Question

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

Answer 1

Answer:

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Answer 2

Given:

Acceleration versus velocity graph of a particle moving in a straight line starting from rest is as shown in diagram.

To find:

Corresponding velocity–time graph.

Calculation:

Let k be slope and c be intercept.

Equation of the acceleration-velocity graph:

$\therefore \: a = - kv + c$

$= > \: \dfrac{dv}{dt} = - kv + c$

$= > \: \dfrac{dv}{dt} = c - kv$

$= > \: \dfrac{dv}{c - kv} = dt$

Integrating on both sides:

$= > \displaystyle \: \int \dfrac{dv}{c - kv} = \int dt$

Putting limits:

$= > \displaystyle \: \int_{0}^{v} \dfrac{dv}{c - kv} = \int_{0}^{t} dt$

$= > \: \dfrac{ - 1}{k} \: ln \bigg \{\dfrac{c - kv}{c} \bigg \} = t$

$= > \: ln \bigg \{\dfrac{c - kv}{c} \bigg \} = - kt$

$= > \: \dfrac{c - kv}{c} = {e}^{ - kt}$

$= > \: 1 - \dfrac{kv}{c} = {e}^{ - kt}$

$= > \: \dfrac{kv}{c} = 1 - {e}^{ - kt}$

$= > \: v = \dfrac{c}{k} \: \bigg \{ 1 - {e}^{ - kt} \bigg \}$

Hence the final equation is :

$\boxed{ \bold{ \red{ \: v = \dfrac{c}{k} \: \bigg \{ 1 - {e}^{ - kt} \bigg \}}}}$