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Finding unknown velocity and time from distance and accelerations

  1. Jan 14, 2014 #1
    the problem goes like this
    YOu are asked to start from rest, go a distance L=105 m and end up at rest again all in minimum time. Your vehicle has a maximum acceleration of a1 and maximum braking of a2. Determint t1, t2 and maximum speed. a1=7.5m/s^2 , a2=-10 m/s^2. The plot of accleration versus time is attached

    I have attempted the solution using the veocity equations but i have 3 unknows the velocity and the times. Tried deriving an expression for accelertionand elocity but not getting it. stuck how to go ahead. Please help
     

    Attached Files:

  2. jcsd
  3. Jan 14, 2014 #2
    Show us what you did so far, please.
     
  4. Jan 14, 2014 #3
    till t1 i found the velocity as 7.5t1 and from t1 to t2 the velocity as -10 (t2-t1). I dont know how to go ahead from this
     
  5. Jan 14, 2014 #4
    For the case of constant acceleration, do you know the basic algebraic equations for velocity and distance as a function of time, given the initial velocity v0 and the initial distance x0? Do you know the relationship between velocity and distance (without time in the equation) for the case of constant acceleration? Please write these equations down for us. These will be the equations that you will be working with.

    Chet
     
  6. Jan 14, 2014 #5
    for constant acceleration, velocity and time is given by v2=v1+at and velocity and distance is given by v2^2=v1^2+2as where s is displacement
     
  7. Jan 14, 2014 #6
    Good. Ok. The one other equation I'd like to see is distance versus time. Question before we begin: Is that 105 m the distance covered during t1, or is it the total distance covered from start to stop?

    Chet
     
  8. Jan 14, 2014 #7
    Well that is what I qm confused about... because what I have posted is exactly the question that I have. .. not able to figure out if its total distance or not
     
  9. Jan 14, 2014 #8
    Of course, it wouldn't be too complicated to do it both ways. Let's first try the version where it's the total distance. Let x = distance traveled during t1 and let (105 - x) = distance traveled during t2.

    Let's first focus on the time interval t1. If a1 is the acceleration during this time interval, how long does it take (t1) for the vehicle to travel from rest at x = 0 to the location x? What is it's velocity when it reaches x?

    Chet
     
  10. Jan 14, 2014 #9
    Velocity at t1 will be either in terms of t1 or distance x
    V(t1)^2=2a1x
    V(t1)=a1t1
     
  11. Jan 14, 2014 #10
    What is t1 in terms of a and x?
    What is V(t1) in terms of a and x?
     
  12. Jan 14, 2014 #11
    T1 wud be sqrt of x/a1
     
  13. Jan 14, 2014 #12
    Actually, it would be
    [tex]t_1=\sqrt{\frac{2x}{a_1}}[/tex]
    The velocity at the end of t1 would be
    [tex]v(t_1)=a_1t_1=\sqrt{2a_1x}[/tex]
    Does this make sense so far? If so, we are ready to start the time interval t2.
     
  14. Jan 14, 2014 #13
    Yep it does... fot the interval t2 the time will be t2-t1 right
     
  15. Jan 14, 2014 #14
    Not really. We are starting the second half of the problem, and we need to use the equation:
    [tex]s=s_0+v_0t+\frac{1}{2}a_2t^2[/tex]
    For this time interval, s0=x and [itex]v_0=\sqrt{2a_1t_1}[/itex] where we restart the clock at the end of t1 to t = 0. Also, the velocity during this time interval is given by: v=v0+a2t

    So, when t = t2, v = 0 and s = 105. Substitute these values into the above equations and solve for t2 and then x.
     
  16. Jan 14, 2014 #15
    But I am getting two unknows now.. t2 and x
     
  17. Jan 14, 2014 #16
    Yes, but you have two equations. Solve the velocity equation for t2 in terms of x, and then substitute that into the displacement equation.

    Chet
     
  18. Jan 14, 2014 #17
    Still not able to get it...seems to be easy but not getting it
     
  19. Jan 14, 2014 #18
    [itex]v_0=\sqrt{2a_1t_1}[/itex] and 0 = v0+a2t2.

    1. Solve for t2
    2. Substitute into 105 = v0t2+0.5a2(t2)2
    3. Solve for x
     
  20. Jan 14, 2014 #19
    But we still have t1 as unkown
     
  21. Jan 14, 2014 #20
    Got the answer.. thank you for your help
     
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