wiz-icon
MyQuestionIcon
MyQuestionIcon
6
You visited us 6 times! Enjoying our articles? Unlock Full Access!
Question

A body travels for 15 sec starting from rest with constant acceleration. If it travels S1,S2 and S3 in the first 5 seconds, second 5 seconds and next five seconds respectively. The relation between S1,S2 and S3 is

A
S1=S2=S3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5S1=3S2=S3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
S1=13S2=15S3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
S1=15S2=13S3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C S1=13S2=15S3
Since the body starts from rest. Therefore u=0.
By using second equation of motion
Distance covered in first 5 seconds will be
S1=12a(5)2=25a2 ------(1)
Distance covered in first 10 seconds will be
S1+S2=12a(10)2=100a2
S2=100a2S1=75a2 ---------(2) Distance covered in first 15 seconds will be
S1+S2+S3=12a(15)2=225a2
S3=225a2S2S1=125a2---------(3)
On comparing (1), (2) and (3) we will get
S1=13S2 and S1=15S3

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