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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