-Plan and carry out an experiment
-Determine breaking distance
-Examine accelerated motion.
What do you see / What do you think?
Tuesday, October 18th
Screen_shot_2011-10-18_at_8.26.06_AM.png
-It looks like the vehicle could not stop in time, in result of going to fast.
-The deer clearly came out of no where.
-It appears that the car slammed on the breaks, causing it to skid and hit the mouse.
-The factors that would determine the stop would be how much weight is in the car, how fast you are going, and how big the car actually is.
Investigation
Tuesday, October 18th
Objective: to determine the effect of initial speed on breaking (stopping) distance.
Vi(m/s)
BreakingDistance(m)
.59m/s
2.8m
.94
5.15m
1.27
11.32
1.06
6.5
.8
4
.97
5
1.23
11
Make Graph (scatter plot)
Smooth curve through the graph.
YAXIS: Breaking Distance vs. XAXIS: Vi(m/s)
Vi=Length of flag/time in gate
Screen_shot_2011-10-19_at_9.21.04_AM.png
Sec5 Investigation Questions:
1.)If initial velocity is doubled by what factor does breaking distance increase.
-If the initial velocity is doubled, our information shows that the breaking distance will quadruple. 2)If initial velocity is tripled by what factor does braking distance increase.
-If the initial velocity is tripled, assuming that the proportion of Vi:BD stays the same, it would be 3^2, so it would be multiplied by 9. 3)If initial velocity is quadrupled (x4) by what factor does braking distance increase.
-If the Initial velocity is quadruples the breaking distance would be multiplied by 16. 4)Do #8 in book on page77
-b) No the breaking distance is going to be different.
-c) The breaking distances were squared of our speed. 5) If the sports car changed its speed to 30 mph what do you expect its braking distance to be? (Hint: if you half the speed by what factor will the braking distance change?)
-Decrease speed is 1/2 --> square of a 1/2. When you square 30 mph (speed), you get 900 miles for your breaking distance. 6) If the touring sedan changed its speed to 30 mph what do you expect its braking distance to be?
-The breaking distance is 900 miles as well because you square the speed (30mph). 7) If each car (sports car and touring sedan) decreased its speed to 15 mph what would their braking distances be? (Hint: if you quarter speed by what factor will the braking distance change?) 15^2 = 225 miles Homework:
Wednesday, October 19th
Physics Words:Negative acceleration: a change in the velocity with respect to time of an object by decreasing speed in the positive direction or increasing speed in the negative direction
Checking Up Questions: 1. If a vehicle is traveling at constant velocity and the comes to a sudden stop, has it undergone negative acceleration or positive acceleration? Explain your answer.
The vehicle has undergone negative acceleration because the car is decreasing speed in a positive direction. 2. Explain how you know that increasing the velocity of an automobile increases the braking distance.
The faster the automobile goes, the more distance it needs to make a complete stop. 3. Why is the term negative acceleration used instead of deceleration? In order to be clear about meanings in physics, the terms positive acceleration and negative acceleration are used instead of acceleration and deceleration. Positive represents one direction and negative represents the opposite direction. (An object could have a negative acceleration by decreasing its speed in the positive direction or increasing its speed in the negative direction.)
Notes: Thursday, October 20th equation 1Vi (2) = ratio of initial velocitiesVi (1)
Honda Civic Stopping Distance:__ Thursday, October 20th
Initial Velocity
Stopping Distance
10 mph
3.42 feet
20 mph
13.67 feet
30 mph
30.75 feet
40 mph
54.67 feet
50 mph
85.42 feet
60 mph
123 feet
70 mph
167.42 feet
80 mph
218.67 feet
90 mph
276.75 feet
100 mph
341.67 feet
4) If Initial Velocity is doubled how does stopping distance change? the stopping distance is multiplied by 4 (use equation 3)
5) If the Initial Velocity is multiplied 4 times how does the stopping distance change? 16 times
6) If the Initial Velocity is halved how does the Stopping Distance Change? the stopping distance is divided by 4 (stopping distance / 4 ) OR multiply by .25
7) If the initial Velocity is quartered how does the stopping distance change? cut it by 1/16 (use equation 2) (10mph)2= times 1/16 (40mph)2
8) What speed would you need to have a stopping distance of a mile? 393.1 mph
Homework Physics to Go pg. 88-89 #1-8Monday, October 24th1.A student measured the braking distance of her automobile and recorded the data in the table. Plot the data on a graph and describe the relationship that exists between initial speed and braking distance. As the initial speed increases, the braking distance increases.
Screen_shot_2011-10-20_at_11.00.17_AM.png
2.Below is a graph of the braking distances in relation to initial speed for two automobiles. Compare qualitatively (without using numbers) the braking distances when each automobile is going at a slow speed and then again at a higher speed. Which automobile is safer? Why? How did you determine what “safer” means in this question?
Screen_shot_2011-10-20_at_10.55.20_AM.png
^Automobile A is safer because it has a slower velocity and a lesser braking distance. The "safer" automobile is the one that would prevent you from getting into an accident if you needed to brake suddenly.
