Guest

A particle move along the parabolic path x = y²+2y+2 in such a way that the y-component of velocity vector remain 5m/s during the motion. The magnitude of the acceleration of the particle is

A particle move along the parabolic path x = y²+2y+2 in such a way that the y-component of velocity vector remain 5m/s during the motion. The magnitude of the acceleration of the particle is

Grade:11

1 Answers

Susmita
425 Points
5 years ago
x=y2+2y+2
Differentiate with respect  to time t to obtain velocity components along x and y direction respectively.
vx=2yvy+2vy
where vx=dx/dt and vy=dy/dt.
Given vy=5m/s.
So vx=10y+10
Differentiate  again with respect  to  time  t  to  obtain  acceleration.
ax=10vy=50m/s2
Please approve the answer if helped.

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free