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Grade: 11
`        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 `
6 months ago

## Answers : (2)

Giriraj Gupta
27 Points
```
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

```
6 months ago
Khimraj
3008 Points
```							x = y2 + 2y +2differentiating wrt to tVx = 2y Vy + 2 VyVy = 5 m/s So Vx = 10y + 10differentiating again with time:d2x/dt2 = a = 10 dy/dt = 10 Vy = 50 m/s2
```
6 months ago
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