Question Statement:
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 $v_0$, the distance travelled by the body before coming to rest for the first time is:
Options:
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(A) $\frac{1}{2C_2} \ln \left(1 + \frac{C_2}{C_1} v_0^2\right)$
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(B) $\frac{1}{2C_2} \ln \left(1 - \frac{C_2}{C_1} v_0^2\right)$
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(C) $\frac{1}{2C_2} \ln (C_1 + C_2 v_0^2)$
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(D) $\frac{1}{2C_2} \ln (1 + C_2 v_0^2)$
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2024 Q.12)**
**1. Physical Formulation & Differential Relations**:
Acceleration $a$ expressed in terms of distance $s$ and velocity $v$ via the chain rule:
a = \frac{dv}{dt} = v \frac{dv}{ds}
Given governing relation:
a = -C_1 - C_2 v^2
Equating the expressions:
v \frac{dv}{ds} = -(C_1 + C_2 v^2)
**2. Separation of Variables & Integration**:
Separating variables to find the stopping distance $S$:
\frac{v \, dv}{C_1 + C_2 v^2} = -ds
Integrating with boundary conditions:
- At $s = 0$, velocity is $v = v_0$
- At $s = S$, the body comes to rest, so $v = 0 $\int_{v_0}^{0} \frac{v \, dv}{C_1 + C_2 v^2} = -\int_{0}^{S} ds = -S
Substitute $u = C_1 + C_2 v^2 \implies du = 2C_2 v \, dv$:
S = \int_{0}^{v_0} \frac{v \, dv}{C_1 + C_2 v^2} = \frac{1}{2C_2} \left[ \ln(C_1 + C_2 v^2) \right]_{0}^{v_0}
Evaluating at the limits:
S = \frac{1}{2C_2} \left[ \ln(C_1 + C_2 v_0^2) - \ln(C_1) \right]
Using logarithm properties $\ln(A) - \ln(B) = \ln\left(\frac{A}{B}\right)$:
S = \frac{1}{2C_2} \ln \left( \frac{C_1 + C_2 v_0^2}{C_1} \right) = \frac{1}{2C_2} \ln \left( 1 + \frac{C_2}{C_1} v_0^2 \right)
**3. Conclusion & Official Answer**:
- **Option (A)**: $\frac{1}{2C_2} \ln \left(1 + \frac{C_2}{C_1} v_0^2\right)$ matches the exact analytical result.
- **Official Answer Key**: **A**