1. ## integration question

Question:

A particle starts from rest and moves in a straight line. Its speed for the first 3 s is proportional to $\displaystyle (6t-t^2)$, where t is the time in seconds from the commencement of motion, and thereafter it travels with uniform speed at the rate it had acquired at the end of the 3rd second. Prove that the distance travelled in the first 3 s is two-thirds of the distance travelled in the next 3 s.

I understand the first 3 seconds is proportional to speed thus
t = 3 $\displaystyle \propto (6t-t^2)$
thus speed $\displaystyle = k(6t-t^2)$
from here i am stuck I dont know what to do

2. ## Re: integration question

$v(t) = c (6t - t^2),~t \in [0,3]$

$v(t) = v(3),~t \in (3, \infty)$

$s(3) = \displaystyle \int \limits_0^3~v(t)~dt = \left . 3c t^2 - \dfrac{ct^3}{3} \right|_{t=0}^{t=3}$

$s(3) = 18c$

$s(6) = s(3) + 3 v(3) = 18c + 3c(18-9) = 45c$

$\dfrac{s(3)}{s(6)-s(3)} = \dfrac{18c}{45c - 18c} = \dfrac{18}{27} = \dfrac{2}{3}$