A subway train starts from rest at a station and accelerates at the rate of 2 metre per second square for 10 seconds. It then runs at a constant speed for 30 sec and decelerates a 4 m/s^2 until it stops at next station.Find the distance between the two stations?
Answers
Answered by
45
see the above images.......
Attachments:
sanman313:
Hi, can you please explain your second sheet in a lil bit elobrated way? 0^2= v^2 - 2.a.s3 how
I have elaborated now.
I have elaborated now.see
edited
thank you!
welcome
Answered by
36
Hey buddy,
◆ Answer-
750 m
◆ Explaination-
For first 10 seconds-
v = u + at
v = 0 + 2×10
v = 20 m/s
Distance here is,
s = ut + 1/2 at^2
s = 0×10 + 1/2 × 2 × 10^2
s = 100 m
For next 30 seconds,
s = vt
s = 20 × 30
s = 600 m
In remaining time-
v^2 = u^2 + 2as
0 = 20^2 + 2(-4)s
s = 50 m
Total distance travelled is
s' = 100 + 600 + 50
s' = 750 m
Therefore, distance between two stations is 750 m.
Hope this helps...
◆ Answer-
750 m
◆ Explaination-
For first 10 seconds-
v = u + at
v = 0 + 2×10
v = 20 m/s
Distance here is,
s = ut + 1/2 at^2
s = 0×10 + 1/2 × 2 × 10^2
s = 100 m
For next 30 seconds,
s = vt
s = 20 × 30
s = 600 m
In remaining time-
v^2 = u^2 + 2as
0 = 20^2 + 2(-4)s
s = 50 m
Total distance travelled is
s' = 100 + 600 + 50
s' = 750 m
Therefore, distance between two stations is 750 m.
Hope this helps...
Similar questions