All categories

3 years ago

The braking distance Dv (in meters) for a certain car moving at velocity v (in meters/second) is given by =Dvv234.

Mathematics

1 answer:

MaRussiya [10]3 years ago

8 0

Answer:

S(t)= $\frac{9t^2}{34}$

Step-by-step explanation:

We are given that

Th braking distance for a certain car moving at velocity v(in m/s) is given by

$D(v)=\frac{v^2}{34}$

The velocity of car B(t) t seconds after starting = $B(t)=3t$

We have to find the value for braking distance S(t) after t seconds.

To find the formula for braking distance we will substitute the value of velocity B(t) in place of v in D(v)

Substitute the value of velocity then we get

Then, we get

The formula for the braking distance S(t) after t seconds= $\frac{(3t)^2}{34}$

The formula for the braking distance S(t) after t seconds= $\frac{9t^2}{34}$

You might be interested in

1) A shirt went from $8 to $10.

bagirrra123 [75]

Answer:

1) 20% change I think its an increase.

If needed I could try to answer the 2nd question but I am sure that I got the first one right.

Step-by-step explanation:

7 0

2 years ago

Find the slope of the line that contains the points (-5, 3) and (-1,-5).

Nutka1998 [239]

Look at the attached picture⤴

Hope this will help u....

8 0

2 years ago

2ab + 4 = d solve for a

Anestetic [448]

Answer:

a = d/2b - 2/b

Step-by-step explanation:

2ab+4=d

2ab=d-4

a= d/2b - 4/2b

8 0

2 years ago

I'LL GIVE YOU BRAINIEST! 1 1/5 · x/3 = 9/10 solve for x

joja [24]

$\huge\text{Hey there!}$

$\huge\textbf{Equation:}$

$\mathbf{1 \dfrac{1}{5} \times \dfrac{x}{3} = \dfrac{9}{10}}$

$\huge\textbf{Solve:}$

$\mathbf{1 \dfrac{1}{5} \times \dfrac{x}{3} = \dfrac{9}{10}}$

$\mathbf{\rightarrow \dfrac{1\times5+1}{5} \times \dfrac{x}{3} = \dfrac{9}{10}}$

$\mathbf{\rightarrow \dfrac{5 + 1}{5} \times \dfrac{x}{3} = \dfrac{9}{10}}$

$\mathbf{\rightarrow \dfrac{6}{5} \times \dfrac{x}{3} = \dfrac{9}{10}}$

$\huge\textbf{Simplify it:}$

$\mathbf{\rightarrow \dfrac{2}{5}x = \dfrac{9}{10}}$

$\huge\textbf{Multiply \boxed{\bf \dfrac{5}{2}} to both sides:}$

$\mathbf{\rightarrow \dfrac{5}{2}\times \dfrac{2}{5}x = \dfrac{5}{2}\times \dfrac{9}{10}}$

$\huge\textbf{Simplify it:}$

$\mathbf{\rightarrow x = \dfrac{5}{2}\times \dfrac{9}{10}}$

$\mathbf{\rightarrow x = \dfrac{5\times9}{2\times10}}$

$\mathbf{\rightarrow x = \dfrac{45}{20}}$

$\mathbf{\rightarrow x = \dfrac{45\div5}{20\div5}}$

$\mathbf{\rightarrow x = \dfrac{9}{4}}$

$\mathbf{x \approx 2 \dfrac{1}{4}}$

$\huge\textbf{Therefore, your answer should be:}$

$\huge\boxed{\mathsf{x = }\frak{2 \dfrac{1}{4}}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

8 0

2 years ago

Read 2 more answers

To three decimal places, find the average value of f(x) = 2x2 + 3 on the interval [0, 2].

salantis [7]

The formula for the average value of a function is $\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx$ where b is the upper bound and a is the lower. For us, this formula will be filled in accordingly. $\frac{1}{2} \int\limits^2_0 {(2x^2+3)} \, dx$ . We will integrate that now: $\frac{1}{2}[ \frac{2x^3}{3}+3x]$ from 0 to 2. Filling in our upper and lower bounds we have $\frac{1}{2}[( \frac{2(2^3)}{3}+3(2))-0]$ which simplifies to $\frac{1}{2}( \frac{16}{3}+6)$ and $\frac{1}{2}( \frac{16}{3}+ \frac{18}{3})= \frac{1}{2}( \frac{34}{3})$ which is 17/3 or 5.667

7 0

3 years ago