3.An automobile is able to stop in 20 m when traveling at 30 mi/h. How much distance will it require to stop when traveling at the following:
a) 15 mi/h? (half of 30 mi/h)(Vi[2]) / (Vi[1]) = (BD[2]) / (BD[1])(30/15)^2 = (20/x)x = 5m
b) 60 mi/h? (twice 30 mi/h)(30/60)^2 = (20/x)x = 80m
c) 45 mi/h? (three times 15 mi/h)(30/45)^2 = (20/x)x = 45m
d) 75 mi/h? (five times 15 mi/h)(30/75)^2 = (20/x)x = 125 m
4. An automobile traveling at 10 m/s requires a braking distance of 30 m. If the driver requires 0.9 s reaction time, what additional distance will the automobile travel before stopping? What is the total stopping distance, including both the reaction distance and the braking distance?
5. Consult the information for the sports car at the end of this chapter. This shows the stopping distance. How far would you expect this automobile to travel until coming to rest when brakes are applied at a speed of 30 mi/h?
6. Use the information for the sedan at the end of this chapter. Find the braking distances for 50 mi/h and 25 mi/h. Draw a graph using the different braking distances. Plot the speeds on the horizontal axis and the braking distances on the vertical axis.
7. Does the braking information for the sedan include the driver's reaction time? If it does not, then how much distance is added to the total braking distance, supporting that the driver has a 1/2 s reaction time? Who should let the consumer know about the 1/2 s reaction time--the information sheet or a drvier training manual?
8. Apply what you learned in this section to write a statement explaining the factors that affect stopping distance. The total stopping distance includes the distance you travel during your reaction time, plus the braking distance. What do you now know about stopping that will make you a safer driver? The factors that affect stopping distance are weather, distractions, reaction time, and weight of car. I now know that it takes more than just reacting to make the car come to a stop, so you have to be more aware of your surroundings and follow the speed limit.
Photo_on_2011-10-24_at_08.03_#2.jpg
Photo_on_2011-10-24_at_08.03.jpg
Active Physics Plus:Monday, October 24th Equations of Motion:d = .5(Vi+Vf)tVf = Vi + atd = Vit + (.5)(a)(t^2)Vf^2 = Vi^2 + 2ad
Average acceleration 1. As the shuttle bus comes to a sudden station to avoid hitting a dog, it accelerates uniformly at -4.1 m/s^2 as it slows from 9.0 m/s to 0 m/s. Find the time interval of acceleration for the bus. a = -4.1 m/s^2t = ?Vi = 9 m/sVf = 0 m/sdon't need dVf = Vi + at0 = 9 + (-4.1)(t)-9 = -4.1tt = 2.2 s
2. At car traveling at 7.0 m/s accelerates uniformly at 2.5 m/s^2 to reach a speed of 12.0 m/s. How long does it take for this acceleration to occur? Vi = 7 m/sVf = 12 m/sa = 2.5 m/s^2t = ?Vf = Vi + at12 = 7 + 2.5tt = 2 s
3. With an average acceleration of -0.50 m/s^2, how long will it take a cyclist to bring a bicycle with an initial speed of 13.5 m/s to a complete stop? a = -0.5 m/s^2Vi = 13.5 m/sVf = 0 m/st = ?Vf = Vi + at0 = 13.5 + -0.5tt = 27 s
4. Turner's treadmill runs with a velocity of -1.2 m/s and speeds up at regular intervals during a half-hour workout. After 25 min, the treadmill has a velocity of -6.5 m/s. What is the average acceleration of the treadmill during this period? Vi = -1.2 m/st = 25 min = 1500 sVf = -6.5 m/sa = ?Vf = Vi + at-6.5 = -1.2 + 1500aa = -0.004 m/s^2
5. Suppose a treadmill has an average acceleration of 4.7 * 10^-3 m/s^2 a. How much does its speed change after 5.0 min?a = 4.7 * 10^-3 m/s^2Vi = 0 m/sVf = ?t = 5 min = 300 sVf = Vi +atVf = 0 + (300)(4.7 * 10^-3)Vf = 1.41 m/s
b. If the treadmill's initial speed is 1.7 m/s, what will its final speed be? Vi = 1.7 m/sVf = ?a = 4.7 * 10^-3 m/s^2t = 300 sVf = Vi + atVf = 1.7 + (4.7 * 10^-3)(300)Vf = 3.11 m/s Displacement with constant uniform acceleration
1. A car accelerates uniformly from rest to a speed of 23.7 km/h in 6.5 s. Find the distance the car travels during this time. t = 6.5sd = ?Vi = 0 m/sVf = 23.7 km/h = (23.6 km / 1 hr) * (1000 m / 1 km) * (1 hr / 60 min) * (1 min / 60 sec) = 6.58 m/sd = .5(Vi+Vf)td = (.5)(0 + 6.58)(6.5)d = 21.385 m
2. When Maggie applies the brakes of her car, the car slows uniformly from 15.0 m/s to 0 m/s in 2.50 s. How many meters before a stop sign must she apply her brakes in order to stop at the sign? Vi = 15 m/sVf = 0 m/st = 2.5 sd = ?d = (.5)(Vi+Vf)td = (.5)(15 + 0)(2.5)d = 18.75 m
3. A jet plane lands with a speed of 100 m/s an can accelerate uniformly at a maximum rate of -5.0 m/s^2 as it comes to rest. Can this plane land at an airport where the runaway is 0.80 km long? Vi = 100 m/sa = -5 m/s^2Vf = 0 m/sd = ?Vf^2 = Vi^2 + 2ad0 = 100^2 + 2(-5)(d)d = 1000 mNo it cannot land.
