The diagram shows the variation of 1 / v (whe | vedclass

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Question

(b)

$\frac{1}{v}$ versus $t$ equation from the given graph would be,

$\frac{1}{v}=\left(3+\frac{1}{\sqrt{3}}\right)-t$

$v=\frac{1}{\left(3+\f

Vedclass · Accepted Answer

$+3$

Vedclass · Answer

(b)

$\frac{1}{v}$ versus $t$ equation from the given graph would be,

$\frac{1}{v}=\left(3+\frac{1}{\sqrt{3}}\right)-t$

$v=\frac{1}{\left(3+\frac{1}{\sqrt{3}}\right)-t}$

$a=\frac{d v}{d t}=\frac{1}{\left[\left(3+\frac{1}{\sqrt{3}}\right)-t\right]^2}$

Putting $t=3\,s$ we get, $a=3\,m / s ^2$.

The diagram shows the variation of $1 / v$ (where, $v$ is velocity of the particle) with respect to time. At time $t=3\,s$ using the details given in the graph, the instantaneous acceleration will be equal to $...........m/s^{2}$

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