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

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

Grade:12th pass

2 Answers

Vikas TU
14149 Points
6 years ago
Let in x and y the eqns. be,
x = t and y^2 + 2y + 2 = t
Now y component of velocity vector is:
2ydy/dt + 2dy/dt + 0 = 1
dy/dt = 1/(2y + 2) = 5
on solvig we get,
10y + 10 = 1
y = -9/10 => -0.9
accln. => -0.5/(y+1)^2
on putting y = -0.9 we get,
accln. = 50 m/s^2
Lakshit
11 Points
6 years ago
{x = y² + 2y + 2} is the constraint equation of motion of the particle, where x and y are time dependent coordinates. Hence we can differentiate the eq w.r.t time, and obtainẊ = 2yẏ + 2ẏ + 0 where ẋ (dot notation)implies dx/dt.̇Since ẏ = 5m/sẊ = 10y +10Again diff. w.r.t timeẍ = 10ẏ = 50m/sAs y component of velocityis constant, y component of accl. is nil, which gives us the final answera = 50i + 0j = 50m/s^2

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