4. A driver in a car traveling at a speed of 78 km/h sees at cat 101 m away on the road/ How long with it take for the car to acceleration uniformly to a stop in exactly 99 m? t = ?d = 99 mVi = 78 km/h = 21.67 m/sVf = 0 km/hd = (.5)(Vi+Vf)t99 = (.5)(21.67 + 0)(t)t = 9.14 s
5. A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3.2 km in 3.5 min. How fast is the car moving after this time? Vi = 6.4 m/sVf = ?d = 3.2 km = 3200 mt = 3.5 min = 210 sd = (.5)(Vi+Vf)t3200 = (.5)(6.4 + Vf)(210)Vf = 24.08 m/s Velocity and displacement with uniform acceleration
1. A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s^2 for 3.6 s. Find the final speed and the displacement of the car during this time. a = 0.92 m/st = 3.6 sd = ?Vi = 23.7 km/h = 6.58 m/sVf = ? = 9.9 m/s(23.7 km / 1 hr) * (1000 m / 1 km) * (1 hr / 60 min) * (1 min / 60 sec) = 6.58 m/sVf = Vi + atVf = 6.58 + (0.92)(3.6)Vf = 9.9 m/sFind displacement (d)d = (.5)(Vi + Vf)(t)d = (.5)( 6.58 + 9.9)(3.6)d = 29.7 m
2. An automobile with an initial speed of 4.30 m/s accelerates uniformly at the rate of 3.0 m/s^2. Find the final speed and the displacement after 5.0 s. Vi = 4.3 m/sa = 3 m/s^2t = 5 sVf = ? = 19.3d = ?Vf = Vi + atVf = 4.3 + (3)(5)Vf = 19.3 m/sd = (.5)(Vi + Vf)(t)d = (.5)(4.3 + 19.3)(5)d = 59 m
3. A car starts from rest and travels for 5.0 s with a uniform acceleration of -1.5 m/s^2. What is the final velocity of the car? How far does the car travel in this time interval? t = 5 sa = -1.5 m/s^2Vi = 0Vf = ? = -7.5 m/sd = ?Vf = Vi + atVf = 0 + (-1.5)(5)Vf = -7.5 m/sd = (.5)(Vi + Vf)(t)d = (.5)(0 + -7.5)(5)d = -18.75 m
4. Adriver of a car traveling at 15.0 m/s applies the brakes, causing a uniform acceleration of -2.0 m/s^2. How long does it take the car to accelerates to a final speed of 10.0 m/s? How far has the car moved during the braking period? a = -2 m/s^2Vi = 15 m/sVf = 10 m/st = ? = 2.5 m/sd = ?Vf = Vi + at10 = 15 + (-2)(t)t = 2.5 sd = (.5)(Vi + Vf)(t)d = (.5)(15 + 10)(2.5) d = 31.25 m Final velocity after any displacement
2. A car traveling at 7.0 m/s accelerates uniformly at the at the rate of 0.8 m/s^2 for a distance of 245 m. a = 0.8 m/s^2d = 245 mVi = 7 m/sVf = ? a. What is tits velocity at the end of the acceleration?Vf^2 = Vi^2 + 2adVf^2 = (7)^2 + (2)(0.8)(245)Vf^2 = 441 m^2/s^2Vf = 21 m/s b. What is its velocity after it accelerates for 125 m?Vf^2 = Vi^2 + 2adVf^2 = (7)^2 + (2)(0.8)(125)Vf = 15.78 m/s c. What is its velocity after it accelerates for 67 m?Vf^2 = Vi^2 + 2adVf^2 = (7)^2 + (2)(0.8)(67)Vf = 12.5 m/s
3. A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s^2. Vi = 0 m/sVf = ? = 15.9 m/sa = 2.3 m/s^2 a. What is the speed of the car after its has traveled 55 m? Vf^2 = Vi^2 + 2adVf^2 = 0 + 2(2.3)(55)Vf = 15.9 m/sb. How long does it take the car to travel 55 m?d = (.5)(Vi + Vf)(t)55 = (.5)(0 + 15.9)(t)t = 6.9 s
4. A certain car is capable of acceleration at a uniform rate of 0.85 m/s^2. What is the magnitude of the car's displacement as it accelerates uniformly from a speed of 83 km/h to one of 94 km/h? a = 0.85 m/s^2Vi = 83 km/h = 23.1 m/sVf = 94 km/h = 26.1 m/sd = ?Vf^2 = Vi^2 + 2ad26.1^2 = 23.1^2 + 2(0.85)(d)d = 86.9 m
5. An aircraft has a liftoff speed of 120 km/h. What minimum uniform acceleration does this require if the aircraft to be airborne after a takeoff run of 240 m? Vi = 0 m/sVf = 120 km/h = 33.33 m/sa = ?d = 240 mVf^2 = Vi^2 + 2ad33.33^2 = 0 + 2(a)(240)a = 2.27 m/s^2
6. A motorboat acceleration uniformly from a velocity of 6.5 m/s to the west to a velocity of 1.5 m/s to the west. If its acceleration was 2.7 m/s^2 to the east, how far did it travel during the acceleration? Vi = 6.5 m/sVf = 1.5 m/sa = 2.7 m/s^2d = ?Vf^2 = Vi^2 + 2ad1.5^2 = 6.5^2 + (2)(2.7)(d)d = -7.41 m
