GATE 2024 AE – Question 12
The acceleration of a body travelling in a straight line is given by $a = -C_1 - C_2v^2$ where $v$ is the velocity, and $C_1$, $C_2$ are positive constants. Starting with an initial positive velocity $v_o$, the distance travelled by the body before coming to rest for the first time is:
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Correct answer: (A) $\frac{1}{2C_2}\ln\left(1 + \frac{C_2}{C_1}v_o^2\right)$
Explanation
Write $a = v\frac{dv}{dx}$, so $\frac{v\,dv}{C_1 + C_2v^2} = -dx$. Integrating from $v_o$ to 0 gives $x = \frac{1}{2C_2}\left[\ln(C_1 + C_2v_o^2) - \ln C_1\right] = \frac{1}{2C_2}\ln\left(1 + \frac{C_2}{C_1}v_o^2\right)$.