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

A particle starts from position x=x0 at time t=0 in positive x direction according tot he give velocity-time (v-t) graph. The position (x) of particle at any time t can be represented as :
1308221_1f0482108e034eaa8c431301eb4abfd9.png

A
x=x0+v0t+12v0t0t2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x=x0+12v0t0t2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
x=x0+12v0t0t2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x=x012v0t0t2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B x=x0+12v0t0t2
Given: x=x0;t=t0

Use equation of motion: v=u+at

v0=(0)+at0

v0=at0

a=v0t0......Eq:01

Now, use equation of motion s=ut+12at2 for distance/position (x) at any time (t);

x=(0)t+12at2

x=12at2

Put the value of 'a' from Eq:01;

x=12(v0t0)t2

Required position of particle at any time (t): x=x0+12(v0t0)t2

Option (B) is correct.












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