Answer:
Step-by-step explanation:
We can assign variables as:
d = original breaking distance
b = original speed of the vehicle
d′ = increased breaking distance
b′ = increased speed of the vehicle
We can look at it this way, originally it was like this:
d=kb2
Now, if we increase the speed to b’ which 200% more than b, we will have
d′=k(b′)2
d′=k(b+2b)2
d′=k(3b)2
d′=9kb2
Obtaining the difference between d and d’, we have
d′−d=9kb2−kb2
d′−d=8kb2
in terms of d
d′−d=8(kb2)
d′−d=8d
Adding d on both sides of the equation,
d′=d+8d
Which means the original breaking distance is increased by 800%. This is similar to the phrase “the speed of the vehicle is increased by 200%” which translates to “b + 2b”.
Hope this answer helps you :)
Have a great day
Mark brainliest
There is a 100% increase in the braking distance after 50% increase in speed.
A body is said to be of speed v m/s if it travels v meter per second.
The braking distance of a car is directly proportional to the velocity of its speed(v m/s)
When the speed of the car is increased by 50% then the speed of the car v m/s becomes (3v/2) m/s.
The braking distance of a car is directly proportional to the square of its speed hence, the braking distance becomes
v²c where c is some constant.
So, the braking distance after 50% of the car speed is increased becomes (3v/2)²c₁, where c₁ is some constant.
Hence, the percentage change in the car's braking distance is
[tex]\frac{\frac{9}{4}-\frac{3}{2}}{\frac{3}{2}}[/tex]
[tex]=\frac{\frac{9-6}{4}}{\frac{3}{2}}\\=1[/tex]
That means there is a 100% increase in the braking distance after 50% increase in speed.
Learn more about speed here-
brainly.com/question/7359669
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