Answer to Question #250118 in Calculus for Alunsina

Question

Cliff left point A at 8 A.M. walking east at 3 kph. Renz left point A at 9 A.M. walking north at 4 kph.

(a) Express the distance between Cliff and Renz as a function of the elapsed time 𝑡 since 8 A.M.

(b) Express the distance between Cliff and Renz as a function of the elapsed time 𝑡 since 9 A.M.

(c) How far apart are Cliff and Renz at 12:00 noon on the same day?

(d) At what time are they 20 km apart? 

Answer 1

Q1) 

 

 

Let “ t “ be in hours.

 

A) As from 8 A.M to 9 A.M cliff was only moving. 

 

During this time distance moved by Cliff in “t” time :

 

D₁= 3t East 

 

Now distance moved by Renz in the remaining “ t-1” hr :

 

D₂= 4(t-1) km North

 

So the next separation would be:

 

D= ( D₁²+D₂²)^0.5= ( 25t²-32t+16)^0.5 for t >1

 

And D= 3t , for t < 1

 

B) As from 8 A.M to 9 A.M cliff was only moving. 

 

During this time distance moved by Cliff in “t” time after 9 Am :

 

D₁= 3 + 3t East , as in 1 hr he already covered 3 km

 

Now distance moved by Renz in the remaining “ t” hr :

 

D₂= 4t km North

 

So the next separation would be:

 

D= ( D₁²+D₂²)^0.5= ( 25t²+18t+9)^0.5

 

C) 

 

Using 2nd formula, t = 3 hr from 9 AM to 12 Noon.

 

D = ( 25(3)²+18(3)+9)^0.5

D= 16.97 km

 

 

 

D) using same formula:

 

D= ( 25t²+18t+9)^0.5

 

20= ( 25t²+18t+9)^0.5

 

Squaring both sides we get 

 

400 = 25t²+18t+9

 

25t²+18t- 391 = 0

On solving we get:

 

t = 3.61 hr or -4.33 hr 

 

So only possible value is 3.61 hr

 

So from 8 AM the total time is :

T = 1+ 3.61 = 4.61 hr