A particle is moving with uniform acceleration along a straight line ABC. Its velocity of at A and C are 5 m/s 25 m/s respectively. If BC - 2 AB, find the ratio of time of travel from A to B to that from B to C

See figure 

${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as} \mathrm{or} {\mathrm{v}}^2-25=2\mathrm{as}....\left(1\right)$
and $(25{)}^2={\mathrm{v}}^2+4\mathrm{as}$
or 625 -  v² = 4 as ..(2)
Dividing eq. (1) by eq. (2), we get
$\frac{{\mathrm{v}}^2-25}{625-{\mathrm{v}}^2}=\frac{1}{2} \mathrm{or} 2{\mathrm{v}}^2-50=625-{\mathrm{v}}^2$
or 3 v² =675 or v² =  225 or V = 15 m/s
For uniform acceleration, [(v - u) / t]should be constant.
$\begin{array}{l}\therefore \frac{15-5}{{\mathrm{t}}_1}=\frac{25-15}{{\mathrm{t}}_2} \mathrm{or} \frac{10}{{\mathrm{t}}_1}=\frac{10}{{\mathrm{t}}_2}\\ \mathrm{or} \frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}=\frac{1}{1}\end{array}$