To determine how the stopping distance [tex]\( s \)[/tex] changes with respect to the car's velocity [tex]\( v \)[/tex], let's analyze the given equation:
[tex]\[
s = \frac{v^2}{a}
\][/tex]
where [tex]\( s \)[/tex] is the stopping distance, [tex]\( v \)[/tex] is the velocity, and [tex]\( a \)[/tex] is a constant representing the acceleration.
1. The relationship shows that [tex]\( s \)[/tex] is directly proportional to [tex]\( v^2 \)[/tex].
2. As [tex]\( v \)[/tex] increases, [tex]\( v^2 \)[/tex] will increase as well.
3. Therefore, since stopping distance [tex]\( s \)[/tex] is directly proportional to the square of the velocity [tex]\( v \)[/tex], any increase in [tex]\( v \)[/tex] will result in an increase in [tex]\( s \)[/tex].
4. Specifically, because [tex]\( s \)[/tex] depends on [tex]\( v^2 \)[/tex], this is a quadratic relationship.
So, as the car's velocity [tex]\( v \)[/tex] increases, the stopping distance [tex]\( s \)[/tex] increases quadratically.
Thus, the correct answer is:
B) Increases quadratically
