If 2 cars are travelling in the same direction and ....

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This is related to movement of indicators but is easy to write with cars.

Two cars travelling in the same direction car A is 10 miles behind car B, Car A is travelling 2x faster than car B. How many miles further down the road will Car A catch up with car B. The answer is 10 miles.

But if car A is travelling 3x faster the answer is 5 miles.

4X faster is a little over 3 miles (rough calculation in my head)

It looks like I should be able to use a formula to calculate that but I am not seeing it.

 

The formula is: 10/(V(a)-V(b))


V(a) = velocity of car A

V(b) = velocity of car b


example car A is 4X faster:


10/(4-1)=3,33

 

Damn yeah why didnt I see that, it seems obvious now you showed me, but I puzzled over it all morning. Thanks for your help.

 
user_123:

The formula is: 10/(V(a)-V(b))


V(a) = velocity of car A

V(b) = velocity of car b


example car A is 4X faster:


10/(4-1)=3,33

Your formula isn't right, as it suppose Vb=1.

The right formula is : d = 10 * Vb / (Va - Vb)

which can be simplified in case where Va = x * Vb (a is x times faster than b), by d = 10 / (x - 1).

EDIT: Va is speed of car a, Vb is speed of car b in this formula.

 

I think the original formula is right because we are calculating on the basis of Va relative to Vb so no matter what Va is, Vb will always be 1 but I agree that your d/(x-1) is the better formula

 
SDC:

I think the original formula is right because we are calculating on the basis of Va relative to Vb so no matter what Va is, Vb will always be 1

Nope, but it's as you wish.
 

if a is 3 times faster than b then how can b be anything but 1 ? b can only be 1x faster than b.

 
user_123: The formula is: 10/(V(a)-V(b))
y1 = m1 x + b1
y2 = m2 x + b2
y1-y2 = (m1-m2)x + (b1-b2) = 0 // at cross
 x = (b2-b1) / (m1-m2)         // when cross
                               // b2-b1 = 10 miles
 
WHRoeder:

And what is m1, m2, (m1-m2) ?
 
SDC:

if a is 3 times faster than b then how can b be anything but 1 ? b can only be 1x faster than b.

Va is velocity of car a, Vb is velocity of car b.

Vb can be 10miles/hour or 100miles/hour, why do you want it to be 1 ? If it's 1mile/hour then Va=x*1mph=2mph (or 3mph or 4 mph). But Vb don't have to be 1mph, it can be 75mph, then Va is 150mph (or 225mph or 300mph).

In all case, if ratio of Va/Vb=2 (=x) then the distance you asked is got by d=10/(x-1)=10/(2-1)=10 miles.

If you have Va and Vb and don't want to calculate the ratio then d=10*Vb/(Va-Vb)=10*75/(150-75)=10 miles.

EDIT: When I wrote velocity above, I mean in fact speed.

 
angevoyageur: And what is m1, m2, (m1-m2) ?
Speed of Car1, of car2, difference in speeds. I was simply trying to show how user_123 came up with the equation 10/(Va-Vb) . Slopes [y=mx+b] if you're talking lines, speed [x(t) = vt + x0] if you're talking cars. Velocity is a vector (speed plus direction) and since they are assumed to be on the same road direction is constant.
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