Solution:

Initial velocity = $v_{i} = 96 kmh^{-1}$ = $96\times\frac{1000}{1000}$ = $\frac{96000}{96000} ms^{-1}$
Final velocity = $v_{f} = 48 kmh^{-1}$ = 48\times $\frac{1000}{3600}$ = $\frac{48000}{3600}ms^{-1}$
Distance = S = 800 m 
Further Distance = S1 =?

First of all, we will find the value of acceleration a 

$2aS = v_{f}^{2} – v_{i}^{2}$
$2 \times a \times800$ = $(\frac{48000}{3600})^{2}$ – $(\frac{96000}{3600})^{2}$ 
1600a = $(\frac{48000}{3600})^{2}$ – $((2 \times \frac{48000}{3600})^{2}$        $(96000 = 2 \times 48000)$
1600a = $(\frac{48000}{3600})^{2} ((1)^{2} – (2)^{2})$        (taking $(\frac{48000}{36000})$ as common)
1600a = $(\frac{48000}{3600})^{2} (1 – 4)$
1600a = $(\frac{48000}{3600})^{2} (-3)$
a = $(\frac{48000}{3600})^{2} \times \frac{3}{1600 }$

Now, we will find the value of further distance S2:
$V_{f}$ = 0              ,           S2 =?
2aS = $v_{f}^{2} – v_{i}^{2}$
$-2 (\frac{48000}{36000})^{2}$ \times $\frac{3}{1600}\times$ S1 = $(0)^{2}$- $(\frac{48000}{36000})^{2}$      
S1 = $(\frac{48000}{36000})^{2}$ \times $(\frac{48000}{36000})^{2}$ \times $\frac{1600}{3}\times 2$ 
S1 = $\frac{1600}{6}$
S2 = 266.66m