Comments on: A train chasing a cat? http://cheapcatfurniture.net/a-train-chasing-a-cat.htm/ Find Cheap Cat Furniture And Cat Scratch Posts Online Wed, 18 Apr 2012 01:22:17 +0000 hourly 1 http://wordpress.org/?v=3.2.1 By: Madhukar http://cheapcatfurniture.net/a-train-chasing-a-cat.htm/comment-page-1/#comment-13604 Madhukar Tue, 02 Mar 2010 13:44:46 +0000 http://cheapcatfurniture.net/a-train-chasing-a-cat.htm#comment-13604 Answer: b) 4 : 1 worked out as under. D / v1 = (3/8) x / v2 ... ( 1 ) ( D + x ) / v1 = (5/8) x / v2 => D / v1 = (5/8) x / v2 - x / v1 ... ( 2 ) From equns. ( 1 ) and ( 2 ), (3/8) x /v2 = (5/8) x / v2 - x / v1 => 3 / (8v2) = 5 (8v2) - 1/ v1 => 1 / v1 = ( 5/8 - 3/8) / v2 => 4v2 = v1 Answer: b) 4 : 1 worked out as under.

D / v1 = (3/8) x / v2 … ( 1 )

( D + x ) / v1 = (5/8) x / v2
=> D / v1 = (5/8) x / v2 – x / v1 … ( 2 )

From equns. ( 1 ) and ( 2 ),

(3/8) x /v2 = (5/8) x / v2 – x / v1
=> 3 / (8v2) = 5 (8v2) – 1/ v1
=> 1 / v1 = ( 5/8 – 3/8) / v2
=> 4v2 = v1

]]> By: Parrocchiano http://cheapcatfurniture.net/a-train-chasing-a-cat.htm/comment-page-1/#comment-13605 Parrocchiano Tue, 02 Mar 2010 13:44:46 +0000 http://cheapcatfurniture.net/a-train-chasing-a-cat.htm#comment-13605 The train is far "x" form the entrance A. The tunnell is AB long. I use the equation [time=space/speed], and i name v1 the train speed, and v2 the cat speed. In the first case x/v1=(3/8)*AB/v2 v1=x*(8/(3AB))*v2 In the second case (x+AB)/v1=5/8*AB/v2 v1=(x+AB)*(8/(5AB))*v2 So x*(8/(3AB))*v2= =(x+AB)*(8/(5AB))*v2 x/3AB=(x+AB)/5AB x/(3AB)-x/(5AB)=1/5 2x/(15AB)=3/15 2x=3AB x=3AB/2 Now i use v1=x*(8/(3AB))*v2 v1=3AB/2*(8/(3AB)*v2 v1=4*v2 b) is correct. I haven't used two variables for the two times: they don't interest in this problem. I just know that in each case train's and cat's times are equal, so i can equiparate their rates space/speed in each case. Sorry for the english, i don't know scientific words, i hope to have been useful! The train is far "x" form the entrance A. The tunnell is AB long.
I use the equation [time=space/speed], and i name v1 the train speed, and v2 the cat speed.

In the first case
x/v1=(3/8)*AB/v2
v1=x*(8/(3AB))*v2

In the second case
(x+AB)/v1=5/8*AB/v2
v1=(x+AB)*(8/(5AB))*v2

So
x*(8/(3AB))*v2=
=(x+AB)*(8/(5AB))*v2

x/3AB=(x+AB)/5AB
x/(3AB)-x/(5AB)=1/5
2x/(15AB)=3/15
2x=3AB
x=3AB/2

Now i use
v1=x*(8/(3AB))*v2

v1=3AB/2*(8/(3AB)*v2
v1=4*v2

b) is correct.

I haven’t used two variables for the two times: they don’t interest in this problem. I just know that in each case train’s and cat’s times are equal, so i can equiparate their rates space/speed in each case.

Sorry for the english, i don’t know scientific words, i hope to have been useful!

]]> By: falzoon http://cheapcatfurniture.net/a-train-chasing-a-cat.htm/comment-page-1/#comment-13606 falzoon Tue, 02 Mar 2010 13:44:46 +0000 http://cheapcatfurniture.net/a-train-chasing-a-cat.htm#comment-13606 Let AB = 8 units in length. Therefore the cat is located at a point 3 units from A and 5 units from B. Using the formula : distance = speed * time, we have : FOR THE CAT : (1) 3 = CS * T1 (CS = cat's speed, T1 = time to get to A) Therefore, T1 = 3 / CS (2) 5 = CS * T2 (T2 = time to get to B) Therefore, T2 = 5 / CS Adding gives : T1 + T2 = 3 / CS + 5 / CS = 8 / CS. FOR THE TRAIN : Let D = distance of train from entrance when whistle blows (3) D = TS * T1 (TS = train speed, T1 is the same, towards A) Therefore, T1 = D / TS (4) D + 3 + 5 = TS * T2 (T2 is the same, towards B) Therefore, T2 = (D + 8) / TS Adding gives : T1 + T2 = D / TS + (D + 8) / TS = (2D + 8) / TS Now we equate both calculations of (T1 + T2) to get : 8 / CS = (2D + 8) / TS Therefore, TS / CS = (2D + 8) / 8 = (D + 4) / 4. Now we have to find D : Note that T2 = (5/3)*T1, because the distances are known and the cat's speed is purportedly constant. In (4), where we have T2 = (D + 8) / TS, we now get : (5/3)*T1 = (D + 8) / TS, so, T1 = 3(D + 8) / (5*TS) Now, equating this T1 with the T1 in (3), we get : D / TS = 3(D + 8) / (5*TS) TS cancels out and we're left with D = 12 units. Substituting D = 12 into TS / CS = (D + 4) /4, as shown above, gives : TS / CS = 16 / 4 = 4 / 1 which is a ratio of 4 : 1, or answer (b). Let AB = 8 units in length.
Therefore the cat is located at a point
3 units from A and 5 units from B.

Using the formula : distance = speed * time, we have :

FOR THE CAT :

(1) 3 = CS * T1 (CS = cat’s speed, T1 = time to get to A)
Therefore, T1 = 3 / CS

(2) 5 = CS * T2 (T2 = time to get to B)
Therefore, T2 = 5 / CS

Adding gives :
T1 + T2 = 3 / CS + 5 / CS = 8 / CS.

FOR THE TRAIN :

Let D = distance of train from entrance when whistle blows

(3) D = TS * T1 (TS = train speed, T1 is the same, towards A)
Therefore, T1 = D / TS

(4) D + 3 + 5 = TS * T2 (T2 is the same, towards B)
Therefore, T2 = (D + 8) / TS

Adding gives :
T1 + T2 = D / TS + (D + 8) / TS = (2D + 8) / TS

Now we equate both calculations of (T1 + T2) to get :

8 / CS = (2D + 8) / TS

Therefore, TS / CS = (2D + 8) / 8 = (D + 4) / 4.

Now we have to find D :

Note that T2 = (5/3)*T1, because the distances are
known and the cat’s speed is purportedly constant.

In (4), where we have T2 = (D + 8) / TS, we now get :
(5/3)*T1 = (D + 8) / TS,
so, T1 = 3(D + 8) / (5*TS)

Now, equating this T1 with the T1 in (3), we get :
D / TS = 3(D + 8) / (5*TS)
TS cancels out and we’re left with D = 12 units.

Substituting D = 12 into TS / CS = (D + 4) /4,
as shown above, gives :
TS / CS = 16 / 4 = 4 / 1
which is a ratio of 4 : 1, or answer (b).

]]>