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Question

A particle moves along parabolic path x=y2+2y+2 in such a way that y component of velocity is constant and equal to 5 m/s during the motion. Magnitude of acceleration of the particle

A
50 m/s2
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B
100 m/s2
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C
102 m/s2
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D
10 m/s2
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Solution

The correct option is A 50 m/s2
Given, Path of projectile
x=y2+2y+2
On differentiating both side w.r.t. time
dxdt=2ydydt+2dydt
vx=2vyy+2vy [vy=dydt]
vx=10y+10 [vy=5 m/s,given]
Again differentiating both side w.r.t. time,
dvxdt=10dydt
ax=50 m/s2
[a=dvdt,vy=dydt=5 m/s, given]
So, magnitude of acceleration of the particle is 50 m/s2.

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