CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A train leaves station 'A' to station 'B'. The train travels straight without any halts between the stations. During the first and last 200 m of its journey, the train has uniform acceleration and retardation both equal to 1ms1 respectively. For the rest of the journey, the train maintains uniform speed Calculate the average speed of the train, given the distance between two stations is 4 km.

A
18211ms1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
9211ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
10011ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1765ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 18211ms1
For the First 200 m,
Initial velocity u=0, Final Velocity v, acceleration a= 1 ms2,
Using v2=u2+2as v2=0+2×1×200 v=20ms1,
Let t1 is time taken to travel first 200 m.
Using v=u+at 20=0+×t1 t1=20s,
For next 3600 m, Speed is constant v=20ms1 ,
Let t2 is time taken to travel 3600 m.
Using s=ut t2=360020=180s,

For the last 200 m , initial velocity u=20ms1, acceleration a=-1 ms2, Final Velocity v=0,
Let t3 is time taken to travel last 200 m
Using v=u+at 0=201×t3 t3=20s,
Average speed =Total DistanceTotal Time
Average speed= 4000t1+t2+t3=400020+180+20= 18211ms1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Acceleration
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon