A ship of mass `3xx10^7kg` initially at rest, is pulled by a force of `5xx10^5N` through a distance of 3m. Assuming that the resistance due to water is negligible, the speed of the ship is

Options

A. `1.5ms^(-1)`
B. `60 ms^(-1)`
C. `0.1ms^(-1)`
D. `5ms^(-1)`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given values
- Mass of the ship, \( m = 3 \times 10^7 \, \text{kg} \)
- Force applied, \( F = 5 \times 10^5 \, \text{N} \)
- Distance moved, \( s = 3 \, \text{m} \)
- Initial velocity, \( u = 0 \, \text{m/s} \) (since the ship is initially at rest)

### Step 2: Calculate the acceleration of the ship
Using Newton's second law of motion, we can calculate the acceleration \( a \) using the formula:
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{5 \times 10^5 \, \text{N}}{3 \times 10^7 \, \text{kg}} = \frac{5}{3} \times 10^{-2} \, \text{m/s}^2 \]
This simplifies to:
\[ a \approx 1.67 \times 10^{-2} \, \text{m/s}^2 \]

### Step 3: Use the kinematic equation to find the final velocity
We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2as \]
Since the initial velocity \( u = 0 \):
\[ v^2 = 0 + 2as \]
Substituting the values of \( a \) and \( s \):
\[ v^2 = 2 \times \left(\frac{5}{3} \times 10^{-2}\right) \times 3 \]
Calculating this gives:
\[ v^2 = 2 \times \frac{5}{3} \times 10^{-2} \times 3 = 2 \times 5 \times 10^{-2} = 10^{-1} \]
Taking the square root to find \( v \):
\[ v = \sqrt{10^{-1}} = \sqrt{0.1} \approx 0.316 \, \text{m/s} \]

### Step 4: Final answer
The speed of the ship after being pulled through a distance of 3 meters is approximately:
\[ v \approx 1 \, \text{m/s} \]