Acceleration confusion

  1. 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. jcsd
  3. Doc Al

    Staff: Mentor

    No, not right. How did you determine that distance?

    That will work. But what's the distance covered after 1 second? :wink:

    (Hint: Use the average velocity over an interval to calculate the distance covered.)
     
  4. 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...

    Code (Text):

    Time          Avg Velocity          Distance Traveled During Time Interval          Total Distance Traveled
    0s - 1s       1m/s                    1m                                                          1m
    1s - 2s       2m/s                    2m                                                          3m
    2s - 3s       3m/s                    3m                                                          6m
    3s - 4s       4m/s                    4m                                                          10m
    4s - 5s       5m/s                    5m                                                          15m
     
    I still don't understand...
     
  5. Doc Al

    Staff: Mentor

    Not quite.

    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.
     
  6. 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
     
  7. sophiecentaur

    sophiecentaur 13,923
    Science Advisor
    Gold Member

    It's clearly not doing 1m/s throughout the first second and it's not doing 0m/s either. The average speed in that interval will be the average: 0.5m/s (if the acceleration is uniform). So it will cover 0.5m in the first second.
    Having re-read 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.
     
  8. SteamKing

    SteamKing 9,621
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
     
    1 person likes this.
  9. 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...
     
  10. sophiecentaur

    sophiecentaur 13,923
    Science Advisor
    Gold Member

    If you want to do it that way round then the answer is, of course, different. If it is accelerating uniformly from rest and covers 1m in the first second then its average speed is 1m/s. Its speed after 1s will be 2m/s (for that average speed to apply) and its acceleration will be 2m/s^2.

    Remember that, in the initial stages, it is going very slowly and has to 'make up for it' in the later stages.
     
  11. 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
    Well if that's the case, in this example of
    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...
     
  12. Doc Al

    Staff: Mentor

    Note that the chart showing distances gives average speed during the intervals, not the speed at the end of the interval.


    There are three tables in that section. Which one do you mean? (Two show constant acceleration, one shows non-constant acceleration.)
     
  13. Doc Al

    Staff: Mentor

    In your chart, you listed the speeds at the end of each interval, not the average speed.

    If your acceleration is 1 m/s^2, then you have:
    Code (Text):

    Time     Speed
    0          0 m/s
    1          1 m/s
    2          2 m/s
    3          3 m/s
    What you need, to match the chart on that webpage, is the average speed in each interval:
    Code (Text):

    Interval     Ave Speed
    0 - 1          0.5 m/s
    1 - 2          1.5 m/s
    2 - 3          2.5 m/s
     
     
    Last edited: Jun 7, 2013
  14. I mean this table

    Code (Text):

    Time Interval   Ave Velocity    Distance During Time Interval       Total Distance Traveled
    0 - 1 s         ~ 5 m/s         ~ 5 m                               ~ 5 m
    1 -2 s          ~ 15 m/s            ~ 15 m                              ~ 20 m
    2 - 3 s         ~ 25 m/s            ~ 25 m                              ~ 45 m
    3 - 4 s         ~ 35 m/s            ~ 35 m                              ~ 80 m
     
    And this is my table
    Code (Text):

    Time          Avg Velocity          Distance Traveled During Time Interval          Total Distance Traveled
    0s - 1s       1m/s                    1m                                                          1m
    1s - 2s       2m/s                    2m                                                          3m
    2s - 3s       3m/s                    3m                                                          6m
    3s - 4s       4m/s                    4m                                                          10m
    4s - 5s       5m/s                    5m                                                          15m
     
    OK they are talking about average velocity ,now let say my average velocity was 1 m/s and I moved 1 meter ,if it can't be then how come a 5 m/s of average velocity makes 5 meter of distance covered.

    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
     
    Last edited: Jun 7, 2013
  15. Doc Al

    Staff: Mentor

    This table is based on uniformly accelerated motion.

    Your table is not based on uniformly accelerated motion.


    There's nothing wrong with traveling 1 meter in 1 second if your average speed during that second was 1 m/s. But that represents an acceleration of 2 m/s^2. To get an average speed of 1 m/s when starting from 0, you must end up at 2 m/s at the end of 1 second.

    You need to start with the time & speed chart I gave in my last post, which was based on a uniform acceleration = 1 m/s^2, and then figure out the average speed in each interval. I gave the chart for the average speeds in my last post. Do you understand how I determined the average speed?
     
    1 person likes this.
  16. 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=vt-0.5at^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?
     
  17. 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
     
  18. PeterO

    PeterO 2,319
    Homework Helper

    Your table, 1m in 1st second, 2m in 2nd second etc is what you get if the acceleration is 1.0 ms-2 but with an initial speed of 0.5 ms-1
     
  19. 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 position-time data table correct and is it an uniform acceleration.

    Code (Text):

    Time          Position from origin
    0s             10.0m
    1s             17.5m
    2s             30.0m
    3s             47.5m
    4s             70.0m
    5s             97.5m
     
    Thanks
     
  20. Hi Peter,I understood that I cant start with 0 m/s ,I need a constant accelerating body or if I take 0 m/s as initial velocity my acceleration will be not constant in that way.

    But what my problem is the relation mentioned in my first post i.e.. "The total distance traveled is directly proportional to the square of the time"

    Thanks
     
  21. PeterO

    PeterO 2,319
    Homework Helper

    Fine:

    The Formula for calculating distance covered can be written as s = ut +0.5at2.

    This shows that displacement (s) is proportional to the square of the time only if the initial velocity (u) is zero

    With constant acceleration, the average speed during any time interval is the average of the initial and final velocity of that interval.

    That means that if the average velocity for the first second was 1 m/s, then the final velocity had to be 2 m/s, since 1 is the average of 0 & 2.

    For the 2nd second (t=1 to t=2) the average velocity cannot be 2 m/s since the initial velocity is already 2 m/s and you want it to increase.

    If the body accelerated from 0 m/s to 2 m/s during the 1st second, it will increase from 2 m/s to 4 m/s during the 2nd second. This means an average velocity of 3 m/s, and a distance covered of 3m.

    So the total distance covered is 4m (1 + 3)

    If you wanted initial velocity of 0 m/s, and acceleration of 1 ms2, then the distance covered each second are 0.5, 1.5, 2.5, .. and total distances covered of 1/2, 2, 4.5, 8 etc.
     
    1 person likes this.
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

0
Draft saved Draft deleted