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Rate and Distance Question

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Rate and Distance Question

by Abdulla » Fri Nov 07, 2008 3:27 pm
I'm confused with two things:

When the traveler travel towards each other why sometimes we equal the two equations together ad solve and the other time we add the two distance and = to the total distance.
i.e
R1= 240 T1= t+10 min. Total Dis=300
R2=160 T2 = t

Rephrasing my question : why we did 240 (t+1/6)+ 160t = 300 rather than
240(t+1/6)=160t and solve ..

When I have to use either way?
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by scoobydooby » Fri Nov 07, 2008 11:38 pm
if they travel toward each other, add the distance covered by each to get the total distance travelled or the distance between them

if they travel in the same direction, one overtakes the other, each must cover the same distance before the slower one is overtaken by the faster one. So one has to equate the distance travelled
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by Abdulla » Fri Nov 07, 2008 11:55 pm
nice, so in this case its always adding the two distances.
OK then what about traveling in the opposite directions?
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Re: Rate and Distance Question

by logitech » Fri Nov 07, 2008 11:57 pm
Abdulla wrote:I'm confused with two things:

When the traveler travel towards each other why sometimes we equal the two equations together ad solve and the other time we add the two distance and = to the total distance.
i.e
R1= 240 T1= t+10 min. Total Dis=300
R2=160 T2 = t

Rephrasing my question : why we did 240 (t+1/6)+ 160t = 300 rather than
240(t+1/6)=160t and solve ..

When I have to use either way?
Towards each other

Lets say the distance between the travelers is D

Traveler 1 , and his speed V1
Traveler 2, and his speed V2

time they will meet T

So when they meet,

traveler 1, travels D1 = V1 * T
traveler 2, travels D2 = V2 * T

Also note that D1 + D2 = D

(V1 * T) + ( V2 * T ) = V T

Here V stands for their relative speed.

if you solve this equation for V:

V = V1+V2

so lets say the distance is 10 miles and V1=2miles/hr and V2=3miles/hr

so they will meet in : 10/(2+3) = 2 hrs

in 2 hrs,

Traveler 1 will travel 2x2 = 4 miles - D1
Traveler 1 will travel 2x3 = 6 miles - D2

Lets say they travel towards same direction


Traveller2...............3miles.......Traveler1......>

And they have 3 miles in between. Since V2>V1

Second guy will catch him at some point, after T hrs

When they travel the same distance, they will meet.

So Traveler 2 will travel

3 miles + X = V2 * T

Traveler 1 will travel

X = V1 * T

If you insert the latter equation into the former one:

3 miles + V1 * T = V2 * T

And if you solve for T:

T = 3 miles / ( V2-V1)

This is why we we subtract the velocities and divide the distance with this difference to find out when they will meet.

in our example:

T = 3/ (3-2) = 3 hrs


[/u]
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by Abdulla » Sat Nov 08, 2008 12:11 am
Thanks logitech..
in my second question i meant opposite directions.
<.................Traveler1 ....... ........Traveler2................>
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by logitech » Sat Nov 08, 2008 12:25 am
Abdulla wrote:Thanks logitech..
in my second question i meant opposite directions.
<.................Traveler1 ....... ........Traveler2................>
When they don't like each other anymore and decide to walk away from each other, the GAP, the DISTANCE, between them will get bigger by time.

So That distance is D

D=D1+D2

D=V1T+V2T

D=T(V1+V2)

Using my example above

D=T(3+2)

D=5T

in other words, the distance will get 5 miles more every hour..
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by Abdulla » Sat Nov 08, 2008 12:27 am
Great explanation..

Thanks ..
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