Question

a vechile starts travelling from point a with an inital velocity 5.0m/s and uniform acceleration 2m/s² and attains a velocity of 25m/s when reached at point b find the distance from b to a

Answer 1

T=?
V=u+at.
25=5+2t
25-5=2t
20=2t
20/2=t
T=10


S=ut+1/2at^2
=50+1/2×2×100
50+100
S=150m
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Answer 2

initial velocity "u"= 5m/s
final velocity "v"=25m/s
acceleration "a"=2m/s^2
so,by third equation of motion. .
v^2=u^2+2 as
=> 25^2=5^2+2×2×s
=>625=25+4s
=> 4s=625-25
=>s=600/4
=>s=150..

so the distance between b to a = 150 m..

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