A particle moves along a parabolic pathx=y^2+2y+2 in such a way that the y component of the velocity vector remains 5m/s during the motion .The magnitude of the acceleration of particle in ms^-2
Giriraj Gupta , 5 Years ago
Grade 11
2 Answers
Giriraj Gupta
Last Activity: 5 Years ago
x = y2 + 2y +2
differentiating the equation with respect to time
thus dx/dt = 2y dy/dt + 2dy/dt
or Vx = 2y Vy + 2 Vy
here Vy = 5 m/s a constant
now Vx = 10y + 10
differentiating again with time:
d2x/dt2 = a = 10 dy/dt = 10 Vy = 50 m/s2
Khimraj
Last Activity: 5 Years ago
x = y2 + 2y +2 differentiating wrt to t Vx = 2y Vy + 2 Vy Vy = 5 m/s So Vx = 10y + 10 differentiating again with time: d2x/dt2 = a = 10 dy/dt = 10 Vy = 50 m/s2
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