The displacement of a particle at the time t | vedclass

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(D) Given $s = (1+t)^{1/2}$.First,find the velocity $v$ by differentiating $s$ with respect to $t$:$v = \frac{ds}{dt} = \frac{1}{2}(1+t)^{-1/2} = \fra

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cube of the velocity

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(D) Given $s = (1+t)^{1/2}$.First,find the velocity $v$ by differentiating $s$ with respect to $t$:$v = \frac{ds}{dt} = \frac{1}{2}(1+t)^{-1/2} = \frac{1}{2\sqrt{1+t}}$.From this,we can see that $\sqrt{1+t} = \frac{1}{2v}$.Now,find the acceleration $a$ by differentiating $v$ with respect to $t$:$a = \frac{dv}{dt} = \frac{1}{2} \cdot \left(-\frac{1}{2}\right)(1+t)^{-3/2} = -\frac{1}{4}(1+t)^{-3/2}$.We can rewrite this as:$a = -2 \cdot \left[ \frac{1}{2}(1+t)^{-1/2} \right]^3$.Since $v = \frac{1}{2}(1+t)^{-1/2}$,we substitute $v$ into the expression for $a$:$a = -2v^3$.Therefore,the acceleration $a$ is proportional to the cube of the velocity $v^3$.

The displacement of a particle at the time $t$ is given by $s = \sqrt{1+t}$. Then its acceleration $a$ is proportional to:

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