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Acceleration confusion ... 
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#1
Jun613, 12:31 PM

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Hi this is my first post...
May be I am doing something wrong... I was thinking about a unit acceleration of 1ms^{2} ,let an object be accelerating 1ms^{2} and at rest the object's velocity was 0ms^{1} and after 5 seconds its velocity was 5ms^{1} ,so it will cover a distance of 15 meters after 5 seconds ,am I right ? If I am right ,then I don't understand the relation given "The total distance traveled is directly proportional to the square of the time",it suppose to give me the distance (15m) if I would squared the time and multiply the total distance covered after 1 second. I got this relation from this site http://www.physicsclassroom.com/Class/1DKin/U1L1e.cfm I think I am doing something wrong... Thanks 


#2
Jun613, 12:47 PM

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(Hint: Use the average velocity over an interval to calculate the distance covered.) 


#3
Jun613, 01:11 PM

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After 1 second it covered 1 meter as the velocity was 1 m/s and after 2 seconds it covered 3 meters as on the second interval of time the velocity was 2 m/s and so on...



#4
Jun613, 01:34 PM

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Acceleration confusion ...
Remember it starts from 0 m/s. After 1 second its speed is 1 m/s, so the average speed during that first second is only 0.5 m/s. Thus it only covers 0.5 meters in the first second. From 1 second to 2 seconds, the speed goes from 1 m/s to 2 m/s, giving an average speed of 1.5 m/s during that interval. And so on. 


#5
Jun713, 02:55 AM

P: 27

Well I feel dumb...may be I am...
But still I don't get it,how can you say 0.5 m/s.... A car is at rest initial velocity 0 m/s ,after one second it covers a distance of 1 meter so the velocity should be 1 m/s during that 1 second interval ... as velocity = displacement/time .Now after the first 1 sec interval , it covers another 2 meters during the next interval of 1 sec ,so its velocity should be 2 m/s during the next interval... Where I am wrong ? And Doc Al thanks for your reply 


#6
Jun713, 03:26 AM

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Having reread your post, I can see you have put your cart before your horse. You need to deal with the speeds before you can calculate the distances. 


#7
Jun713, 03:33 AM

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That's where your problem lies. If a body starts at rest and undergoes constant acceleration of 1 m/s^2, the velocity of 1 m/s is reached only after 1 sec. has elapsed. Remember, the average velocity over a certain interval is the [(initial velocity) + (final velocity)] / 2, and since the body starts at rest, initial velocity = 0 by definition.
Take a closer look at this interval of 1 sec. What is the velocity after 0.5 sec has elapsed? After 0.25 sec has elapsed? The velocity of the body at 0.25 sec. or 0.5 sec. is clearly not 1 m/s. The velocity of the body is continuously increasing from 0 m/s until it reaches 1 m/s at t = 1 sec. During this interval of 1 sec., the body does not travel 1 meter. It cannot unless the velocity remained constant at 1 m/s for the entire interval. Looking at the average velocity, which is (0 + 1)/2 = 0.5 m/s, the distance traveled during the first second of acceleration is 0.5 m. 


#8
Jun713, 03:37 AM

P: 27

Well forget everything, let say a car is at rest ,it starts moving after one second it covers 1 meter ,after 2 seconds it covers 3 meter ,after 3 seconds it covers 6 meter.
I want to know what is the acceleration,velocity in each time interval of the car.... I will go crazy very soon... 


#9
Jun713, 03:59 AM

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Remember that, in the initial stages, it is going very slowly and has to 'make up for it' in the later stages. 


#10
Jun713, 04:17 AM

P: 27

Well I could understand what you guys are trying to say ,a body from rest just cant take a speed of 1 m/s only after 1 sec it will come to that speed,in the first interval I am doing it wrong... but what about this example in this site
http://www.physicsclassroom.com/Class/1DKin/U1L1e.cfm You see the table under the section "The Meaning of Constant Acceleration" I know it says about a free falling object,lets say my car was moving with a speed of 1 m/s rather than being at rest and plug in the rest... what I am doing different from that example @sophiecentaur http://www.physicsclassroom.com/Class/1DKin/U1L1e.cfm You see the table under the section "The Meaning of Constant Acceleration" the acceleration shouldn't be constant. Thanks for all the help guys... 


#11
Jun713, 04:34 AM

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#12
Jun713, 04:55 AM

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If your acceleration is 1 m/s^2, then you have:



#13
Jun713, 05:08 AM

P: 27

I mean this table
What I am confused is what difference I have made .... from that above example... I moved 1 m in the first interval with an average velocity of 1 m/s They moved 5 m in the first interval with an average velocity of 5 m/s I moved 2 m in the second interval with an average velocity of 2 m/s They moved 15 m in the first interval with an average velocity of 15 m/s They are increasing with 10 m/s and I am increasing with 1 m/s ,except that I don't see any much difference ,well I know my one is not correct as the math is not fitting in. I am not able to visualize it,well I will keep thinking on this... Thanks again 


#14
Jun713, 06:12 AM

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#15
Jun713, 06:31 AM

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you are trying to use constant velocity equation v=s/t instead of a constant acceleration equation
here are the variables s=displacement, t=time, u=initial velocity v=final velocity a=acceleration here are the constant acceleration equations s=ut+0.5at^2 v=u+at s=(v+u/2)*t v^2=u^2+2as s=vt0.5at^2 when constant acceleration is involved these equations can be used. if constant velocity then v=s/t can be used. so take a car that is at rest and then accelerates at 1ms^2 for 5 seconds we have this situation  note initial velocity will be zero s=0*t+0.5*1*5 the ut cancels due to u=0 so we have s=0.5*1*5 s=2.5 meters. I hope this has helped? 


#16
Jun813, 04:24 AM

P: 27

OK guys I have been asking so many stupid questions ,so I will take a pause here and stop bothering you people,and try to understand it by myself and then I will post further questions (hope they will be not stupid enough)
Thanks for all your help guys 


#17
Jun1113, 08:57 AM

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#18
Jun1213, 04:15 AM

P: 27

OK I am back with my questions...well here is a new question (related to this thread) I just want to know is it an uniform acceleration or not...
An object moving along a straight path with initial velocity of 5 m/s and a constant acceleration of 5 m/s^2 ,suppose that the object is at a distance of 10 meter from the origin (think about X axis) at time t = 0 ,then is this positiontime data table correct and is it an uniform acceleration.



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