Someone help me pls pls pls

The velocity function, in feet per second, is given for a particle moving along a straight line. Find (a) the displacement and (

AveGali [126]

Answer:

(a) 2 feet.

(b) 2 feet.

Step-by-step explanation:

We have been given that the velocity function $v(t)=\frac{1}{\sqrt{t}}$ in feet per second, is given for a particle moving along a straight line.

(a) We are asked to find the displacement over the interval $1\leq t\leq 4$ .

Since velocity is derivative of position function , so to find the displacement (position shift) from the velocity function, we need to integrate the velocity function.

$\int\limits^b_a {v(t)} \, dt$

$\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$

$\int\limits^4_1 {\frac{1}{t^{\frac{1}{2}}} \, dt$

$\int\limits^4_1 t^{-\frac{1}{2}} \, dt$

Using power rule, we will get:

$\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}}\right] ^4_1$

$\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}}\right] ^4_1$

$\left[2t^{\frac{1}{2}}\right] ^4_1$

$2(4)^{\frac{1}{2}}-2(1)^{\frac{1}{2}}=2(2)-2=4-2=2$

Therefore, the total displacement on the interval $1\leq t\leq 4$ would be 2 feet.

(b). For distance we need to integrate the absolute value of the velocity function.

$\int\limits^b_a |{v(t)|} \, dt$

$\int\limits^4_1 |{\frac{1}{\sqrt{t}}}| \, dt$

Since square root is not defined for negative numbers, so our integral would be $\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$ .

We already figured out that the value of $\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$ is 2 feet, therefore, the total distance over the interval $1\leq t\leq 4$ would be 2 feet.

7 0

3 years ago

A video posted on the internet has gone viral, and the total number of views is increasing by 25% every hour. If the video curre

sergeinik [125]

$\qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &155000\\ r=rate\to 25\%\to \frac{25}{100}\dotfill &0.25\\ t=hours\dotfill &6\\ \end{cases} \\\\\\ A=155000(1 + 0.25)^{6}\implies A=155000(1.25)^6\implies A\approx 591278$

8 0

2 years ago

Need help ASAP please

Rom4ik [11]

Answer:

What do you need help with?

Step-by-step explanation:

5 0

4 years ago

What is the equation of the graph below

Delicious77 [7]

Answer:

y = csc (x) + 2

Step-by-step explanation:

From the graph, we can derive the parent function y = csc(x). Notice how there are asymptotes at x = 2πk and x = π + 2πk, which is where csc(x) is undefined.

Finally, we can see a vertical shift of 2 which we can see from the mid-line of the graph which is at y = 3.

5 0

3 years ago

A stick is broken into two pieces, at a uniformly random break point. Find the CDF and average of the length of the longer piece

Oksanka [162]

Answer:

P(L ≤ l) =P (1-l ≤ U ≤ l)= l- ( 1 - l ) = 2 l - 1

Step-by-step explanation:

let assume that stick has length 1.Random variable L that make length of a longer piece and random variable U that mark point .See that L < l means that

U≤ l and 1-U ≤l

P(L ≤ l) =P (1-l ≤ U ≤ l)= l- ( 1 - l ) = 2 l - 1

this means 1-l≤U≤l

so we have

if we have L [1/2,1]

then apply the formula we have E(L)=3/4

5 0

4 years ago