Rate distance btwn two points two hours later?

Two cars start moving from the same point, one travels south at 60 mi/h and the other travels west at 25 mi/h. at what rate is the distance between the cars increasing two hours later?

um if possible please explain to me how to get the answer...not just the answer.

Rate distance btwn two points two hours later?

Distance between the two cars is

D = sqrt[(60t)虏 + (25t)虏]

D = sqrt(3600t虏 + 625t虏)

D = sqrt(4225 t虏)

D = sqrt(4225) t = 65t

dD/dt = 65 miles per hour

Rate distance btwn two points two hours later?

v = d/t

v = dt

v = rate or velocity in mph

d = distance

t = time

car travelling south the distance travelled is

d1 = 60t

car travelling west the distance travelled is

d2 = 25t

the distance between cars c can be obtained using pythagorean theorem

c^2 = d1^2 + d2^2

or

c = (d1^2 + d2^2)^1/2

c =((60t)^2 + (25t)^2)^1/2

c = =(3600t^2 + 625t^2)^1/2

Get the derivative of c with time

using the chain rule

let u= 3600t^2 + 625t^2

du/dt = 7200t + 1250t

c = u^(1/2)

dc/du = (1/2)u^(-1/2)

then dc/dt = (dc/du)(du/dt)

dc/dt = (1/2)u^(-1/2)(7200t +1250t) or

dc/dt = (7200t +1250t)/(2sqrt(u))

dc/dt = 8450t/(2sqrt(3600t^2 + 625t^2))

dc/dt = 8450t/(2sqrt( 4225t^2)) = 8450t/(2t(sqrt(4225)))

t cancels out since the two cars are travelling at a constant speed, the distance between them also changes at a constant rate,

dc/dt = 8450/(2(65)) = 8450/130

dc/dt = 65 mph

Rate distance btwn two points two hours later?

Give first position is (0,0) on xy-asix.

At two hours later

the first car position was at (0 mi,-120 mi)

the second car position was at (-50 mi,0 mi)

1) the distance between the cars is squareroot of (120^2+50^2)

= 130 mi. in two hours ...... ANSWER.

2) distance rate per hour is 130/2 = 65 mi/hr. ...... ANSWER.

Rate distance btwn two points two hours later?

A................B

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.

.

.

C .

(m making one of the cars go east,instead of west coz otherwise the figure wudnt come out right (the concept is same)

let car x be travelling east @ 25 m/hr., n y be travelling south at 60m/hr.

after 2 hrs, distance AB will be = 25x2=50

n distance AC = 60x2= 120.

join AC..we hav a right angle triange ABC., therefore

BCxBC= ABxAB + ACx AC (pythagoras theorem)

therefore, BCxBC = 50x50 + 120x120

BCx BC= 16900

hence, BC= 130miles.

since after 2 hrs the distance between the cars is 130 miles, it is increasing at the rate of 65m/hr.

Rate distance btwn two points two hours later?

-----------N

W-1------|--------E -----1 is showing position of car 1

-------- ---|

-----------2 -----2 is showing position of car 2

-----------S--------

see the directions, one car is moving to south 60 mi/h

in two hours it will travel 60 脳 2 = 120 km.

another car is moving to west 60 mi/h

in two hours it will travel 25 脳 2 = 50 km.

It forms a right angled triangle having sides 120 and 50 km

you have to find the diagonal of the triangle.

(diagonal)虏 = (side 1)虏 + (side 2)虏

= 120虏 + 50虏

= 14400 + 2500

= 16900

diagonal = 鈭?6900 = 130

Distance between two cars = 130 miles.

Rate distance btwn two points two hours later?

Howdy %26amp; I sure hope all is well. :) As for your question:

First off, as the cars move away from each other, they create a triangle between them, composed of:

* The distance from the starting point to how far West the West-moving car got to 2 hours later.

* The distance from the starting point to how far South the South-moving car got to 2 hours later.

