Question

A train starting from rest atttains a velocity of 108km/h in 2 minutes. Assuming the acceleration to be uniform find the value of ( I) acceleration (ii) distance covered

Answer 1

u=0m/s t= 10 mins=600 s (10×60s)
v=108km/h=(108×5/18)m/s=30m/s
a=?
S=?

From 1st equation of motion,
v=u+at
→v-u=at
→v-u/t=a
→a=v-u/t
→a=30-0/600
→a=0.05m/s^2

Now,
From 2nd equation of motion,
S=ut+1/2at^2
→S=0×600+1/2×0.05×(600)^2
→S=9000m
Hence,
the acceleration is
0.05m/s^2
and the distance travelled is 9000 m

Hope this will help you :-)

Answer 2

Vf = vi + at
A = vf - vi ÷t
A = 30 m/s - 0 ÷ 120 sec
A = 0.25 m/s2



2 as = vf2 - vi2
2× 0.25× s = (30)2
0.5 × s = 900
S = 900÷ 0.5
S = 1800 meter


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