Hello,

I've hit rough terrain again, unfortunately. Before I begin I've posted the question I'm stuck on. I've solved partibut partiiinvolves parti, respectively.

Q: A body starts at $\displaystyle 2\ ms^{-1}$ and moves with constant acceleration for 4 metres. If it then has a velocity of $\displaystyle 6\ ms^{-1}$ find:

i)how long it will take to move the next 4 m? [A = 0.56 s]

ii)how far it will move in the next 3 seconds? [A = 36 m]

For partiof the question:

Let $\displaystyle u = 2\ ms^{-1} $,$\displaystyle v = 6\ ms^{-1} $ and $\displaystyle s = 4m $

Then, $\displaystyle s = \left(\frac{u + v}{2}\right)\cdot t \Rightarrow t = \left(\frac{2s}{u + v}\right) = \left(\frac{(2)(4)}{6 + 2}\right) = 1\ second $

Now, $\displaystyle v = u + at $ therefore, $\displaystyle a \Rightarrow \left(\frac{v - u}{t}\right) = \left(\frac{6 - 2}{1}\right) = 4\ ms^{-2} $

$\displaystyle v = \sqrt{u^{2} + 2as} = \sqrt{2^{2} + [(2)(4)(8)]} = 2\sqrt{17}\ ms^{-1}\ after\ 8m $

From $\displaystyle v = u + at $, $\displaystyle t \Rightarrow \left(\frac{v - u}{a}\right) = \left(\frac{2\sqrt{17} - 6}{4}\right) = \left(\frac{-3 + \sqrt{17}}{2}\right) \approx 0.56\ seconds $

For partiiof the question:

Let $\displaystyle s = ut + \frac{1}{2}at^{2} $, then $\displaystyle s = [(2\sqrt{17})(3)] + \left[\left(\frac{1}{2}\right)(4)(3^{2})\right] \approx 42.73\ m \neq 36\ m $

I don't understand why this does not equate to 36 m? Is the initial velocity for the next 3 seconds $\displaystyle 2\sqrt{17} $?