A train starting from rest with a uniform acceleration of 0.2m/s² for five minutes calculate the final velocity acquired the distance in ts time
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u = 0
a = 0.2 m/s^2
t= 5 min = 5×60 = 300 s
v=?
Using first equation of motion, we can find it,
v-u=at
v-0 = 0.2 × 300
v = 60 m/s
s=?
Using third equation of motion, we can find it,
s= 1/2 at^2 + ut
s= 1/2 × 0.2 × 300×300
s= 9000 m
__________________
Hope it helps...!!!
___________________
u = 0
a = 0.2 m/s^2
t= 5 min = 5×60 = 300 s
v=?
Using first equation of motion, we can find it,
v-u=at
v-0 = 0.2 × 300
v = 60 m/s
s=?
Using third equation of motion, we can find it,
s= 1/2 at^2 + ut
s= 1/2 × 0.2 × 300×300
s= 9000 m
__________________
Hope it helps...!!!
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