Mini Challenge Plan Thursday, October 27th Nicky: making keynote, storyline, equation 2 (with Kate)Mel: graph 1, equation 1, responsibilities Kate: graph 2, equation 2 (with Nicky)
Allison: graph 3, equation 3
Storyline: Allison was driving home from the football game when the following happened: 1. (NICKY) Allison was driving us home from the football game and we were approaching an intersection at 20 m/s. The light turned yellow when she was 35 meters away. Does she have enough time to stop if she has an acceleration of -2 m/s? A= -2 m/sD= 50 mT= ?Vi= 20 m/sVf= 0 m/sEquation: vf=vi+at
2.(ALLISON) Allison is driving at 70 m/s and is approaching a curve that has a speed limit of 11 m/s. She has 15 meters to brake. What should her deceleration be? A= ?D= 15 mT= XVi= 70 m/sVf= 11 m/sEquation: vf^2= Vi^2+2ad
3. (MEL) Allison has a reaction time of .5 seconds, which includes moving her foot from gas to brake. They are going through a school zone at 11 m/s and the car has a maximum deceleration of -5 m/s^2. What is Allison’s reaction distance, braking distance, and total stopping distance? A= -5 m/s^2T= .5 sD= ?Equation: v= d/t
Graphs:Distance vs Time (Allison)Velocity vs Time (Kate and Nicky)Braking Distance vs Initial Velocity (Mel)
Equations:Braking DistanceDecelerationv=d/t
Total Stopping Distance ActivityFriday, October 28th
1) Draw a diagram of a car moving down a road and braking for an object.
tots_stop_distamce.jpg
2) Assuming we are traveling forward with a positive velocity, then we would have a negative acceleration when we are braking.
Calculating Reaction Distance 1) Finding reaction distance using diagram abovev=d/t (<------ this is the one without acceleration because during this time there is no acceleration)reaction distance=v/reaction time 2) v=10m/s, t=1sd=10m 3) v=20m/s, t=1sd=20m 4) An increase in speed changes the distance because the vehicle would travel farther during that time. A decrease in speed would mean a shorter distance.One second is a pretty short reaction time, and it may be this large because of decisions (whether to stop or not at the yellow light), age, or intoxication.
Calculating Braking Distance 1)vf^2=vi^2 +2a(braking distance)0 = Vi^2 + 2(a)(braking distance)d = -(vi)^2/2a 5) d = -(51^2)/2(-11)d = 118.23 m 6) d = -(51^2)/2(-24)d = 54.19 7) Better brakes have a larger acceleration. Better brakes make for a shorter braking distance.
Create a problemWednesday, November 2nd -Solve for dr, db TSD.-Embed givens in problem.-Don't embed givens that don't need to be mentioned.
My Problem: A teenager speeds up to a yellow light. What would the reaction time, braking distance , and total stopping distance be for the teenager be if They are traveling at 20 m/s (44 mph) and have a reaction time of .4s, and a deceleration of -5 m/s^2 tr = .4 sVi = 20 m/sa= -5 m/s^2V= d/t20 m/s (.4 s) = dreaction/ .4s (.4s)dreaction = 8 mVf^2 = Vi^2 +2ad-(20 m/s)^2 / -10 = dbrakingdbraking = 40 mTSD = 48 m.
Alex's Problem: Your driving in your brand new customized car that Bill Trembath bought you . You're going down your street and have a reaction time of .5 seconds and traveling at 25 m/s with a deceleration of -6 m/s^2. What is your reaction distance , braking distance, and total stopping distance? tr = .5 secondsVi = 25 m/sa= -6 m/s^2V= d/t25 m/s (.5 s) = d/ (.5 seconds) (.5 s)dreaction = 12.5 m Vf^2= Vi^2 + 2ad-(25 m/s)^2 / -12 = dbraking52.08 m= dbrakingTSD = 64.58 m
Learning OutcomesTuesday, October 18th
-Plan and carry out an experiment
-Determine breaking distance
-Examine accelerated motion.
