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

# A particle starts from the origin of coordinates at time t=0 and moves in the xy plane with a constant acceleration α in the y-direction. It's equation of motion is y=βx2. It's velocity component in the x- direction is

A
variable
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2αβ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
α2β
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
α2β
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D √α2βay=α⇒dvydt=α⇒vy=∫αdt=αt+constantAt t=0, vy=0⇒constant=0⇒vy=αt⇒dydt=αt ............(1)⇒2βxdxdt=αt⇒2βxvx=αtvx=αt2βx ............(2)From (1)y=∫αtdt=αt22+c′At t=0, y=0⇒c′=0⇒y=αt22βx2=αt22⇒t=√2βx2α ....(3)Substituting t from (3) into (2)vx=α√2βx2α2βx=√α2β

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Speed and Velocity
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program