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Sportscar minimum time

  1. Sep 5, 2016 #1
    1. The problem statement, all variables and given/known data
    A sportscar, Electro-Fiasco I, can accelerate uniformly to 100 km/h in 3.5 s. Its maximum braking rate cannot exceed 0.7g. What is the minimum time required to go 1 km, assuming it begins and ends at rest?

    2. Relevant equations
    SUVAT equations

    3. The attempt at a solution
    I think that I solved the problem, but I want some confirmation before I go with my solution.

    Let ##r_m## be the point at which the car begins to decelerate.
    Let ##r_f## be the final point. = 1000 m
    Let ##a_i## be the beginning acceleration. = 7.93 m/s^2
    Let ##a_f## be the final acceleration. -0.7g

    First I derived the following three equations:

    1. ##r_m = \frac{1}{2} a_f t_m^2##
    2. ##r_f - r_m = (a_f t_m)t_f - \frac{1}{2}(a_i) t_f^2##
    3. ##t_m = \frac{a_i}{a_f}t_f##

    Then I combined these to find the following expression ##\displaystyle t_f = \sqrt{\frac{r_f}{\frac{1}{2} a_i + \frac{a_f^2}{2 a_f}}}##

    I then plugged all my numbers in to find that final time is 12.5
    Then the initial time is 10.82

    And so we add these together to find the minimum time 23.32 seconds.

    I need a sanity check. Is this the right answer?
     
  2. jcsd
  3. Sep 5, 2016 #2

    TSny

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    Your approach looks good, but it appears that you have some of the subscripts on the accelerations switched in all of the above equations.

    I believe these values for the times are correct.
     
  4. Sep 5, 2016 #3
    Is ##\displaystyle t_f = \sqrt{\frac{r_f}{\frac{1}{2} a_i + \frac{a_i^2}{2 a_f}}}## the correct formula?

    Also, is there a better way to do this problem? All of the calculations gave me a headache...
     
  5. Sep 5, 2016 #4

    Simon Bridge

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    Verify your reasoning with a velocity-time diagram.
    That way you do not need to rely on getting the right equations: everything boils down to geometry of triangles.

    You are saying that at time ##t_m## the car is at displacement ##r_m## from the origin, and starts to decelerate ... so the acceleration changes from ##a_i## to ##-a_f## this right?
    So for ##0<t<t_m##, the acceleration is ##a_i## ?
    Is that what eq1 says?

    Check derivation for the others too.
     
  6. Sep 5, 2016 #5

    TSny

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    Nice!
     
  7. Sep 5, 2016 #6

    TSny

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    Looks like the subscripts on the accelerations are incorrect. It appears that when you are setting up your equations, you are taking ##a_i## to be the final acceleration and ##a_f## to be the initial acceleration.

    Simon has a clever approach.

    Or you can try to see if invoking the equation ##v_2^2 = v_1^2 + 2a\Delta x## helps shorten the calculation.
     
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