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IR  speed [#permalink]
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22 Jul 2013, 15:56
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A car is traveling on a straight stretch of roadway, and the speed of the car is increasing at a constant rate. At time 0 seconds, the speed of the car is V 0 meters per second; 10 seconds later, the front bumper of the car has traveled 125 meters and the speed of the car is V 10 meters per second. In the table below, select values of V 0 and V 10 that are together consistent with the information provided. Make only two selections, one in each column. My doubt is: The question indicates that the speed of the car is increasing at a constant rate. So, is it talking about an arithmetic sequence, or a geometric sequence? According to the OE, the average speed is \(\frac{1}{2}*( V0 + V10)\). So, it seems that it is talking about an arithmetic sequence because that's the way we use to calculate the average in an arithmetic sequence. Please confirm. However, I remember that, when a question mentions that something is increasing at a constant rate, we must multiply the first value by a constant, we shouldn't add. Please, your help.OA:
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Re: IR  speed [#permalink]
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31 Jul 2013, 09:52
danzig wrote: A car is traveling on a straight stretch of roadway, and the speed of the car is increasing at a constant rate. At time 0 seconds, the speed of the car is V 0 meters per second; 10 seconds later, the front bumper of the car has traveled 125 meters and the speed of the car is V 10 meters per second. In the table below, select values of V 0 and V 10 that are together consistent with the information provided. Make only two selections, one in each column. My doubt is: The question indicates that the speed of the car is increasing at a constant rate. So, is it talking about an arithmetic sequence, or a geometric sequence? According to the OE, the average speed is \(\frac{1}{2}*( V0 + V10)\). So, it seems that it is talking about an arithmetic sequence because that's the way we use to calculate the average in an arithmetic sequence. Please confirm. However, I remember that, when a question mentions that something is increasing at a constant rate, we must multiply the first value by a constant, we shouldn't add. Please, your help.OA: t=0 speed V0 t=10 " V10 speed is raised at const rate. SO, V0, V1 ,....V10 are in AM Let V10= V0 + (101)X .....X is increment for every sec Since d=SXt 125= (V0(1sec) +V1(1sec) +...V10(1sec So, 125= V0+ V0+1X + V0+2X +...V0+9X = 10V0 + (1+2+3+..9)X= 10V0 + (9*5)X But, 9X=V10V0 So, 125 = 5V0+5V10.................V0+v10=25 (5+20)



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Re: IR  speed [#permalink]
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21 Sep 2013, 21:39
mbmanoj wrote: danzig wrote: A car is traveling on a straight stretch of roadway, and the speed of the car is increasing at a constant rate. At time 0 seconds, the speed of the car is V 0 meters per second; 10 seconds later, the front bumper of the car has traveled 125 meters and the speed of the car is V 10 meters per second. In the table below, select values of V 0 and V 10 that are together consistent with the information provided. Make only two selections, one in each column. My doubt is: The question indicates that the speed of the car is increasing at a constant rate. So, is it talking about an arithmetic sequence, or a geometric sequence? According to the OE, the average speed is \(\frac{1}{2}*( V0 + V10)\). So, it seems that it is talking about an arithmetic sequence because that's the way we use to calculate the average in an arithmetic sequence. Please confirm. However, I remember that, when a question mentions that something is increasing at a constant rate, we must multiply the first value by a constant, we shouldn't add. Please, your help.OA: t=0 speed V0 t=10 " V10 speed is raised at const rate. SO, V0, V1 ,....V10 are in AM Let V10= V0 + (101)X .....X is increment for every sec Since d=SXt 125= (V0(1sec) +V1(1sec) +...V10(1sec So, 125= V0+ V0+1X + V0+2X +...V0+9X = 10V0 + (1+2+3+..9)X= 10V0 + (9*5)X But, 9X=V10V0 So, 125 = 5V0+5V10.................V0+v10=25 (5+20) Doubt : I agree with "Speed is raised at const rate. SO, V0, V1 ,....V10 are in AM" But not with "Let V10= V0 + (101)X .....X is increment for every sec," which will give V1=V0 for t=1 seconds i.e. V1=V0+X, V2=V0+2X, V3 = V0+3X, so.. V10 = V0+10X Please comment.



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Re: IR  speed [#permalink]
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23 Oct 2013, 11:56
danzig wrote: My doubt is: The question indicates that the speed of the car is increasing at a constant rate. So, is it talking about an arithmetic sequence, or a geometric sequence? According to the OE, the average speed is \(\frac{1}{2}*( V0 + V10)\). So, it seems that it is talking about an arithmetic sequence because that's the way we use to calculate the average in an arithmetic sequence. Please confirm. However, I remember that, when a question mentions that something is increasing at a constant rate, we must multiply the first value by a constant, we shouldn't add. Please, your help.
