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Velocity, Speed, and Time

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Driving along a crowded freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase your speed by 4.5 mi/h, the time to go one mile decreases by 10 s. What was your original speed?


    2. Relevant equations

    V=D/T

    3. The attempt at a solution

    V1T1 = 1 mi
    v2t2= i mile
    v2= v1 + change of V
    t2= t1 - change of t
    I then used equations 3 and 4 to eliminate v2 and t2 in equation 2.
    (v1 + change of V)(t1 - change of t) = 1 mi

    I know that i need to get rid of t1 now, and i should probably use the first equation (v1t1 = 1 mi) to do it. I just dont know how to integrate the two of them.
    Thank you in advance
     
  2. jcsd
  3. Feb 10, 2013 #2

    tms

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    You have two equations and two unknowns: [itex]v_1[/itex] and [itex]t_1[/itex]. Solve them in the usual way: solve one for one of the unknowns and substitute the result in the other equation.
     
  4. Feb 10, 2013 #3
    That's what i thought i was doing. by putting in v1t1=1mile into (v1 + change in v)(t1-change in t). but would that be (v1 + change in v)(v1 - change in t)? or (v1 + change in v)((1/v1) - change in t)?
     
  5. Feb 10, 2013 #4

    mfb

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    Staff: Mentor

    The second one (if you add units), as V1T1=1mile is equivalent to T1=1mile/V1.
     
  6. Feb 10, 2013 #5
    Ah ok. so then i just simplify that equation to find v1.
    which would be.. (v12)(Δt) + (ΔvΔt) - (Δv)(1 mi) = 0

    so then i do quadratic?
     
  7. Feb 10, 2013 #6

    mfb

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    That is a good idea, indeed. But the units of the last equation are wrong, there is some error in the calculation.
     
  8. Feb 10, 2013 #7
    (Δt)v12 + (Δv)(Δt)v1 - (Δv)(1mi) = 0

    fixed???
     
  9. Feb 10, 2013 #8

    mfb

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    Now the units match. I didn't check the equations, but you can check the result yourself afterwards with known v1 and t1.
     
  10. Feb 10, 2013 #9
    i got the right answer.
    thank you
     
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