PLEAS HELP I NEED HELP SO BAD ANSWER FAST AND CORRECTLY PLSSSSSS

Two drivers, Alison and Kevin, are participating in a drag race. Beginning from a standing start, they each proceed with a const

Vilka [71]

Answer:

Alison wins against Kevin by 0.93 s

Step-by-step explanation:

Alison covers the last 1/4 of the distance in 3 seconds, at a constant acceleration $a_a$ , we have the following equation of motion

$s/4 = a_at_a^2/2$

where s (m) is the total distance, ta = 3 s is the time

$s = 4a_a3^2/2 = 18a_a$

$a_a = s/18$

Similarly, Kevin overs the last 1/3 of the distance in 4 seconds, at a constant acceleration $a_k$ , we have the following equation of motion:

$s/3 = a_kt_k^2/2$

tk = 4 s is the time

$s = 3a_k4^2/2 = 24a_k$

$a_k = s/24$

Since $a_a = s/18$ we can conclude that $a_a > a_k$ , so Alison would win.

The time it takes for Alison to cover the entire track

$s = a_aT_a^2/2$

$T_a^2 = 2s/a_a = 2s/(s/18) = 36$

$T_a = \sqrt{36} = 6 s$

The time it takes for Kevin to cover the entire track

$s = a_kT_k^2/2$

$T_k^2 = 2s/a_k = 2s/(s/24) = 48$

$T_a = \sqrt{48} = 6.93 s$

So Alison wins against Kevin by 6.93 - 6 = 0.93 s

4 0

2 years ago

Henry is using cardboard to build a box in the shape of a rectangular prism. A net of the box is shown.

Brut [27]

Answer:240 square feet

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Which are correct representations of the inequality –3(2x – 5) < 5(2 – x)? Check all that apply.

k0ka [10]

Answer:

C. $-6x+15$

A. First Picture

Step-by-step explanation:

We are given the inequality, $-3(2x-5)$ .

On simplifying the inequality, we get,

$-3(2x-5)$

i.e. $-6x+15$

i.e. $-x$

i.e. $x>5$

So, we get the correct representations of the inequality are,

C. $-6x+15$

A. the First Picture

5 0

2 years ago

Read 2 more answers

Select the correct expressions. 1 Identify each expression that represents the slope of a tangent to the curve y= * +1 at any po

Olegator [25]

The expressions which represents the slope of a tangent to the curve $y=\frac{1}{x+1}$ at any point (x, y) are:

$f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}$

<h3>The slope of a tangent to the curve.</h3>

Mathematically, the slope of a tangent line to the curve is given by this equation:

$f'(x) =limh \rightarrow 0\frac{f(x+h)-f(x)}{h}$

Given the function:

$f(x)=y=\frac{1}{x+1}$

When (x + h), we have:

$f(x+h)=y=\frac{1}{x+h+1}$

Next, we would find the derivative of f(x):

$f'(x) =limh \rightarrow 0\frac{\frac{1}{x+h+1} -\frac{1}{x+1}}{h}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -(x+h+1)}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -x-h-1}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}\\\\f'(x) = \frac{-1}{(x+1)(x+1)} \\\\f'(x) = \frac{-1}{(x+1)^2}$