Speed v of a particle moving along a straight | vedclass

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(B) Given the equation $v^2 = 108 - 9x^2$.Differentiating with respect to $t$:$2v \frac{dv}{dt} = -18x \frac{dx}{dt}$Since $v = \frac{dx}{dt}$,we have

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The magnitude of the acceleration at a distance $3 \, m$ from the fixed point is $27 \, m/s^2$.

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(B) Given the equation $v^2 = 108 - 9x^2$.Differentiating with respect to $t$:$2v \frac{dv}{dt} = -18x \frac{dx}{dt}$Since $v = \frac{dx}{dt}$,we have $2v a = -18xv$,which simplifies to $a = -9x$.This is the characteristic equation of Simple Harmonic Motion $(a = -\omega^2 x)$,where $\omega^2 = 9$,so $\omega = 3 \, rad/s$.At $x = 3 \, m$,the magnitude of acceleration is $|a| = |-9(3)| = 27 \, m/s^2$. Thus,option $B$ is correct.The motion is simple harmonic about $x = 0$ (the equilibrium position).Maximum displacement (amplitude $A$) occurs when $v = 0$:$0 = 108 - 9x^2 \implies x^2 = 12 \implies x = \sqrt{12} \, m \approx 3.46 \, m$. Thus,option $D$ is incorrect.

Speed $v$ of a particle moving along a straight line,when it is at a distance $x$ from a fixed point on the line is given by $v^2 = 108 - 9x^2$ (all quantities in $S.I.$ units). Then:

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