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Grade: 12
        
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
one year ago

Answers : (1)

Susmita
425 Points
							
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.
one year ago
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