Discussion
The acceleration of a body travelling in a straight line is given by \(a = -C_{1} - C_{2}v^{2}\) 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:
Answer: A
The distance is \(\frac{1}{2C_{2}}\ln\left(1 + \frac{C_{2}}{C_{1}}v_{o}^{2}\right)\).
Write the acceleration in terms of distance, \(a = v\,dv/dx\), and separate the variables:
\[ \begin{aligned} v\frac{dv}{dx} &= -(C_{1} + C_{2}v^{2}) \\ dx &= -\frac{v\,dv}{C_{1} + C_{2}v^{2}} \end{aligned} \]
Integrate from \(v_{o}\) down to rest:
\[ \begin{aligned} s &= \frac{1}{2C_{2}}\left[\ln(C_{1} + C_{2}v^{2})\right]_{0}^{v_{o}} \\ &= \frac{1}{2C_{2}}\ln\frac{C_{1} + C_{2}v_{o}^{2}}{C_{1}} \end{aligned} \]
which is the first option. The minus sign version has no physical meaning here, \(\ln(C_{1} + C_{2}v_{o}^{2})\) forgets the lower limit, and \(\ln(1 + C_{2}v_{o}^{2})\) drops the division by \(C_{1}\).
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