An object starts moving from rest on a horizontal ground such that position vector of object at time t w-r-t- to its starting points is given by r- - 4t2- - 2t2 - 8t4- where - and - are the unit vectors along x and y axes- The equation of trajectory of the object will be

The correct option is A 2y = x x^2 Given, r⃗=4t^2î+(2t^2 8t^4)ĵ From the above, #160; x=4t^2 t^2=x4 . . . . . . . . . . . . .(1) y=2t^ ...

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Standard XII

Mathematics

Equation of Perpendicular from a Point on a Line

An object sta...

Question

An object starts moving from rest on a horizontal ground such that position vector of object at time t w.r.t. to its starting points is given by $\to r = 4 t^{2}^i + (2 t^{2} - 8 t^{4})^j$ , where $^i$ and $^j$ are the unit vectors along x and y axes. The equation of trajectory of the object will be

2 y = x - x^{2}

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y = x + x^{2}

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y = 2 x - x^{2}

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2 y = x - 4 x^{2}

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Solution

The correct option is A $2 y = x - x^{2}$
Given,
$\to r = 4 t^{2}^i + (2 t^{2} - 8 t^{4})^j$
From the above,
$x = 4 t^{2}$
$t^{2} = \frac{x}{4}$ . . . . . . . . . . . . .(1)
$y = 2 t^{2} - 8 t^{4}$ . . . . . . . . . . . . . . (2)
Substitute equation (1) in equation (2), we get
$y = 2 \frac{x}{4} - 8 \frac{x^{2}}{16}$
$y = \frac{x}{2} - \frac{x^{2}}{2}$
$2 y = x - x^{2}$
The correct option is A.

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An object starts moving from rest on a horizontal ground such that position vector of object at time t w.r.t. to its starting points is given by →r=4t2^i+(2t2−8t4)^j, where ^i and ^j are the unit vectors along x and y axes. The equation of trajectory of the object will be

The correct option is A 2y=x−x2Given,→r=4t2^i+(2t2−8t4)^jFrom the above, x=4t2t2=x4. . . . . . . . . . . . .(1)y=2t2−8t4. . . . . . . . . . . . . . (2)Substitute equation (1) in equation (2), we gety=2x4−8x216y=x2−x222y=x−x2The correct option is A.