Dear Danzig & Mbmanoj & Chakdum, This is tricky. When a car is increasing at constant acceleration, then it is NOT a sequence, either arithmetic or geometric. Thinking about the motion in terms of a sum of what happens in each second is not helpful, and in particular, thinking of it as a sum of constantmotion chunks each second is DEAD WRONG. If it goes from, say 5 m/s to 15 m/s in 10 seconds, then the acceleration is 2 m/s^2, but that does not mean: 5 m/s for the duration of the 1st second, 7 m/s for the duration of the 2nd second, etc. That is a complete misunderstanding of the nature of acceleration. Instead, the formula given in the OE, average velocity = (vo + vf)/2, is always correct for constant acceleration. Here's one way to think about that formula. Think about the graph of speed vs. time. The speed is continuously increasing from (vo) to (vf). Attachment:
constant acceleration v vs. t.JPG [ 16.64 KiB  Viewed 3937 times ]
The diagonal dark green line is the graph of the speed vs. time for this object. The slope of this line is the acceleration. On a speed vs. time graph, the area under the curve equals the distance traveled, so that brickred region should have an area equal to the total distance traveled. That brickred region is a trapezoid, and Area of a Trapezoid = (average of the parallel bases)*(height) Here, the parallel bases are the two vertical segments  the one on the left has a length of (vo) and the one on the right has a length of (vf), so we average those two. The trapezoid is flipped on its edge, so the "height" (i.e. the distance between the two parallel segments) is the horizontal length at the bottom  10 s. Thus Area = (10 s)*(vo + vf)/2 = distance traveled Now, think about average velocity (AV). We know that this relates total distance (DT) and total time (TT) of any trip. (DT) = (AV)*(TT) DT = 10*(AV) But from the equation above, from the area of the trapezoid, we know DT = 10*(vo + vf)/2 Comparing those two makes immediately clear: AV = (vo + vf)/2 This formula has nothing to do with a sequence of any kind. It comes from the area of a trapezoid! For this problem, it is a very useful shortcut: DT = 125 TT = 10 125 = 10*(vo + vf)/2 125 = 5*(vo + vf) 25 = (vo + vf) So we just need to numbers that have a sum of 25. Does all this make sense? Mike
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Re: IR  speed [#permalink]
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28 Sep 2014, 05:57
mikemcgarry wrote: danzig wrote: My doubt is: The question indicates that the speed of the car is increasing at a constant rate. So, is it talking about an arithmetic sequence, or a geometric sequence? According to the OE, the average speed is \(\frac{1}{2}*( V0 + V10)\). So, it seems that it is talking about an arithmetic sequence because that's the way we use to calculate the average in an arithmetic sequence. Please confirm. However, I remember that, when a question mentions that something is increasing at a constant rate, we must multiply the first value by a constant, we shouldn't add. Please, your help.
Dear Danzig & Mbmanoj & Chakdum, This is tricky. When a car is increasing at constant acceleration, then it is NOT a sequence, either arithmetic or geometric. Thinking about the motion in terms of a sum of what happens in each second is not helpful, and in particular, thinking of it as a sum of constantmotion chunks each second is DEAD WRONG. If it goes from, say 5 m/s to 15 m/s in 10 seconds, then the acceleration is 2 m/s^2, but that does not mean: 5 m/s for the duration of the 1st second, 7 m/s for the duration of the 2nd second, etc. That is a complete misunderstanding of the nature of acceleration. Instead, the formula given in the OE, average velocity = (vo + vf)/2, is always correct for constant acceleration. Here's one way to think about that formula. Think about the graph of speed vs. time. The speed is continuously increasing from (vo) to (vf). Attachment: constant acceleration v vs. t.JPG The diagonal dark green line is the graph of the speed vs. time for this object. The slope of this line is the acceleration. On a speed vs. time graph, the area under the curve equals the distance traveled, so that brickred region should have an area equal to the total distance traveled. That brickred region is a trapezoid, and Area of a Trapezoid = (average of the parallel bases)*(height) Here, the parallel bases are the two vertical segments  the one on the left has a length of (vo) and the one on the right has a length of (vf), so we average those two. The trapezoid is flipped on its edge, so the "height" (i.e. the distance between the two parallel segments) is the horizontal length at the bottom  10 s. Thus Area = (10 s)*(vo + vf)/2 = distance traveled Now, think about average velocity (AV). We know that this relates total distance (DT) and total time (TT) of any trip. (DT) = (AV)*(TT) DT = 10*(AV) But from the equation above, from the area of the trapezoid, we know DT = 10*(vo + vf)/2 Comparing those two makes immediately clear: AV = (vo + vf)/2 This formula has nothing to do with a sequence of any kind. It comes from the area of a trapezoid! For this problem, it is a very useful shortcut: DT = 125 TT = 10 125 = 10*(vo + vf)/2 125 = 5*(vo + vf) 25 = (vo + vf) So we just need to numbers that have a sum of 25. Does all this make sense? Mike Awesome explanation! Thank you!
