Question

A boat of mass \(700\,\text{kg}\) is travelling at a speed of \(24\,\text{ms}^{-1}\) when its engine is shut off. The magnitude of frictional force \(f\) between boat and water is given as \(f=35v\), where \(v\) is the speed in \(\text{ms}^{-1}\) and \(f\) is in newton. The speed of boat becomes \(6\,\text{ms}^{-1}\) in time

• A. \(18\,\text{s}\)
• B. \(36\,\text{s}\)
• C. \(34\,\text{s}\)
• D. \(28\,\text{s}\)

Hint:

When resistive force is proportional to velocity, use \(m\frac{dv}{dt}=-kv\). This gives exponential decay of speed with time.

Answer 1

The Correct Option is D

Step 1: Write the equation of motion using frictional force.
The frictional force acts opposite to the motion of the boat.
Given, \[ f=35v \] Using Newton's second law, \[ m\frac{dv}{dt}=-35v \] Since \[ m=700\,\text{kg}, \] we get \[ 700\frac{dv}{dt}=-35v \]

Step 2: Separate the variables.
\[ \frac{dv}{dt}=-\frac{35}{700}v \] \[ \frac{dv}{dt}=-\frac{1}{20}v \] So, \[ \frac{dv}{v}=-\frac{1}{20}dt \]

Step 3: Apply the limits.
Initially, \[ v=24\,\text{ms}^{-1} \] Finally, \[ v=6\,\text{ms}^{-1} \] Let the required time be \(t\).
Thus, \[ \int_{24}^{6}\frac{dv}{v} = -\frac{1}{20}\int_0^t dt \] \[ \left[\ln v\right]_{24}^{6} = -\frac{t}{20} \] \[ \ln 6-\ln 24=-\frac{t}{20} \] \[ \ln\left(\frac{6}{24}\right)=-\frac{t}{20} \] \[ \ln\left(\frac{1}{4}\right)=-\frac{t}{20} \] \[ -\ln 4=-\frac{t}{20} \] Therefore, \[ t=20\ln 4 \]

Step 4: Approximate the value.
Using \[ \ln 4\approx 1.386, \] we get \[ t=20(1.386) \] \[ t=27.72\,\text{s} \] \[ t\approx 28\,\text{s} \]

Step 5: Final conclusion.
Hence, the speed of the boat becomes \(6\,\text{ms}^{-1}\) in \[ \boxed{28\,\text{s}} \]