A ship of mass `3 xx 10^(7)` kg initially at rest is pulled by a force of ` 5 xx 10^(4)` N through a distance of 3 m. Assuming that the resistance due to water is neglible, what will be the speed of ship ?

Options

A. 1.5 m/sec
B. 60 m/sec
C. 0.1 m/sec
D. 5m/sec

Explanation

To solve the problem step by step, we will follow the principles of Newton's second law of motion and the equations of motion.

### Step 1: Identify the given values
- Mass of the ship, \( m = 3 \times 10^7 \) kg
- Force acting on the ship, \( F = 5 \times 10^4 \) N
- Distance pulled, \( s = 3 \) m
- Initial speed, \( u = 0 \) m/s (the ship is initially at rest)

### Step 2: Calculate the acceleration of the ship
Using Newton's second law of motion, we know that:
\[ F = m \cdot a \]
Where \( a \) is the acceleration. Rearranging this gives us:
\[ a = \frac{F}{m} \]

Substituting the given values:
\[ a = \frac{5 \times 10^4 \, \text{N}}{3 \times 10^7 \, \text{kg}} = \frac{5}{3} \times 10^{-3} \, \text{m/s}^2 \]

### Step 3: Use the equation of motion to find the final speed
We will use the third equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final speed
- \( u \) is the initial speed (which is 0)
- \( a \) is the acceleration we calculated
- \( s \) is the distance

Substituting the known values:
\[ v^2 = 0^2 + 2 \left(\frac{5}{3} \times 10^{-3}\right)(3) \]

Calculating further:
\[ v^2 = 2 \times \frac{5}{3} \times 10^{-3} \times 3 \]
\[ v^2 = 10^{-2} \]

### Step 4: Calculate the final speed
Taking the square root of both sides:
\[ v = \sqrt{10^{-2}} = 0.1 \, \text{m/s} \]

### Conclusion
The speed of the ship after being pulled through a distance of 3 m is \( 0.1 \, \text{m/s} \).