Explanation

(b) $\therefore\ v = |t - 2|ms^{- 1}$
$v = t - 2,$ when $t > 2s$
$v = 2 - t$, when $t < 2s$
$\therefore\ a = \frac{dv}{dt} = 1ms^{- 1}$ when $t > 2s$
$a = - 1ms^{- 1}$ when $t < 2$s
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In the direction of motion from $A$ to $C$, bee decelerates but for $C$ to $B$, bee accelerates.
Let $AC = s_{1},BC = s_{2}$
$u_{A} = 2ms^{- 1},t = 0$
$u_{c} = 0\ $at $t = 4$s
$\therefore\ s_{1} = \left( \frac{u_{A} + u_{C}}{2} \right)t_{1}$
$s_{2} = \left( \frac{u_{C} + u_{B}}{2} \right)t_{2}$
$\therefore\ s = s_{1} + s_{2} = \left( \frac{2 + 0}{2} \right)2 + \left( \frac{0 + 2}{2} \right)2 = 4$m