Question

A kite 300m high and there are 500m of chord out. if wind moves the kite horizontally at the rate of 5 km/ hrs. directly away from the person who is flying it

Answer 1

Answer:

4 K.m/ Hr.

Step-by-step explanation:

Hey Mate,

Given,

x = 400 m

y = 300 m

r = 500 m

dx/dt = 5 K.m/ Hrs

dy/dt = 0

To find ,

dr/dt = ?

X² + Y² = R²

2x (dx/dt) + 2y (dy/dt) = 2r (dr/dt)

{ 2 × (400/1000) × 5 } + { 2 × (300/1000) × 0 } = {2 × (300/1000) × dr/dt}

dr/dt = (4/5) × 5

dr/dt = 4 K.m/ Hr.

Answer 2

"Answer: $\pi /2$

Given that  

x = 400 m , y  = 300 m ,  r = 500 m

$\frac { dx }{ dt } \quad =\quad 5kmph\quad =\quad 5\times \frac { 5 }{ 18 } \quad =\quad \frac { 25 }{ 18 }$

$\frac { dy }{ dt } \quad =\quad 0$

$\frac { dr }{ dt } \quad =\quad ?$

X2 + Y2 = R2

$2x(\frac { dx }{ dt } )\quad =\quad 2y(\frac { dy }{ dt } )\quad =\quad 2r(\frac { dr }{ dt } )$

$2\times 400\times (\frac { 25 }{ 18 } )\quad =\quad 2\times 500(\frac { dr }{ dt } )$

$\frac { dr }{ dt } \quad =\quad 4\quad kmph$

For a 2nd order system, if the peak time is 2 times the rise time (tp=2tr), then the system would be?

Solution:

Given that  , peak time is 2 times the rise time (tp=2tr)

Peak time expression  $tp = \pi/(wn\sqrt1 - E2)$

Rise time expression  $tr = ( \pi - \theta ) / (wn\sqrt1-E2)$

  tp = 2 tr

$\pi/(wn\sqrt1-E2) = 2 (  \pi - \theta ) / (wn\sqrt1-E2)$

$\pi = 2 ( \pi - \theta )$

$\pi = 2 \pi  - 2 \theta$

$2 \theta = \pi$

$\theta = \pi/2$ ( phase difference) "