A particle moves in the x-y plane according to the law x - kt- y - kt 1 - - t where k and - are positive constants and t is time- The trajectory of the particle is-

The correct option is C y=x α x^2k We have x=kt ; t=xk Also y=kt (1 α t #160; ) #160; We put #160; t=xk , thus we get y=kxk (1 ...

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Standard XII

Mathematics

Algebra of Limits

A particle mo...

Question

A particle moves in the $x - y$ plane according to the law $x = k t, y = k t (1 - α t)$ where $k$ and $α$ are positive constants and $t$ is time. The trajectory of the particle is:

y = k x

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y = x - \frac{α x^{2}}{k}

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y = - \frac{α x^{2}}{k}

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y = α x

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Solution

The correct option is C $y = x - \frac{α x^{2}}{k}$
We have
$x = k t; t = \frac{x}{k}$
Also
$y = k t (1 - α t)$
We put $t = \frac{x}{k}$ , thus we get
$y = \frac{k x}{k} (1 - \frac{α x}{k}); y = x - \frac{α x^{2}}{k}$

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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:

The correct option is C y=x−αx2kWe havex=kt;t=xkAlsoy=kt(1−αt) We put t=xk, thus we gety=kxk(1−αxk);y=x−αx2k

A particle moves in the $x - y$ plane according to the law $x = k t, y = k t (1 - α t)$ where $k$ and $α$ are positive constants and $t$ is time. The trajectory of the particle is:

The correct option is C $y = x - \frac{α x^{2}}{k}$
We have
$x = k t; t = \frac{x}{k}$
Also
$y = k t (1 - α t)$
We put $t = \frac{x}{k}$ , thus we get
$y = \frac{k x}{k} (1 - \frac{α x}{k}); y = x - \frac{α x^{2}}{k}$