What do you see / What do you think?
Tuesday, October 18th
-It looks like the vehicle could not stop in time, in result of going to fast.
-The deer clearly came out of no where.
-It appears that the car slammed on the breaks, causing it to skid and hit the mouse.
-The factors that would determine the stop would be how much weight is in the car, how fast you are going, and how big the car actually is.
Investigation
Tuesday, October 18th
Objective: to determine the effect of initial speed on breaking (stopping) distance.
Smooth curve through the graph.
YAXIS: Breaking Distance vs. XAXIS: Vi(m/s)
Vi=Length of flag/time in gate
Sec5 Investigation Questions:
1.) If initial velocity is doubled by what factor does breaking distance increase.
-If the initial velocity is doubled, our information shows that the breaking distance will quadruple.
2) If initial velocity is tripled by what factor does braking distance increase.
-If the initial velocity is tripled, assuming that the proportion of Vi:BD stays the same, it would be 3^2, so it would be multiplied by 9.
3) If initial velocity is quadrupled (x4) by what factor does braking distance increase.
-If the Initial velocity is quadruples the breaking distance would be multiplied by 16.
4) Do #8 in book on page77
-b) No the breaking distance is going to be different.
-c) The breaking distances were squared of our speed.
5) If the sports car changed its speed to 30 mph what do you expect its braking distance to be? (Hint: if you half the speed by what factor will the braking distance change?)
-Decrease speed is 1/2 --> square of a 1/2. When you square 30 mph (speed), you get 900 miles for your breaking distance.
6) If the touring sedan changed its speed to 30 mph what do you expect its braking distance to be?
-The breaking distance is 900 miles as well because you square the speed (30mph).
7) If each car (sports car and touring sedan) decreased its speed to 15 mph what would their braking distances be? (Hint: if you quarter speed by what factor will the braking distance change?)
15^2 = 225 miles
Homework:
Wednesday, October 19th
Physics Words: Negative acceleration: a change in the velocity with respect to time of an object by decreasing speed in the positive direction or increasing speed in the negative direction
Checking Up Questions:
1. If a vehicle is traveling at constant velocity and the comes to a sudden stop, has it undergone negative acceleration or positive acceleration? Explain your answer.
The vehicle has undergone negative acceleration because the car is decreasing speed in a positive direction.
2. Explain how you know that increasing the velocity of an automobile increases the braking distance.
The faster the automobile goes, the more distance it needs to make a complete stop.
3. Why is the term negative acceleration used instead of deceleration?
In order to be clear about meanings in physics, the terms positive acceleration and negative acceleration are used instead of acceleration and deceleration. Positive represents one direction and negative represents the opposite direction. (An object could have a negative acceleration by decreasing its speed in the positive direction or increasing its speed in the negative direction.)
Notes:
Thursday, October 20th
equation 1Vi (2) = ratio of initial velocitiesVi (1)
equation 2Braking Distance (2) = ratio of braking distancesBraking Distance (1)
equation 3
(Vi (2))2 = braking distance (2)
(Vi (1))2 _braking distance (1)
Honda Civic Stopping Distance:__
Thursday, October 20th
4) If Initial Velocity is doubled how does stopping distance change?
the stopping distance is multiplied by 4 (use equation 3)
5) If the Initial Velocity is multiplied 4 times how does the stopping distance change?
16 times
6) If the Initial Velocity is halved how does the Stopping Distance Change?
the stopping distance is divided by 4 (stopping distance / 4 )
OR
multiply by .25
7) If the initial Velocity is quartered how does the stopping distance change?
cut it by 1/16 (use equation 2)
(10mph)2= times 1/16
(40mph)2
8) What speed would you need to have a stopping distance of a mile?
393.1 mph
Homework Physics to Go pg. 88-89 #1-8Monday, October 24th1. A student measured the braking distance of her automobile and recorded the data in the table. Plot the data on a graph and describe the relationship that exists between initial speed and braking distance.
As the initial speed increases, the braking distance increases.
2. Below is a graph of the braking distances in relation to initial speed for two automobiles. Compare qualitatively (without using numbers) the braking distances when each automobile is going at a slow speed and then again at a higher speed. Which automobile is safer? Why? How did you determine what “safer” means in this question?
^Automobile A is safer because it has a slower velocity and a lesser braking distance. The "safer" automobile is the one that would prevent you from getting into an accident if you needed to brake suddenly.
3. An automobile is able to stop in 20 m when traveling at 30 mi/h. How much distance will it require to stop when traveling at the following:
a) 15 mi/h? (half of 30 mi/h)(Vi[2]) / (Vi[1]) = (BD[2]) / (BD[1])(30/15)^2 = (20/x)x = 5m
b) 60 mi/h? (twice 30 mi/h)(30/60)^2 = (20/x)x = 80m
c) 45 mi/h? (three times 15 mi/h)(30/45)^2 = (20/x)x = 45m
d) 75 mi/h? (five times 15 mi/h)(30/75)^2 = (20/x)x = 125 m
4. An automobile traveling at 10 m/s requires a braking distance of 30 m. If the driver requires 0.9 s reaction time, what additional distance will the automobile travel before stopping? What is the total stopping distance, including both the reaction distance and the braking distance?
