Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
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
[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
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.