numerical

Asked by  | 22nd Jun, 2008, 01:47: PM

Expert Answer:

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.

 

Answered by  | 22nd Jul, 2008, 10:32: PM

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