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Question

A particle moves along a parabolic path x=y2+y+1 in such a way that the y-component of the velocity vector remain 4ms−1 during the motion. The magnitude of the acceleration of the prtile is

A
4ms2
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B
ms2
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C
32ms2
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D
50ms2
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Solution

The correct option is D 50ms2
Let in x and y the equations be,
x=t and y2+2y+2=t
Now y component of velocity vector is :
2ydy/dt+2dy/dt+0=1
dy/dt=1/(2y+2)=5
On solving we get,
10y+10=1
y=9/10=0.9
acceleration =0.5/(y+1)2
on putting y=0.9, we get
acceleration =50 m/s2

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