A car moving with a velocity of 10m/s can be stopped by the application of a constant force F In a distance of 20m. If the velocity of the car is 30m/s. It can be stopped by this force in

Options

A. `(20)/(3)m`
B. 20 m
C. 60 m
D. 180 m

Explanation

To solve the problem, we will use the equations of motion. The key equation we will use is:

\[ v^2 = u^2 + 2as \]

where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration (or deceleration in this case), and
- \( s \) is the distance.

### Step-by-Step Solution:

**Step 1: Determine the deceleration when the car is moving at 10 m/s.**

Given:
- Initial velocity \( u = 10 \, \text{m/s} \)
- Final velocity \( v = 0 \, \text{m/s} \) (since the car stops)
- Distance \( s = 20 \, \text{m} \)

Using the equation of motion:

\[ 0 = (10)^2 + 2a(20) \]

This simplifies to:

\[ 0 = 100 + 40a \]

Rearranging gives:

\[ 40a = -100 \implies a = -\frac{100}{40} = -2.5 \, \text{m/s}^2 \]

**Step 2: Use the deceleration to find the stopping distance when the car is moving at 30 m/s.**

Now, we need to find the stopping distance when the initial velocity \( u = 30 \, \text{m/s} \) and the same deceleration \( a = -2.5 \, \text{m/s}^2 \).

Using the same equation of motion:

\[ 0 = (30)^2 + 2(-2.5)s \]

This simplifies to:

\[ 0 = 900 - 5s \]

Rearranging gives:

\[ 5s = 900 \implies s = \frac{900}{5} = 180 \, \text{m} \]

### Final Answer:
The car can be stopped by the same force in a distance of **180 meters**.

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