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Sportscar acceleration problem

  1. Nov 28, 2007 #1
    A sportscar can accelerate uniformly to [tex] 120 \frac{mi}{h} [/tex] in [tex] 30 \ s [/tex]. Its maximum braking rate cannot exceed [tex] 0.7g [/tex]. What is the minimum time required to go [tex] \frac{1}{2} \ mi [/tex], assuming that it begins and ends at rest.

    So [tex] 120 = 0 + 30a, \ a = 4 [/tex].

    Then [tex] \frac{1}{2} = 2t^{2} [/tex]. But I know I have to incorporate the braking rate.

    How would I proceed from here?
  2. jcsd
  3. Nov 28, 2007 #2


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    Staff: Mentor

    Make sure that the accelerations are in the same units.

    120 mph = 176 ft/s and the acceleration from 0 to 176 ft/s in 30 s is 5.867 ft/s2, and the magnitude of deceleration is 0.7*9.81 ft/s2 = 6.867 ft/s2.

    Now over 0.5 mile or 2640 ft, if the car accelerates over distance d ft, then it must decelerate over distance (2640 - d) ft, and one must find d such that t is minimized, or the average speed is maximized since v(avg) = 2640 ft/t, where t is the time to travel 1/2 mile.
  4. Nov 28, 2007 #3
    So then we have [tex] at_1 - 0.7gt_2 = 0 [/tex] and [tex] 2640 = \frac{1}{2}(at_{1}^{2} - 0.7gt_{2}^{2}) [/tex] and solve for [tex] t_1 [/tex] and [tex] t_2 [/tex]?

    So then [tex] t = t_1 + t_2 [/tex]?
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