5. Consult the information for the sports car at the end of this chapter. This shows the stopping distance. How far would you expect this automobile to travel until coming to rest when brakes are applied at a speed of 30 mi/h?
6. Use the information for the sedan at the end of this chapter. Find the braking distances for 50 mi/h and 25 mi/h. Draw a graph using the different braking distances. Plot the speeds on the horizontal axis and the braking distances on the vertical axis.
7. Does the braking information for the sedan include the driver's reaction time? If it does not, then how much distance is added to the total braking distance, supporting that the driver has a 1/2 s reaction time? Who should let the consumer know about the 1/2 s reaction time--the information sheet or a drvier training manual?
8. Apply what you learned in this section to write a statement explaining the factors that affect stopping distance. The total stopping distance includes the distance you travel during your reaction time, plus the braking distance. What do you now know about stopping that will make you a safer driver? The factors that affect stopping distance are weather, distractions, reaction time, and weight of car. I now know that it takes more than just reacting to make the car come to a stop, so you have to be more aware of your surroundings and follow the speed limit.
Active Physics Plus:Monday, October 24th
Equations of Motion:d = .5(Vi+Vf)tVf = Vi + atd = Vit + (.5)(a)(t^2)Vf^2 = Vi^2 + 2ad
Average acceleration
1. As the shuttle bus comes to a sudden station to avoid hitting a dog, it accelerates uniformly at -4.1 m/s^2 as it slows from 9.0 m/s to 0 m/s. Find the time interval of acceleration for the bus.
a = -4.1 m/s^2t = ?Vi = 9 m/sVf = 0 m/sdon't need dVf = Vi + at0 = 9 + (-4.1)(t)-9 = -4.1tt = 2.2 s
2. At car traveling at 7.0 m/s accelerates uniformly at 2.5 m/s^2 to reach a speed of 12.0 m/s. How long does it take for this acceleration to occur?
Vi = 7 m/sVf = 12 m/sa = 2.5 m/s^2t = ?Vf = Vi + at12 = 7 + 2.5tt = 2 s
3. With an average acceleration of -0.50 m/s^2, how long will it take a cyclist to bring a bicycle with an initial speed of 13.5 m/s to a complete stop?
a = -0.5 m/s^2Vi = 13.5 m/sVf = 0 m/st = ?Vf = Vi + at0 = 13.5 + -0.5tt = 27 s
4. Turner's treadmill runs with a velocity of -1.2 m/s and speeds up at regular intervals during a half-hour workout. After 25 min, the treadmill has a velocity of -6.5 m/s. What is the average acceleration of the treadmill during this period?
Vi = -1.2 m/st = 25 min = 1500 sVf = -6.5 m/sa = ?Vf = Vi + at-6.5 = -1.2 + 1500aa = -0.004 m/s^2
5. Suppose a treadmill has an average acceleration of 4.7 * 10^-3 m/s^2
a. How much does its speed change after 5.0 min?a = 4.7 * 10^-3 m/s^2Vi = 0 m/sVf = ?t = 5 min = 300 sVf = Vi +atVf = 0 + (300)(4.7 * 10^-3)Vf = 1.41 m/s
b. If the treadmill's initial speed is 1.7 m/s, what will its final speed be?
Vi = 1.7 m/sVf = ?a = 4.7 * 10^-3 m/s^2t = 300 sVf = Vi + atVf = 1.7 + (4.7 * 10^-3)(300)Vf = 3.11 m/s
Displacement with constant uniform acceleration
1. A car accelerates uniformly from rest to a speed of 23.7 km/h in 6.5 s. Find the distance the car travels during this time.
t = 6.5sd = ?Vi = 0 m/sVf = 23.7 km/h = (23.6 km / 1 hr) * (1000 m / 1 km) * (1 hr / 60 min) * (1 min / 60 sec) = 6.58 m/sd = .5(Vi+Vf)td = (.5)(0 + 6.58)(6.5)d = 21.385 m
2. When Maggie applies the brakes of her car, the car slows uniformly from 15.0 m/s to 0 m/s in 2.50 s. How many meters before a stop sign must she apply her brakes in order to stop at the sign?
Vi = 15 m/sVf = 0 m/st = 2.5 sd = ?d = (.5)(Vi+Vf)td = (.5)(15 + 0)(2.5)d = 18.75 m
3. A jet plane lands with a speed of 100 m/s an can accelerate uniformly at a maximum rate of -5.0 m/s^2 as it comes to rest. Can this plane land at an airport where the runaway is 0.80 km long?
Vi = 100 m/sa = -5 m/s^2Vf = 0 m/sd = ?Vf^2 = Vi^2 + 2ad0 = 100^2 + 2(-5)(d)d = 1000 mNo it cannot land.
