Originally Posted by

**Air** **Question:**

A particle moves along a straight line starts at time $\displaystyle t=0$ seconds with a velocity $\displaystyle 4 \ ms^{-1}$. At any subsequent time $\displaystyle t$ seconds the acceleration of the particle is $\displaystyle (6t-8) \ ms^{-2}$. Find:

(a) The distance moves before first coming to instantaneous rest,

(b) the total time $\displaystyle T$ seconds taken by the particle to return to the starting point,

(c) the greatest speed of the particle for $\displaystyle 0 \le t \le T$.

**My Answers (Without Method):**

(a) $\displaystyle x = \frac{32}{27} \ m$

(b) $\displaystyle T = 2 \ s$

**My Problem in this question:**

I don't understand part (c). Here is my method:

$\displaystyle v = 3t^2 - 8t + 4$

$\displaystyle \therefore$ Greatest speed when $\displaystyle \frac{\mathrm{d}v}{\mathrm{d}t} = 0$. (At turning point?)

$\displaystyle \therefore \frac{\mathrm{d}v}{\mathrm{d}t} = 6t-8=0 \implies t = \frac43$

$\displaystyle \therefore v = 3\left(\frac43\right)^2 - 8\left(\frac43\right) + 4 = -\frac43$

$\displaystyle \text{Speed} = |v| = \frac43$ **?!!! **

How would part (c) be done? The correct answer is $\displaystyle 4 \ ms^{-1}$. Thanks in advance.