Question

When a particle's displacement $\mathrm{x}$ at the time $\mathrm{t}$ is given by $\mathrm{x}={\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}$, where $\mathrm{A},\mathrm{B},\mathrm{and}\mathrm{C}$ are constants and $\mathrm{v}$ is the particle's velocity, the number $4\mathrm{Ax}-{\mathrm{v}}^2$is found

• A. $4\mathrm{AC}+{\mathrm{B}}^2$
• B. $4\mathrm{AC}–{\mathrm{B}}^2$
• C. $2\mathrm{AC}–{\mathrm{B}}^2$
• D. $2\mathrm{AC}+{\mathrm{B}}^2$

Answer 1

The correct option is B

$4\mathrm{AC}–{\mathrm{B}}^2$

Step: 1 GIven data:

Displacement of the particle $\left(\mathrm{x}\right)={\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}$.

Step: 2 Formula used:

Velocity = Change in position with respect to time = $\frac{\mathrm{dx}}{\mathrm{dt}}$.

Step: 3 Calculate the value of $4\mathrm{Ax}-{\mathrm{v}}^2$ :

$\mathrm{x}={\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{d}\left({\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}\right)}{\mathrm{dt}}=2\mathrm{At}+\mathrm{B}$

Therefore, $\mathrm{v}=2\mathrm{At}+\mathrm{B}$.

$$\begin{gathered} {\mathrm{v}}^2={\left(2\mathrm{At}+\mathrm{B}\right)}^2 \\ {\mathrm{v}}^2=\left(4{\mathrm{A}}^2{\mathrm{t}}^2+4\mathrm{ABt}+{\mathrm{B}}^2\right) \end{gathered}$$

Therefore,

$$\begin{gathered} 4\mathrm{Ax}=4{\mathrm{A}}^2{\mathrm{t}}^2+4\mathrm{ABt}+4\mathrm{AC} \\ {\mathrm{v}}^2–4Ax={\mathrm{B}}^2–4\mathrm{AC} \\ \Rightarrow 4\mathrm{AX}–{\mathrm{v}}^2=4\mathrm{AC}–{\mathrm{B}}^2 \end{gathered}$$

Hence option B is correct.