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Question

A particle moves along the parabolic path x=y2+2y+2 in such a way that y-component of velocity vector remains 5 ms−1 during the motion. The magnitude of the acceleration of the particle is

A
50 ms2
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B
100 ms2
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C
102ms2
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D
0.1 ms2
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Solution

The correct option is A 50 ms2
Since velocity in y is constant, thus acceleration in y=0. Thus magnitude of acceleration is given by the x-component of acceleration.
Now, x=y2+2y+2.
Differentiate with time to get,
vx=2yvy+2vy=10y+10
Again differentiate with time to get,
ax=10vy
Since vy=5, thus ax=50m/s2

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