Question

5) A train the begins its journey from a station with a
uniform acceleration of 0.2 m/s^2 -and moves for 1 minuets
Next it moves for a further 5 min at uniform velocity. It then
applies brakes producing a retardation of 0.3m/s^2 & stops at the nest station. What is the distance between the two station?

Answer 1

A= 0.2m/s^2
T= 1 min= 60sec
U= 0m/s
We will apply,
S= ut+1/2at^2= 0 x 60 + 1/2 x 0.2 x 60= 6m
Now, we find it’s velocity
v=u+at= 0 + 0.2 x 60= 12m/s

Now, it moves further for 5mins at uniform velocity
So, t= 5mins= 300 sec
V= 12m/s
Uniform velocity= distance/time
12= distance/300
3600m= distance

Now, retardation of 0.3m/s^2
V=0m/s (it stops after reaching)
U= 12m/s
A= -0.3m/s^2 ( retardation)
V=u+at
➡️ 0= 12 + (-0.3t)
➡️ -12= -0.3t
➡️ -12/-0.3= t
➡️ 12 x 10/3= t
➡️ 40sec= t
Now,
S=ut+1/2at^2= 12 x 40 + 1/2 x 3/10 x 1600= 480 + 240= 720m

Total distance= 6 + 3600 + 720= 4326m

Answer 2

Explanation:

See my answer. I think you will understand.