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Marizza181 [45]
3 years ago
9

While passing a slower car on the highway, i accelerate uniformly from 12 m/s to 24 m/s in a time of 10.0s. How far do you trave

l during this time and what is the magnitude of your acceleration
Mathematics
1 answer:
Taya2010 [7]3 years ago
6 0
You say that there is uniform acceleration so:

vf-vi=at  (final velocity minus initial velocity is equal to acceleration times time)

We know vf, vi, and t so we can solve for acceleration:

24-12=a10

12=10a

a=1.2

That is the acceleration, we will need to integrate with respect to time twice...

v=⌠a dt

v=at+vi  , we know a=1.2m/s^2 and vi=12m/s

v=1.2t+12, 

x=⌠1.2t+12 dt

x=1.2t^2/2+12t+xo, we can just let xo=0 for this problem...

x(t)=0.6t^2+12t

Now we know that this acceleration lasts for 10 seconds so the distance traveled in that time is:

x(10)=0.6(10^2)+12(10)

x(10)=60+120

x(10)=180 meters
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one of the fastest cars in the world was recently measured covering 5726 meters in 50 seconds what is the unit rate of speed of
IRISSAK [1]

ANSWER

114.52 meters per second

EXPLANATION

We have that the car covers a distance of 5726 meters in 50 seconds.

To find the speed, we have to use the formula for speed which is:

\text{Speed = }\frac{dis\tan ce}{time}

So, the unit rate of speed of the car is:

\begin{gathered} \text{Speed = }\frac{5726}{50} \\ \text{Speed = 114.52 meters per second} \end{gathered}

That is the unit rate of speed of the car in meters per second.

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1 year ago
A line with a slope of 7 and an x-intercept of 4 passes through which of the following points?
saw5 [17]
A I think but I’m not postitive
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3 years ago
У = 2х – 1<br> у = 3х + 2<br> Using substitution method
sergij07 [2.7K]

Answer:

x = -3, y = -7

Step-by-step explanation:

Eliminate the equal sides of each equation and combine.

Solve 2x - 1 = 3x + 2 for x.

-x - 1 = 2

-x = 3

x = -3

Evaluate y when x = -3

y = 3 (-3) + 2

y = -7

(-3, -7)

<em>good luck, i hope this helps :)</em>

7 0
3 years ago
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Help me with differentation and integration please!!
Marina86 [1]

Answer:

See below

Step-by-step explanation:

\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x

Recall

\dfrac{d}{dx}\tan x=\sec^2

Using the chain rule

\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}

such that u = \tan x

we can get a general formulation for

y = \tan^n x

Considering the power rule

\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}

we have

\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x

therefore,

\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x

Now, once

\sec^2 x - 1= \tan^2x

we have

3\tan^2x \sec^2x =  3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x

Hence, we showed

\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x

================================================

For the integration,

$\int \sec^4 x\, dx $

considering the previous part, we will use the identity

\boxed{\sec^2 x - 1= \tan^2x}

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx $

Considering u = \tan x

and then du=\sec^2x\ dx

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx  = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0
2 years ago
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