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Re: IR  speed [#permalink]
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24 Nov 2017, 09:59
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mikemcgarry, Would you mind briefly commenting my following logic? We are told that the speed of the car is increasing at a constant rate with respect to time, that time is 10 seconds and that the car has traveled 125 meters. Therefore, I conclude that my Speed at V0 will be lower than 12.5 meters (125/10 = 12.5 meters / second). Then there is only one choice for V0 in the table. V0 = 5.Then I know I need the rate to constantly increase until V10 and need to get to 125 meters. (V10 + V0)/2 = 12.5 meters/s (V10 + 5)/2 = 12.5 V10 + 5 = 12.5*2 V10 = 25  5 = 20 Thanks! mikemcgarry wrote: danzig wrote: My doubt is: The question indicates that the speed of the car is increasing at a constant rate. So, is it talking about an arithmetic sequence, or a geometric sequence? According to the OE, the average speed is \(\frac{1}{2}*( V0 + V10)\). So, it seems that it is talking about an arithmetic sequence because that's the way we use to calculate the average in an arithmetic sequence. Please confirm. However, I remember that, when a question mentions that something is increasing at a constant rate, we must multiply the first value by a constant, we shouldn't add. Please, your help.
Dear Danzig & Mbmanoj & Chakdum, This is tricky. When a car is increasing at constant acceleration, then it is NOT a sequence, either arithmetic or geometric. Thinking about the motion in terms of a sum of what happens in each second is not helpful, and in particular, thinking of it as a sum of constantmotion chunks each second is DEAD WRONG. If it goes from, say 5 m/s to 15 m/s in 10 seconds, then the acceleration is 2 m/s^2, but that does not mean: 5 m/s for the duration of the 1st second, 7 m/s for the duration of the 2nd second, etc. That is a complete misunderstanding of the nature of acceleration. Instead, the formula given in the OE, average velocity = (vo + vf)/2, is always correct for constant acceleration. Here's one way to think about that formula. Think about the graph of speed vs. time. The speed is continuously increasing from (vo) to (vf). Attachment: constant acceleration v vs. t.JPG The diagonal dark green line is the graph of the speed vs. time for this object. The slope of this line is the acceleration. On a speed vs. time graph, the area under the curve equals the distance traveled, so that brickred region should have an area equal to the total distance traveled. That brickred region is a trapezoid, and Area of a Trapezoid = (average of the parallel bases)*(height) Here, the parallel bases are the two vertical segments  the one on the left has a length of (vo) and the one on the right has a length of (vf), so we average those two. The trapezoid is flipped on its edge, so the "height" (i.e. the distance between the two parallel segments) is the horizontal length at the bottom  10 s. Thus Area = (10 s)*(vo + vf)/2 = distance traveled Now, think about average velocity (AV). We know that this relates total distance (DT) and total time (TT) of any trip. (DT) = (AV)*(TT) DT = 10*(AV) But from the equation above, from the area of the trapezoid, we know DT = 10*(vo + vf)/2 Comparing those two makes immediately clear: AV = (vo + vf)/2 This formula has nothing to do with a sequence of any kind. It comes from the area of a trapezoid! For this problem, it is a very useful shortcut: DT = 125 TT = 10 125 = 10*(vo + vf)/2 125 = 5*(vo + vf) 25 = (vo + vf) So we just need to numbers that have a sum of 25. Does all this make sense? Mike
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Re: IR  speed [#permalink]
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27 Nov 2017, 11:13
Hadrienlbb wrote: mikemcgarry, Would you mind briefly commenting my following logic? We are told that the speed of the car is increasing at a constant rate with respect to time, that time is 10 seconds and that the car has traveled 125 meters. Therefore, I conclude that my Speed at V0 will be lower than 12.5 meters (125/10 = 12.5 meters / second). Then there is only one choice for V0 in the table. V0 = 5.Then I know I need the rate to constantly increase until V10 and need to get to 125 meters. (V10 + V0)/2 = 12.5 meters/s (V10 + 5)/2 = 12.5 V10 + 5 = 12.5*2 V10 = 25  5 = 20 Thanks! Dear Hadrienlbb, My friend, I would say that yours is a brilliant and wonderfully elegant solution! Kudos! Mike
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