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

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(B) From the graph,the equation of the straight line is given by the slope-intercept form: $\frac{1}{v} = mt + c$.The slope $m = 	an(180^\circ - 45^\

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$+3$

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(B) From the graph,the equation of the straight line is given by the slope-intercept form: $\frac{1}{v} = mt + c$.The slope $m = 	an(180^\circ - 45^\circ) = -	an(45^\circ) = -1$.At $t = 3\,s$,$\frac{1}{v} = \frac{1}{\sqrt{3}}$.Using the equation $\frac{1}{v} = -t + c$,we substitute the values at $t = 3\,s$:$\frac{1}{\sqrt{3}} = -3 + c \implies c = 3 + \frac{1}{\sqrt{3}}$.Thus,the equation is $\frac{1}{v} = -t + (3 + \frac{1}{\sqrt{3}})$.Rearranging for $v$: $v = \frac{1}{(3 + \frac{1}{\sqrt{3}}) - t}$.The acceleration $a$ is given by $a = \frac{dv}{dt}$.Using the chain rule: $a = \frac{d}{dt} [(3 + \frac{1}{\sqrt{3}}) - t]^{-1} = -1 \cdot [-1] \cdot [(3 + \frac{1}{\sqrt{3}}) - t]^{-2} = \frac{1}{[(3 + \frac{1}{\sqrt{3}}) - t]^2}$.At $t = 3\,s$,the term $[(3 + \frac{1}{\sqrt{3}}) - t] = \frac{1}{\sqrt{3}}$.Therefore,$a = \frac{1}{(1/\sqrt{3})^2} = \frac{1}{1/3} = 3\,m/s^2$.

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

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