4. A driver in a car traveling at a speed of 78 km/h sees at cat 101 m away on the road/ How long with it take for the car to acceleration uniformly to a stop in exactly 99 m?
t = ?d = 99 mVi = 78 km/h = 21.67 m/sVf = 0 km/hd = (.5)(Vi+Vf)t99 = (.5)(21.67 + 0)(t)t = 9.14 s
5. A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3.2 km in 3.5 min. How fast is the car moving after this time?
Vi = 6.4 m/sVf = ?d = 3.2 km = 3200 mt = 3.5 min = 210 sd = (.5)(Vi+Vf)t3200 = (.5)(6.4 + Vf)(210)Vf = 24.08 m/s
Velocity and displacement with uniform acceleration
1. A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s^2 for 3.6 s. Find the final speed and the displacement of the car during this time.
a = 0.92 m/st = 3.6 sd = ?Vi = 23.7 km/h = 6.58 m/sVf = ? = 9.9 m/s(23.7 km / 1 hr) * (1000 m / 1 km) * (1 hr / 60 min) * (1 min / 60 sec) = 6.58 m/sVf = Vi + atVf = 6.58 + (0.92)(3.6)Vf = 9.9 m/sFind displacement (d)d = (.5)(Vi + Vf)(t)d = (.5)( 6.58 + 9.9)(3.6)d = 29.7 m
2. An automobile with an initial speed of 4.30 m/s accelerates uniformly at the rate of 3.0 m/s^2. Find the final speed and the displacement after 5.0 s.
Vi = 4.3 m/sa = 3 m/s^2t = 5 sVf = ? = 19.3d = ?Vf = Vi + atVf = 4.3 + (3)(5)Vf = 19.3 m/sd = (.5)(Vi + Vf)(t)d = (.5)(4.3 + 19.3)(5)d = 59 m
3. A car starts from rest and travels for 5.0 s with a uniform acceleration of -1.5 m/s^2. What is the final velocity of the car? How far does the car travel in this time interval?
t = 5 sa = -1.5 m/s^2Vi = 0Vf = ? = -7.5 m/sd = ?Vf = Vi + atVf = 0 + (-1.5)(5)Vf = -7.5 m/sd = (.5)(Vi + Vf)(t)d = (.5)(0 + -7.5)(5)d = -18.75 m
4. A driver of a car traveling at 15.0 m/s applies the brakes, causing a uniform acceleration of -2.0 m/s^2. How long does it take the car to accelerates to a final speed of 10.0 m/s? How far has the car moved during the braking period?
a = -2 m/s^2Vi = 15 m/sVf = 10 m/st = ? = 2.5 m/sd = ?Vf = Vi + at10 = 15 + (-2)(t)t = 2.5 sd = (.5)(Vi + Vf)(t)d = (.5)(15 + 10)(2.5) d = 31.25 m
Final velocity after any displacement
2. A car traveling at 7.0 m/s accelerates uniformly at the at the rate of 0.8 m/s^2 for a distance of 245 m.
a = 0.8 m/s^2d = 245 mVi = 7 m/sVf = ?
a. What is tits velocity at the end of the acceleration?Vf^2 = Vi^2 + 2adVf^2 = (7)^2 + (2)(0.8)(245)Vf^2 = 441 m^2/s^2Vf = 21 m/s
b. What is its velocity after it accelerates for 125 m?Vf^2 = Vi^2 + 2adVf^2 = (7)^2 + (2)(0.8)(125)Vf = 15.78 m/s
c. What is its velocity after it accelerates for 67 m?Vf^2 = Vi^2 + 2adVf^2 = (7)^2 + (2)(0.8)(67)Vf = 12.5 m/s
3. A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s^2.
Vi = 0 m/sVf = ? = 15.9 m/sa = 2.3 m/s^2
a. What is the speed of the car after its has traveled 55 m?
Vf^2 = Vi^2 + 2adVf^2 = 0 + 2(2.3)(55)Vf = 15.9 m/sb. How long does it take the car to travel 55 m?d = (.5)(Vi + Vf)(t)55 = (.5)(0 + 15.9)(t)t = 6.9 s
4. A certain car is capable of acceleration at a uniform rate of 0.85 m/s^2. What is the magnitude of the car's displacement as it accelerates uniformly from a speed of 83 km/h to one of 94 km/h?
a = 0.85 m/s^2Vi = 83 km/h = 23.1 m/sVf = 94 km/h = 26.1 m/sd = ?Vf^2 = Vi^2 + 2ad26.1^2 = 23.1^2 + 2(0.85)(d)d = 86.9 m
5. An aircraft has a liftoff speed of 120 km/h. What minimum uniform acceleration does this require if the aircraft to be airborne after a takeoff run of 240 m?
Vi = 0 m/sVf = 120 km/h = 33.33 m/sa = ?d = 240 mVf^2 = Vi^2 + 2ad33.33^2 = 0 + 2(a)(240)a = 2.27 m/s^2
6. A motorboat acceleration uniformly from a velocity of 6.5 m/s to the west to a velocity of 1.5 m/s to the west. If its acceleration was 2.7 m/s^2 to the east, how far did it travel during the acceleration?
