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. Assume that the resistance due to water is nigligible, the speed of the ship is

Options

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

Answer

To solve the problem step by step, we will use the given data and apply the relevant physics concepts.

### Given:
- Mass of the ship, \( m = 3 \times 10^7 \) kg
- Force applied, \( F = 5 \times 10^4 \) N
- Distance moved, \( s = 3 \) m
- Initial speed, \( u = 0 \) m/s (since the ship is initially at rest)

### Step 1: Calculate the acceleration of the ship
Using Newton's second law of motion, we know that:
\[ F = m \cdot a \]
Where:
- \( F \) is the force applied,
- \( m \) is the mass of the ship,
- \( a \) is the acceleration.

Rearranging the equation to find acceleration:
\[ a = \frac{F}{m} \]

Substituting the 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 2: Use the third 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,
- \( a \) is the acceleration,
- \( s \) is the distance moved.

Since the initial speed \( u = 0 \):
\[ v^2 = 0 + 2as \]
Substituting the values of \( a \) and \( s \):
\[ v^2 = 2 \left(\frac{5}{3} \times 10^{-3}\right)(3) \]
Calculating further:
\[ v^2 = 2 \times \frac{5}{3} \times 10^{-3} \times 3 = 10^{-2} \]

### Step 3: Calculate the final speed
Taking the square root to find \( v \):
\[ v = \sqrt{10^{-2}} = 0.1 \, \text{m/s} \]

### Final Answer:
The speed of the ship is \( 0.1 \, \text{m/s} \).

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