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Question

A particle moves in the x-y plane with velocity vx=8t2 and vy=2. If it passes through the point x= 14 and y= 4 at t= 2s, the equation of the is:

A
x=y2y+2
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B
x=y22
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C
x=y2+y6
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D
None of these
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Solution

The correct option is A x=y2y+2
vx=8t2 or, dxdt=8t2
Integrate to get, x=4t22t+c
Given, at t=2, x=14, thus 14=4(4)2(2)+c or c=2
Thus, x=4t22t+2
Similarly, vy=2 or dydt=2
Integrate to get, y=2t+c
Given, at t=2, y=4, thus 2(2)+c=4 or c=0
Thus, y=2t (or t=y2)
Put this t in x to get x=4(y2)22(y2)+2=y2y+2

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