numerical

For the first car:

v=u-at

or 0=15-a(5)

or a=3 ms^-2

So using v²-u²=2as

s=(15)² /6=37.5 m

For the second car:

v=u-at

or 0=5/6 -a(10)

or a=1/12 ms^-2

So using v²-u²=2as

s=(5/6)² /(1/6)=4.16 m

 

Hence second car travels further after application of brakes.

Plese post other parts of question separately.