Vi = 6.5 m/sVf = 1.5 m/sa = 2.7 m/s^2d = ?Vf^2 = Vi^2 + 2ad1.5^2 = 6.5^2 + (2)(2.7)(d)d = -7.41 m
Mini Challenge Plan
Thursday, October 27th
Nicky: making keynote, storyline, equation 2 (with Kate)Mel: graph 1, equation 1, responsibilities Kate: graph 2, equation 2 (with Nicky)
Allison: graph 3, equation 3
Storyline: Allison was driving home from the football game when the following happened:
1. (NICKY) Allison was driving us home from the football game and we were approaching an intersection at 20 m/s. The light turned yellow when she was 35 meters away. Does she have enough time to stop if she has an acceleration of -2 m/s?
A= -2 m/sD= 50 mT= ?Vi= 20 m/sVf= 0 m/sEquation: vf=vi+at
2.(ALLISON) Allison is driving at 70 m/s and is approaching a curve that has a speed limit of 11 m/s. She has 15 meters to brake. What should her deceleration be?
A= ?D= 15 mT= XVi= 70 m/sVf= 11 m/sEquation: vf^2= Vi^2+2ad
3. (MEL) Allison has a reaction time of .5 seconds, which includes moving her foot from gas to brake. They are going through a school zone at 11 m/s and the car has a maximum deceleration of -5 m/s^2. What is Allison’s reaction distance, braking distance, and total stopping distance?
A= -5 m/s^2T= .5 sD= ?Equation: v= d/t
Graphs:Distance vs Time (Allison)Velocity vs Time (Kate and Nicky)Braking Distance vs Initial Velocity (Mel)
Equations:Braking DistanceDecelerationv=d/t
Total Stopping Distance ActivityFriday, October 28th
1) Draw a diagram of a car moving down a road and braking for an object.
Calculating Reaction Distance
1) Finding reaction distance using diagram abovev=d/t (<------ this is the one without acceleration because during this time there is no acceleration)reaction distance=v/reaction time
2) v=10m/s, t=1sd=10m
3) v=20m/s, t=1sd=20m
4) An increase in speed changes the distance because the vehicle would travel farther during that time. A decrease in speed would mean a shorter distance.One second is a pretty short reaction time, and it may be this large because of decisions (whether to stop or not at the yellow light), age, or intoxication.
Calculating Braking Distance
1)vf^2=vi^2 +2a(braking distance)0 = Vi^2 + 2(a)(braking distance)d = -(vi)^2/2a
5) d = -(51^2)/2(-11)d = 118.23 m
6) d = -(51^2)/2(-24)d = 54.19
7) Better brakes have a larger acceleration. Better brakes make for a shorter braking distance.
Prep for Chapter Challenge: Friday, October 28th
1) tr=.5s, vi=11, a=-4, dr=?, db=?, TSD = ?11 = d/.5 (v=reaction distance/reaction time)reaction distance = 5.5 m0 = 121 + 2(-4)dbraking distance = 15.13 mtotal distance = 20.63 m
2) vi=27, dr=?, db =?, TDP =?, a=-4, t=.527=d/.5reaction distance = 13.5 m0 = 729 + 2(-4)dbraking distance = 91.13 mtotal distance = 104.63 m
3) vi=27, t=1, dr=?,db=?, TSD =?, a=-4reaction distance = 27 m0=729 + 2(-4)dbraking distance = 91.13 mtotal distance = 118.13 m
4) t=.5, vi=27, dr=?, db=?, TSD=?, a=-227=d/.5reaction distance = 13.5 m0 = 729 + 2(-2)dbraking distance = -182.25 mtotal distance = 195.75 m
Create a problemWednesday, November 2nd
-Solve for dr, db TSD.-Embed givens in problem.-Don't embed givens that don't need to be mentioned.
My Problem:
A teenager speeds up to a yellow light. What would the reaction time, braking distance , and total stopping distance be for the teenager be if They are traveling at 20 m/s (44 mph) and have a reaction time of .4s, and a deceleration of -5 m/s^2
tr = .4 sVi = 20 m/sa= -5 m/s^2V= d/t20 m/s (.4 s) = dreaction/ .4s (.4s)dreaction = 8 mVf^2 = Vi^2 +2ad-(20 m/s)^2 / -10 = dbrakingdbraking = 40 mTSD = 48 m.
Alex's Problem:
Your driving in your brand new customized car that Bill Trembath bought you . You're going down your street and have a reaction time of .5 seconds and traveling at 25 m/s with a deceleration of -6 m/s^2. What is your reaction distance , braking distance, and total stopping distance?
tr = .5 secondsVi = 25 m/sa= -6 m/s^2V= d/t25 m/s (.5 s) = d/ (.5 seconds) (.5 s)dreaction = 12.5 m
Vf^2= Vi^2 + 2ad-(25 m/s)^2 / -12 = dbraking52.08 m= dbrakingTSD = 64.58 m