Answer
$a). 25m,$
$b). Car B.$
Work Step by Step
a). $S=ut+\frac{1}{2}at^{2}$
after 10s, position of car A = $ 2.5\times10+0.5\times3\times100=25+150=175m$
after 10s, position of car B = $5\times10+0.5\times3\times100=50+150=200m$
So, difference in position of the two cars$= 200-175=25m$
b). car A, $v=u+at=2.5+3\times10=32.5 m/s$
car B, $v=u+at=5+3\times10=35m/s$
So, car B is faster at the end of 10s.
