The displacement x of a particle at time t is given by $x = A t^{2} + B t + C$ where A, B, C are constants and v is velocity of a particle, then the value of $4 A x - v^{2}$ is:

4 A C + B^{2}

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4 A C - B^{2}

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2 A C - B^{2}

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2 A C + B^{2}

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Solution

The correct option is C $4 A C - B^{2}$
$x = A t^{2} + B t + C \dots (i)$
$\Rightarrow v = 2 A t + B$
$\Rightarrow v^{2} = 4 A^{2} t^{2} + 4 A B t + B^{2} \dots (i i)$
{By multiplying $4 A$ in Eq. (i) }
$4 A x = 4 A^{2} t^{2} + 4 A B t + 4 A C \dots (i i i)$
{from Eq. (i) and Eq. (iii)}
$\Rightarrow v^{2} - 4 A x = B^{2} - 4 A C$
$\Rightarrow 4 A x - v^{2} = 4 A C - B^{2}$

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The displacement x of a particle at time t is given by x=At2+Bt+C where A, B, C are constants and v is velocity of a particle, then the value of 4Ax−v2 is:

The correct option is C 4AC−B2x=At2+Bt+C…(i)⇒v=2At+B⇒v2=4A2t2+4ABt+B2…(ii){By multiplying 4A in Eq. (i) }4Ax=4A2t2+4ABt+4AC…(iii){from Eq. (i) and Eq. (iii)}⇒v2−4Ax=B2−4AC⇒4Ax−v2=4AC−B2

The displacement x of a particle at time t is given by $x = A t^{2} + B t + C$ where A, B, C are constants and v is velocity of a particle, then the value of $4 A x - v^{2}$ is: