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The Equation for the Path of Projectile

The initial v...

Question

The initial velocity of a particle moving along $x -$ axis is $u (a t t = 0 a n d x = 0)$ and its acceleration a is given by $a = k x$ . Which of the following equation is correct between its velocity $(v)$ and position $(x)$ ?

v^{2} - u^{2} = 2 k x

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v^{2} = u^{2} + 2 k x^{2}

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v^{2} = u^{2} + k x^{2}

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v^{2} + u^{2} = 2 k x

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Solution

The correct option is C $v^{2} = u^{2} + k x^{2}$
a = kx
$a = \frac{d v}{d t} = \frac{d v}{d x} \times \frac{d x}{d t}$
Vdv = adx
= kx dx
$\frac{V^{2}}{2} = \frac{k x^{2}}{2} + C$
at $t = 0 c = 0 v e l o c i t y = u$
So
$C = \frac{U^{2}}{2} V^{2} = u^{2} + k x^{2}$
Hence option (C) is correct

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The initial velocity of a particle moving along x−axis is u (at t=0 and x=0) and its acceleration a is given by a=kx. Which of the following equation is correct between its velocity (v) and position (x)?

The correct option is C v2=u2+kx2a = kx a=dvdt=dvdx×dxdtVdv = adx = kx dxV22=kx22+Cat t=0c=0velocity=uSo C=U22V2=u2+kx2 Hence option (C) is correct

The initial velocity of a particle moving along $x -$ axis is $u (a t t = 0 a n d x = 0)$ and its acceleration a is given by $a = k x$ . Which of the following equation is correct between its velocity $(v)$ and position $(x)$ ?

The correct option is C $v^{2} = u^{2} + k x^{2}$
a = kx
$a = \frac{d v}{d t} = \frac{d v}{d x} \times \frac{d x}{d t}$
Vdv = adx
= kx dx
$\frac{V^{2}}{2} = \frac{k x^{2}}{2} + C$
at $t = 0 c = 0 v e l o c i t y = u$
So
$C = \frac{U^{2}}{2} V^{2} = u^{2} + k x^{2}$
Hence option (C) is correct