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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: wildareal on October 02, 2010, 09:49:58 pm

Title: Lift Question
Post by: wildareal on October 02, 2010, 09:49:58 pm: In a tall building the lift passes the 50th floor with a velocity of -8m/s and an acceleration of 1/9(t-5)m/s^2. If each floor spans a distance of 6m, find at which floor the lift will stop.
Title: Re: Lift Question
Post by: Martoman on October 02, 2010, 10:11:44 pm: $a = \frac{1}{9}(t-5)$
$v = \int \frac{1}{9}(t-5) dt$

$v = \frac{1}{9}(\frac{1}{2} t^2-5t) +c$

t = 0, v = -8

c = -8

We want to know where the lift stops, ie when its final velocity is 0 so:

v = 0,

$0 = \frac{1}{9}(\frac{1}{2} t^2-5t) - 8$

$144 = t^2-10t$

$(t+8)(t-18) = 0$

t = 18 only as t cannot equal a negative number ie -8

So you want to find the total distance covered between t = 0, 18

This is the integral of the velocity function but we need to check by graphing if any negative areas will muck us up.

A graph shows its below the x axis for the interval [0,18] (lovely) so we just take the integral then the negative of it or change the signs

$\int_{18}^{0}\frac{1}{9}(\frac{1}{2}t^2-5t)-8dt = 126$

Now each floor spans 6 meters so 126/6 = 21

assuming its going down then 50-21 = 29th floor.