%26amp;

* The diagonal between the 2 cars.

I%26#039;ll try to draw you a diagram, but I know the system will just mangle it. In case it works, X is how far West one car got, Y is how far South the other car got %26amp; Z is the diagonal between the 2 cars.

%26gt;---X---

%26gt;\~~~~|

%26gt; \~~~ |

%26gt;%26gt;\~~~|

%26gt;%26gt; \~~ |

%26gt;%26gt;%26gt;\~~|

%26gt;%26gt;Z \~ | Y

%26gt;%26gt;%26gt;%26gt;\~|

%26gt;%26gt;%26gt;%26gt; \ |

%26gt;%26gt;%26gt;%26gt;%26gt;\|

What you%26#039;re looking for is dZ/dt. So, according to the Pythagoras theorem we have:

X^2 + Y^2 = Z^2

Differentiating with respect to time t we get:

(1) 2X(dX/dt) + 2Y(dY/dt) = 2Z(dZ/dt)

To simplify, we divide both sides of the equation by 2 (both sides, so as to preserve the equality, of course, as I%26#039;m sure you already know. :) ). We get:

2X(dX/dt)/2 + 2Y(dY/dt)/2 = 2Z(dZ/dt)/2

Simplifying yields:

(2) X(dX/dt) + Y(dY/dt) = Z(dZ/dt)

We are given in the question that:

dX/dt = 25 mi/h

%26amp;

dY/dt = 60 mi/h

Now, for the car going West:

In 1 hour it travels 25 miles

In 2 hours it travels X

That is:

1 hour ---%26gt; 25 miles

2 hours ---%26gt; X

Cross-multiplying yields:

X*1 hour = 2 hours*25 miles

To Solve for X, we divide both sides of the equation by 1 hour. We get:

X*1 hour/1 hour = 2 hours*25 miles/1 hour

Simplifying yields:

X = 50 miles

Similarly, for the South-moving car, we have:

1 hour ---%26gt; 60 miles

2 hours ---%26gt; Y

Cross-multiplying yields:

Y*1 hour = 2 hours*60 miles

To Solve for Y, we divide both sides of the equation by 1 hour. We get:

Y*1 hour/1 hour = 2 hours*60 miles/1 hour

Simplifying yields:

Y = 120 miles

Now, after 2 hours, X is 50 miles %26amp; Y is 120 miles. So Z is:

X^2 + Y^2 = Z^2

(50 miles)^2 + (120 miles)^2 = Z^2

2500 miles^2 + 14400 miles^2 = Z^2

16900 miles^2 = Z^2

Now, to solve for Z, we take the square root of both sides by raising them both to the power of 1/2. We get:

(16900 miles^2)^(1/2) = (Z^2)^(1/2)

Simplifying yields:

130 miles = Z

Thus, after 2 hours, the diagonal distance between them is 130 miles.

Now, armed with the information we calculated %26amp; what%26#039;s given in the question, we substitute them into (2). We get:

X(dX/dt) + Y(dY/dt) = Z(dZ/dt)

(50 miles)(25 mi/h) + (120 miles)(60 mi/h) = (130 miles)(dZ/dt)

Simplifying yields:

1250 (miles^2)/h + 7200 (miles^2)/h = (130 miles)(dZ/dt)

Simplifying further yields:

8450 (miles^2)/h = (130 miles)(dZ/dt)

Finally, to solve for (dZ/dt), we divide both sides of the equation by 130 miles. We get:

[8450 (miles^2)/h]/(130 miles) = (130 miles)(dZ/dt)/(130 miles)

Simplifying yields:

65 mi/h = (dZ/dt)

Therefore, after 2 hours, the distance between the 2 cars is increasing at the rate of 65 mi/h.

I sure hope that was as clear %26amp; as helpful as you needed it to be. :) Good luck, take care %26amp; have a great day. :)

Cheers! :)