The acceleration versus velocity graph of a p | vedclass

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(D) The given graph is a straight line with a negative slope,representing the relation $a = -k v + c$. Since the particle starts from rest ($v=0$ at $

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(D) The given graph is a straight line with a negative slope,representing the relation $a = -k v + c$. Since the particle starts from rest ($v=0$ at $t=0$),the intercept $c$ must be positive,so $a = a_0 - k v$.Substituting $a = \frac{d v}{d t}$,we get $\frac{d v}{d t} = a_0 - k v$.Rearranging the terms,$\frac{d v}{a_0 - k v} = d t$.Integrating both sides,$-\frac{1}{k} \ln(a_0 - k v) = t + C$.At $t=0, v=0$,so $C = -\frac{1}{k} \ln(a_0)$.This leads to $\ln\left(\frac{a_0 - k v}{a_0}\right) = -k t$,which simplifies to $v(t) = \frac{a_0}{k}(1 - e^{-k t})$.This equation represents a curve that starts from the origin and approaches a constant terminal velocity asymptotically. This corresponds to the shape shown in the last option.

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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