Question

The acceleration of a body travelling in a straight line is \(a=-C_1-C_2v^2\) where \(v\) is the velocity and \(C_1,C_2 > 0\). Starting with an initial positive velocity \(v_0\), the distance travelled before coming to rest for the first time is:

• A. \(\displaystyle \frac{1}{2C_2}\ln\!\Big(1+\frac{C_2}{C_1}v_0^2\Big)\)
• B. \(\displaystyle \frac{1}{2C_2}\ln\!\Big(1-\frac{C_2}{C_1}v_0^2\Big)\)
• C. \(\displaystyle \frac{1}{2C_2}\ln(C_1+C_2v_0^2)\)
• D. \(\displaystyle \frac{1}{2C_2}\ln(1+C_2v_0^2)\)

Hint:

In 1-D motion with \(a(v)\), switch to the relation \(a=v\,dv/dx\). This converts the problem into a single integral in \(v\) with limits \(v_0\to 0\). Always make the log's argument dimensionless, e.g., \(\ln\!\big(1+\frac{C_2}{C_1}v_0^2\big)\).

Answer 1

The Correct Option is A

Step 1: Use \(a=v\,\dfrac{dv}{dx}\).
Since \(a=\dfrac{dv}{dt}\) and \(v=\dfrac{dx}{dt}\), we have \[ a=v\frac{dv}{dx}. \] Given \(a=-C_1-C_2v^2\), so \[ v\frac{dv}{dx}=-C_1-C_2v^2 \;\Rightarrow\; \frac{dx}{dv}=-\frac{v}{C_1+C_2v^2}. \] Step 2: Integrate from \(v=v_0\) to \(v=0\).
Distance to first rest: \[ x=\int_{t_0}^{t_*} v\,dt=\int_{v_0}^{0}\frac{dx}{dv}\,dv = -\int_{v_0}^{0}\frac{v}{C_1+C_2v^2}\,dv =\int_{0}^{v_0}\frac{v}{C_1+C_2v^2}\,dv. \] Step 3: Evaluate the integral.
Let \(u=C_1+C_2v^2\Rightarrow du=2C_2v\,dv\). Then \[ \int\frac{v}{C_1+C_2v^2}\,dv =\frac{1}{2C_2}\int \frac{du}{u} =\frac{1}{2C_2}\ln u =\frac{1}{2C_2}\ln(C_1+C_2v^2). \] Thus, \[ x=\left.\frac{1}{2C_2}\ln(C_1+C_2v^2)\right|_{0}^{v_0} =\frac{1}{2C_2}\Big[\ln(C_1+C_2v_0^2)-\ln C_1\Big] =\frac{1}{2C_2}\ln\!\Big(1+\frac{C_2}{C_1}v_0^2\Big). \] \(\boxed{\displaystyle x=\frac{1}{2C_2}\ln\!\Big(1+\frac{C_2}{C_1}v_0^2\Big)}\)