figure 2-21. Caption: Example 2-11. An air bag deploying on impact. Solution: Assume the acceleration is constant; the car goes from 100 km/h to zero in a distance of about 1 m (the crumple zone). This takes a time t = 0.07 s, so the air bag has to inflate faster than this. The seat belt keeps the driver in position, and also assures that the driver decelerates with the car, rather than by hitting the dashboard. (1) (2) (3) (4)

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. Estimate how 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 acceleration a or distance x either! Estimate x = 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 !!! Figure 2-21. Caption: Example 2-11. An air bag deploying on impact. Solution: Assume the acceleration is constant; the car goes from 100 km/h to zero in a distance of about 1 m (the crumple zone). This takes a time t = 0.07 s, so the air bag has to inflate faster than this. The seat belt keeps the driver in position, and also assures that the driver decelerates with the car, rather than by hitting the dashboard. (1) (2) (3) (4)

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

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 distance depends on 1) the driver’s reaction time, 2) the car’s initial speed, 3) the car’s acceleration.

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 distance depends on 1) the driver’s reaction time, 2) the car’s initial speed, 3) the car’s acceleration.

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) Figure 2-23. Caption: Example 2–12. Graphs of (a) v vs. t and (b) x vs. t. Solution: We are given the initial and final speeds and the acceleration, and are asked for the distance. Substituting in the appropriate equation gives 16.0 m; adding the 7.0 m the car traveled before the driver hit the brakes gives a total distance of 23.0 m. x = x0 + v0t + (½)at2 x(t) x = v0t

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

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 !

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 speed v0S = 42 m/s (& also that the acceleration aP of the police car is constant!), ESTIMATE how long it takes the police car to overtake the speeder. Then ESTIMATE the police car’s speed at that moment & decide if the assumptions were reasonable. Figure 2-24. Solution: First, assume the speeder continues at a constant speed, and that the police car’s acceleration is constant. If a car can go from 0 to 100 km/h in 5 sec, this is an acceleration of 5.6 m/s2. Using this, we find that the police car’s speed when it catches up to the speeder is 84 m/s, about 300 km/h. Not a good idea. Probably the acceleration of the police car is not constant, and maybe the speeder slows down.

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 speed v0S = 42 m/s (& also that the acceleration aP of the police car is constant!), ESTIMATE how long it takes the police car to overtake the speeder. Then ESTIMATE the police car’s speed at that moment & decide if the assumptions were reasonable. Figure 2-24. Solution: First, assume the speeder continues at a constant speed, and that the police car’s acceleration is constant. If a car can go from 0 to 100 km/h in 5 sec, this is an acceleration of 5.6 m/s2. Using this, we find that the police car’s speed when it catches up to the speeder is 84 m/s, about 300 km/h. Not a good idea. Probably the acceleration of the police car is not constant, and maybe the speeder slows down. Note! Before working this problem, we need to work another problem, which will give us an ESTIMATE of the acceleration aP of 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 + aPt with 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 ESTIMATE for the acceleration aP

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 speed v0S = 42 m/s & that aP = 5.6 m/s2. ESTIMATE how long it takes the police car to overtake the speeder. Then ESTIMATE the police car’s speed at that time & decide if the assumptions were reasonable. Figure 2-24. Solution: First, assume the speeder continues at a constant speed, and that the police car’s acceleration is constant. If a car can go from 0 to 100 km/h in 5 sec, this is an acceleration of 5.6 m/s2. Using this, we find that the police car’s speed when it catches up to the speeder is 84 m/s, about 300 km/h. Not a good idea. Probably the acceleration of the police car is not constant, and maybe the speeder slows down. Solution: The speeder moves at constant speed v0S = 42 m/s so, 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.

Not only unreasonable, but also very dangerous! The Problem also asks: ESTIMATE the 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 x versus t & v versus t curves are shown here: More reasonable v versus t curves are: Figure 2-24. Solution: First, assume the speeder continues at a constant speed, and that the police car’s acceleration is constant. If a car can go from 0 to 100 km/h in 5 sec, this is an acceleration of 5.6 m/s2. Using this, we find that the police car’s speed when it catches up to the speeder is 84 m/s, about 300 km/h. Not a good idea. Probably the acceleration of the police car is not constant, and maybe the speeder slows down.

A jet lands on an aircraft carrier at velocity 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.

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

x = x0 + v0t + (½)at2 = 0 + (63)(2) + (½)(-31.5)(2)2 = 63 m 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) Use x = x0 + v0t + (½)at2 , which gives x = x0 + v0t + (½)at2 = 0 + (63)(2) + (½)(-31.5)(2)2 = 63 m