a man crosses a 90 m long straight track with a uniform acceleration in 6 s . if his initial velocity is 3m/s ,then he leaves the track with velocity
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Distance traveled my man(s)=90m
time taken (t)=6s
initial velocity (u)=3m/s
find acceleration (a)= ?
use s=ut+1/2at^2
90=3.6+1/2a.36
90=18+18a
72=18a
a=4 m/s^2
find final velocity (v)= ?
use v=u+at
3+4.6
3+24
v=27 m/s
time taken (t)=6s
initial velocity (u)=3m/s
find acceleration (a)= ?
use s=ut+1/2at^2
90=3.6+1/2a.36
90=18+18a
72=18a
a=4 m/s^2
find final velocity (v)= ?
use v=u+at
3+4.6
3+24
v=27 m/s
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