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How do I find intial acceleration?

  1. Feb 12, 2012 #1
    You accelerate at a m/s2 for 4.0 s, then coast for 8.0 seconds, then
    decelerate to a standstill in 4.0 s. In total, you travel 144 meters. What was your
    initial acceleration?

    Please show your work as i need to learn from you.
     
  2. jcsd
  3. Feb 12, 2012 #2

    ojs

    User Avatar

    Well, we don't usually do the work for the students here but we try to nudge them in the right direction.

    The right direction here is how would you write up an equation for the distance traveled as a function of the acceleration a (what is the deceleration as a function of a? can you tell that by the numbers given?).
     
  4. Feb 12, 2012 #3
    this is what i did ΔX = v_0t+.5at^2

    Given the total Traveled ΔX= 144m v_0=0 t=4s

    My equation looks something like this

    144m=0+(4s)+.5a(4s)^2

    144m=8s^2a

    a=18m/s^2

    but i think this is wrong because i took the total (144m) traveled. I don't know how to get the acceleration at 0m and 4s.
     
  5. Feb 12, 2012 #4

    ojs

    User Avatar

    Ok, that's a start.

    But your equation assumes that the whole 144m are traveled in the initial 4 seconds which is not true, the 144 m are traveled in 16 seconds but with varying speed (except between 4 and 12 seconds) so you have to break the distance down into three parts and each part gets its own term.

    Can you see how you get the velocity at 4 seconds (and thus at 12 seconds also) as a function of a?
     
  6. Feb 12, 2012 #5
    I surely don't understand how to break the distance into 3 parts. Unless of course to divide:

    144/3 = 48 m?
     
  7. Feb 12, 2012 #6

    ojs

    User Avatar

    Ok then, what you have to do here is write up 3 terms that all show the distance traveled as a function of a.

    The first 4 seconds the distance traveled is 0.5at^2, then you have to add a second term that shows the distance traveled during the next 8 seconds as a function of a and lastly a term that describes the distance traveled during the last 4 seconds.

    so it will be a function like this one here:

    144m = 0.5a*(t1)^2 + v1*t2 + (v1_0*t3 + 0.5a3*(t3)^2)

    where a is our initial acceleration, t1 is the initial 4 seconds, v1 is the speed after 4 seconds, t2 is the 8 second interval and the last part in the parenthesis is the distance traveled during the last 4 seconds denoted by t3, v1_0 is the initial velocity of that part (how is that related to v1?) and a3 is the deceleration (how is that related to a?).

    Can you continue from here?
     
  8. Feb 12, 2012 #7

    ojs

    User Avatar

    Ok then, what you have to do here is write up 3 terms that all show the distance traveled as a function of a.

    The first 4 seconds the distance traveled is 0.5at^2, then you have to add a second term that shows the distance traveled during the next 8 seconds as a function of a and lastly a term that describes the distance traveled during the last 4 seconds.

    so it will be a function like this one here:

    144m = 0.5a*(t1)^2 + v1*t2 + (v1_0*t3 + 0.5a3*(t3)^2)

    where a is our initial acceleration, t1 is the initial 4 seconds, v1 is the speed after 4 seconds, t2 is the 8 second interval and the last part in the parenthesis is the distance traveled during the last 4 seconds denoted by t3, v1_0 is the initial velocity of that part (how is that related to v1?) and a3 is the deceleration (how is that related to a?).

    Can you continue from here?
     
  9. Feb 12, 2012 #8
    What formula is that?
     
  10. Feb 12, 2012 #9

    ojs

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    That is the well known formula

    x = x_0 + v_0*t + 0.5*a*t^2

    used in all three parts of the journey.

    The first term in my first equation is the last part of this formula, that describes the distance traveled during the first 4 seconds.
    The second term in my first equation is the second part of this formula, it describes the distance traveled during the 8 seconds that object has constant speed.
    The third term in my first equation (the one inside the parenthesis) is the distance traveled during the last 4 seconds with v1 being the initial speed when it starts to decelerate and a3 being the deceleration.
     
  11. Feb 12, 2012 #10

    ojs

    User Avatar

    That is the well known formula

    x = x_0 + v_0*t + 0.5*a*t^2

    used in all three parts of the journey.

    The first term in my first equation is the last part of this formula, that describes the distance traveled during the first 4 seconds.
    The second term in my first equation is the second part of this formula, it describes the distance traveled during the 8 seconds that object has constant speed.
    The third term in my first equation (the one inside the parenthesis) is the distance traveled during the last 4 seconds with v1 being the initial speed when it starts to decelerate and a3 being the deceleration.
     
  12. Feb 12, 2012 #11
    Thank you ojs.

    my professor never gave us this formula. Hence, i have a problem with this question and similar questions as this.
     
  13. Apr 13, 2012 #12
    Hi guys, just wondering if anyone could help me out?
     
  14. Apr 13, 2012 #13
    The distance covered by each crate is 20m. Assuming that each crate starts at rest, calculate the speed of each crate at the end of the 20m journey.

    I have used v2 = u2 + 2as

    u = start speed = 0ms-1
    s = distance travelled = 20m
    a = acceleration = 2.4148ms-2
    v2 = final speed ms-1

    how would i show the final speed? v2 = 9.8281 ms-1 or
    v = 96.592 ms-1
     
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