wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A train accelerating uniformly from rest attains a maximum speed of 40 ms1 in 20 s. It travels at this speed for 20 s and is brought to rest by uniform retardation in further 40 s. What is the magnitude average velocity during this period?

A
80/3 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
40 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
25 ms1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
30 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 25 ms1
Using first equation of motion
v=u+at
40=(20)a1 (u=0)
a1=2 m/s2
S1=12a1t2
S1=400 m
Thereafter it travels with constant speed of 40 m/s, hence distance travelled will be,
S2=40×20=800 m
Further it travels with uniform retardation until final velocity becomes 0 Putting t=40 s,u=40 m/s and v=0 in first equation of motion, we get
0=40+(a2)t
40=(40)a2
a2=1 m/s2
retardation = 1 m/s2
Putting a=1 m/s2 in third equation of motion, we get
02402=2(1)S3
S3=800 m
Now average velocity =Total displacementtime
=400+800+80020+20+40=25 m/s

flag
Suggest Corrections
thumbs-up
64
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Motion Under Constant Acceleration
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon