A train starting from stationary position and moving with uniform accelaration attains a speed of 36km/hour in 10 minutes.what is the distance travelled?
vamshee:
3000m
Answers
Answered by
39
Given,
initial velocity(u)=0
final velocity(v)=36km/hr=36*5/18(m/s)=10m/s
time(t)=10 min=600 sec
distance(s)=?
from 1st equation of motion,
v=u+at
10=a(600)
a=10/600
a=0.16m/(s^2)
from second equation of motion,
s=ut+1/2*a*(t^2)
=0+1/2(0.16)[(600)^2]
=28800m
=28.8km
therefore,distance=28.8 km
initial velocity(u)=0
final velocity(v)=36km/hr=36*5/18(m/s)=10m/s
time(t)=10 min=600 sec
distance(s)=?
from 1st equation of motion,
v=u+at
10=a(600)
a=10/600
a=0.16m/(s^2)
from second equation of motion,
s=ut+1/2*a*(t^2)
=0+1/2(0.16)[(600)^2]
=28800m
=28.8km
therefore,distance=28.8 km
Answered by
17
u=0
v=36 km/hr=36 X 5/18=10 m/sec
t=10 min=600 sec
v=u+at
a=10/600 = 1/60 m/sec²
s=ut+1/2at²
= 1/2 X 1/60 X 600 X 600
= 30000mts
v=36 km/hr=36 X 5/18=10 m/sec
t=10 min=600 sec
v=u+at
a=10/600 = 1/60 m/sec²
s=ut+1/2at²
= 1/2 X 1/60 X 600 X 600
= 30000mts
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