When speed of a vehicle becomes n times, with the application of same stopping force, its stopping distance becomes

Options

A. `n^(2)` times
B. `n^(3)` times
C. `(1)/(n^(2))` times
D. `sqrt(n)` times

Answer

To solve the problem of how the stopping distance of a vehicle changes when its speed is increased by a factor of \( n \) while applying the same stopping force, we can follow these steps:

### Step 1: Understand the relationship between speed, acceleration, and stopping distance
When a vehicle is moving at an initial speed \( u \) and comes to a stop under a constant deceleration (or stopping force), we can use the equation of motion:
\[ v^2 = u^2 + 2a x \]
where:
- \( v \) is the final velocity (0 when the vehicle stops),
- \( u \) is the initial velocity,
- \( a \) is the acceleration (negative since it is deceleration),
- \( x \) is the stopping distance.

### Step 2: Apply the equation for the initial speed
For the initial speed \( u \):
\[ 0 = u^2 + 2a x \]
Rearranging gives:
\[ x = -\frac{u^2}{2a} \]

### Step 3: Increase the speed to \( n \times u \)
Now, if the speed of the vehicle is increased to \( n \times u \), we can apply the same equation of motion:
\[ 0 = (n u)^2 + 2a x' \]
where \( x' \) is the new stopping distance.

### Step 4: Substitute and rearrange for the new stopping distance
Substituting \( (n u)^2 \) into the equation gives:
\[ 0 = n^2 u^2 + 2a x' \]
Rearranging this equation gives:
\[ x' = -\frac{n^2 u^2}{2a} \]

### Step 5: Relate the new stopping distance to the original stopping distance
From our earlier expression for \( x \):
\[ x = -\frac{u^2}{2a} \]
Now, we can express \( x' \) in terms of \( x \):
\[ x' = n^2 \left(-\frac{u^2}{2a}\right) = n^2 x \]

### Conclusion
Thus, when the speed of the vehicle becomes \( n \) times the original speed, the stopping distance becomes \( n^2 \) times the original stopping distance.

### Final Answer
The stopping distance becomes \( n^2 \) times the original stopping distance.

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