
how can this be wrong?
a particle is moving along a straight line with constant acceleration. in an interval of T seconds it moves D metres; in the next interval of 3T seconds it moves 9D metres. how far does it move in a further interval of T seconds?
so I calculated the acceleration pretty simply, the velocity increased by 2D/T over 3T, so the constant acceleration is 2/3DT^2. using this I calculated the distance travelled over the next interval of T seconds as being 0.5(3 + 3 + 2/3) = 20/6 D metres
the book says 5D metres. where did i go wrong?

Dear furor celtica,
From $\displaystyle S=ut+\frac{1}{2}at^2 $
$\displaystyle D=uT+\frac{1}{2}aT^2 $
$\displaystyle 10D=4uT+\frac{1}{2}a(4T)^2$
$\displaystyle 10D=4(D\frac{1}{2}aT^2)+\frac{1}{2}a(4T)^2$
By simplification, $\displaystyle a=\frac{D}{T^2}$ and $\displaystyle u=\frac{D}{2T}$
Therefore the distance traveled in the last T seconds,
$\displaystyle S=(\frac{D}{2T}\times{5T})+\frac{1}{2}\frac{D}{T^2 }(5T)^210T$
By simplification, $\displaystyle S=5D$
So the book is correct. You haven't considered the fact that the particle could have an initial velocity which I had taken as u.