A particle moves in the x−y plane according to the law x=kt,y=kt(1−αt) where k and α are positive constants and t is time. The trajectory of the particle is:
A
y=kx
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B
y=x−αx2k
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C
y=−αx2k
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D
y=αx
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Solution
The correct option is Cy=x−αx2k We have x=kt;t=xk Also y=kt(1−αt) We put t=xk, thus we get y=kxk(1−αxk);y=x−αx2k