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The braking distance (dm) of a car is directly
proportional to the square of its speed (v m/s)
When the speed of the car is increased by
50%, what is the percentage change in its
braking distance?

Sagot :

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.

What is speed?

A body is said to be of speed v m/s if it travels v meter per second.

How to solve it?

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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