A particle moves along the parabolic path y=x2 +2x +2 in such a way that x component of velocity vector remain 5 ms-1 during the motion .the magnitude of acceleration of the particle in ms-2 is

Question

A particle moves along the parabolic path y=x2 +2x +2 in such a way that x component of velocity vector remain 5 ms-1 during the motion .the magnitude of acceleration of the particle in ms-2 is (A) 50 (B) 200 (C) 10√2 (D) 0.1

Tardigrade Admin · Accepted Answer

compare with y = tan θ x - (g/2u2 cos2 θ) x2

Q. A particle moves along the parabolic path $y = x^{2} + 2 x + 2$ in such a way that x component of velocity vector remain $5 m s^{- 1}$ during the motion .the magnitude of acceleration of the particle in $m s^{- 2}$ is

50

200

$102$

0.1

compare with y = tan $θ$ x - $\frac{g}{2 u ^{2} co s ^{2} θ} x^{2}$

Q. A particle moves along the parabolic path y=x2+2x+2 in such a way that x component of velocity vector remain 5ms−1 during the motion .the magnitude of acceleration of the particle in ms−2 is

50

200

102​

0.1

compare with y = tan θ x - 2u2cos2θg​x2

Q. A particle moves along the parabolic path $y = x^{2} + 2 x + 2$ in such a way that x component of velocity vector remain $5 m s^{- 1}$ during the motion .the magnitude of acceleration of the particle in $m s^{- 2}$ is

$102$

compare with y = tan $θ$ x - $\frac{g}{2 u ^{2} co s ^{2} θ} x^{2}$