1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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]?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook