If the speed of a car increases 4 times, the stopping distance ( in m) for this will increase by _________ .

Answer

To solve the problem, we need to understand the relationship between the speed of a car and its stopping distance. The stopping distance is the distance a car travels before it comes to a complete stop after the brakes are applied.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The stopping distance (s) of a vehicle is related to its speed (v) by the equation derived from the equations of motion. Specifically, we can use the formula:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity (which is 0 when the car stops)
- \( u \) = initial velocity
- \( a \) = acceleration (which is negative in the case of deceleration)
- \( s \) = stopping distance

2. **Setting Up the Initial Conditions**:
Since the car is initially at rest, we can assume:
- Initial speed \( u = 0 \)
- Let the initial speed be \( v_1 \) and the stopping distance be \( s_1 \).

3. **Considering the Increased Speed**:
According to the problem, the speed of the car increases 4 times. Therefore, the new speed \( v_2 \) can be expressed as:
\[ v_2 = 4v_1 \]

4. **Applying the Relationship**:
From the stopping distance formula, we can express the stopping distances in terms of the speeds:
\[ v_1^2 \propto s_1 \quad \text{and} \quad v_2^2 \propto s_2 \]
Therefore, we can write:
\[ \frac{v_1^2}{v_2^2} = \frac{s_1}{s_2} \]

5. **Substituting for \( v_2 \)**:
Substitute \( v_2 = 4v_1 \) into the equation:
\[ \frac{v_1^2}{(4v_1)^2} = \frac{s_1}{s_2} \]
Simplifying this gives:
\[ \frac{v_1^2}{16v_1^2} = \frac{s_1}{s_2} \]
This reduces to:
\[ \frac{1}{16} = \frac{s_1}{s_2} \]

6. **Finding the Stopping Distance Ratio**:
Rearranging gives:
\[ s_2 = 16s_1 \]
This means that the stopping distance increases by a factor of 16 when the speed of the car increases 4 times.

### Final Answer:
The stopping distance will increase by **16 times**.