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Applied maths...

  • 01-10-2015 6:16pm
    #1
    UhmMaths?!
    Registered Users, Registered Users 2 Posts: 12 ✭


    I'm stuck on a question guys:

    Q2 page34

    A car can move with acceleration 3a and deceleration 5a. Find in terms of k, the time taken to cover a distance of 90ak^2 (k is squared) from rest to rest.

    i) subject to speed limit of 15ak
    ii) subject to no speed limit.

    Help is much appreciated guys!


Comments

  • #2
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    i)
    How long will the car take to get from 0 to 15ak? (v = u + at1 with "v" = 15ak, "u" = 0, "a" = 3a, solve for t1)
    How far will the car travel in this time (s = ut + ½at², again "u"=0, "a" = 3a, t is the t1 you found above.

    How long will the car take to get from 15ak to 0? (v = u + at2 with "v" = 0, "u" = 15ak, "a" = -5a, solve for t2)
    How far will the car travel in this time (s = ut + ½at², again "u"=15ak, "a" = -5a, t is the t2 you found above.

    Work out the remainder of the above two distances from 90ak².
    How long (at 15ak) will it take to travel this distance?

    ii)
    For this case, the car travels while accelerating at 3a to some point x (we don't really care where this is, just how long it take to get to this speed) and then decelerates (at -5a) to rest at 90ak².
    Let the time it spends to this max point be t. It will have reached speed 3at. At this point, to decelerate from 3at to zero at -5a will take 3t/5. (This makes sense; if you accelerate at only 3/5ths of the speed you brake at, it will only take you 3/5ths of the time to decelerate back to rest / zero.

    Anyway, what do we now know?
    The car has been travelling from rest to some speed 3at for t seconds and then from 3at to rest in 3t/5 seconds. How far has it gone during this time?
    Break it down into two periods:
    Under acceleration - it travels s1 = ut + ½at² with "u" = 0, "a" = 3a and "t" = 3at
    Under deceleration - it travels s2 = ut + ½at² with "u" = 3at, "a" = -5a and "t" = 3t/5

    The sum of the two distances s1 and s2 should be 90ak². Set them equal and solve for t in terms of k


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