The acceleration versus velocity graph of a p | vedclass

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(C) The given acceleration-velocity $(a-v)$ graph is a straight line with a negative slope and a positive intercept on the $a$-axis. It can be represe

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(C) The given acceleration-velocity $(a-v)$ graph is a straight line with a negative slope and a positive intercept on the $a$-axis. It can be represented by the equation $a = -kv + c$,where $k$ and $c$ are positive constants.Since the particle starts from rest,at $t = 0$,$v = 0$. From the graph,at $v = 0$,$a = a_0$ (a positive value). Thus,the equation is $a = a_0 - kv$.We know that $a = \frac{dv}{dt}$,so:$\frac{dv}{dt} = a_0 - kv$Rearranging the terms for integration:$\int_{0}^{v} \frac{dv}{a_0 - kv} = \int_{0}^{t} dt$Integrating both sides:$-\frac{1}{k} [\ln(a_0 - kv)]_{0}^{v} = t$$\ln\left(\frac{a_0 - kv}{a_0}ight) = -kt$Taking the exponential of both sides:$1 - \frac{kv}{a_0} = e^{-kt}$$v(t) = \frac{a_0}{k}(1 - e^{-kt})$This equation represents a curve that starts from the origin ($v=0$ at $t=0$) and approaches a terminal velocity $v_{max} = \frac{a_0}{k}$ as $t 	o \infty$. The slope of the $v-t$ graph,which is acceleration,decreases as velocity increases,which matches the given $a-v$ graph. This corresponds to the graph shown in option $C$.

The acceleration versus velocity graph of a particle moving in a straight line starting from rest is as shown in the figure. The corresponding velocity-time graph would be:

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