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 kinematic equation that relates initial velocity, final velocity, acceleration, and distance traveled. The equation we will use is:

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

Where:
- \( v \) = final velocity (0 m/s when the car stops)
- \( u \) = initial velocity
- \( a \) = acceleration (which will be negative since it is deceleration)
- \( s \) = distance traveled

### Step 1: Calculate the acceleration when the car is moving at 10 m/s

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

Using the kinematic equation:
\[ 0 = (10)^2 + 2a(20) \]

This simplifies to:
\[ 0 = 100 + 40a \]

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

### Step 2: Use the calculated acceleration to find the stopping distance at 30 m/s

Now, we need to find the stopping distance when the initial velocity is \( u = 30 \, \text{m/s} \).

Using the same kinematic equation:
\[ v^2 = u^2 + 2as \]

Where:
- \( v = 0 \, \text{m/s} \)
- \( u = 30 \, \text{m/s} \)
- \( a = -2.5 \, \text{m/s}^2 \)

Substituting the values:
\[ 0 = (30)^2 + 2(-2.5)s \]

This simplifies to:
\[ 0 = 900 - 5s \]

Rearranging gives:
\[ 5s = 900 \]
\[ s = \frac{900}{5} = 180 \, \text{m} \]

### Conclusion

The car moving at 30 m/s can be stopped by the same force in a distance of **180 meters**.

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