Physics, 11.05.2021 19:40, shavis123

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) What is the velocity of the top of the ladder when the base is given below

Answers: 1

Show answers

Answers

answered: ians1544

This question is incomplete, the complete question is;

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) What is the velocity of the top of the ladder when the base is given below

7 feet away from the wall

15 feet away from the wall

20 feet away from the wall

- When x is 7 feet away from the wall, $\frac{dr}{dt}$ = -7/12 ft/s

- When x is 15 feet away from the wall, $\frac{dr}{dt}$ = -3/2 ft/s

- When x is 20 feet away from the wall, $\frac{dr}{dt}$ = -8/3 ft/s

Explanation:

Given the data in the question;

Let ft and ft represent the distance of the top of ladder from the ground and the distance of the bottom of ladder from the wall respectively.

Now using the Pythagoras Theorem ( right angled triangle ).

(hypotenuse)² = ( adjacent )² + ( opposite )²

r² = ( 25 )² - x²

r² = 625 - x² ------- let this be equation 1

now, we differentiate with respect to t

2r $\frac{dr}{dt}$ = 0 - 2x $\frac{dx}{dt}$

r $\frac{dr}{dt}$ = -x $\frac{dx}{dt}$

$\frac{dr}{dt}$ = -x $\frac{dx}{dt}$ × 1/r

$\frac{dr}{dt}$ = -x/r $\frac{dx}{dt}$

Now, from equation, r² = 625 - x²

r = √( 625 - x² )

Then

$\frac{dr}{dt}$ = -x / √( 625 - x² ) $\frac{dx}{dt}$

given that, The base of the ladder is pulled away from the wall at a rate of 2 feet per second i.e $\frac{dx}{dt}$ = 2 ft/s

so;

$\frac{dr}{dt}$ = ( -x / √( 625 - x² ) ) × 2

$\frac{dr}{dt}$ = -2x / √( 625 - x² )

∴

When x is 7 feet away from the wall;

$\frac{dr}{dt}$ = -2(7) / √( 625 - (7)² )

$\frac{dr}{dt}$ = -14 / √( 625 - 49 )

$\frac{dr}{dt}$ = -14 / √576

$\frac{dr}{dt}$ = -14 / 24

we simplify

$\frac{dr}{dt}$ = -7/12 ft/s

Therefore, When x is 7 feet away from the wall, $\frac{dr}{dt}$ = -7/12 ft/s

When x is 15 feet away from the wall

$\frac{dr}{dt}$ = -2(15) / √( 625 - (15)² )

$\frac{dr}{dt}$ = -30 / √( 625 - 225 )

$\frac{dr}{dt}$ = -30 / √400

$\frac{dr}{dt}$ = -30 / 20

we simplify

$\frac{dr}{dt}$ = -3/2 ft/s

Therefore, When x is 15 feet away from the wall, $\frac{dr}{dt}$ = -3/2 ft/s

When x is 20 feet away from the wall

$\frac{dr}{dt}$ = -2(20) / √( 625 - (20)² )

$\frac{dr}{dt}$ = -40 / √( 625 - 400 )

$\frac{dr}{dt}$ = -40 / √225

$\frac{dr}{dt}$ = -40 / 15

we simplify

$\frac{dr}{dt}$ = -8/3 ft/s

Therefore, When x is 20 feet away from the wall, $\frac{dr}{dt}$ = -8/3 ft/s

answered: Guest

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.

answered: Guest

it doesn't. it was imagined up in the middle ages.