cars leave the same city

[sanju09]

Two cars leave the same city traveling on the same road in the same direction. The second car leaves one hour after the �rst. How long will it take the second car to catch up with the �rst one?

(1) The second car is traveling 10 miles per hour faster than the �rst one.

(2) The second car averages 60 miles per hour.


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[neelgandham]

Two cars leave the same city traveling on the same road in the same direction. The second car leaves one hour after the �rst. How long will it take the second car to catch up with the �rst one?

(1) The second car is traveling 10 miles per hour faster than the �rst one.

(2) The second car averages 60 miles per hour.

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Answer C

[sanju09]

Answer C

[spoiler]correct! explanation?[/spoiler]

[neelgandham]

Sorry Sirjee (Assuming you are Sanjay ) ! Here is my explanation..

Two cars leave the same city traveling on the same road in the same direction. The second car leaves one hour after the �rst. How long will it take the second car to catch up with the �rst one?

Say the speed of first car is A, speed of second car is B, total distance traveled by the cars before they meet is X,

Then time taken by A = Time taken by B + 1 hour => X/A = X/B + 1 => X(B-A) = A*B. So the question can be rephrased to X/B = ?

(1) The second car is traveling 10 miles per hour faster than the �rst one.
=> B-A =10 = > X(10) = A*B - Doesn't answer the question !

(2) The second car averages 60 miles per hour.
B = 60mph Doesn't answer the question !

(1)and (2) X(10) = A*B => X(10) = 60*A => X/A = 6 => X/B =5 (As, X/A = X/B + 1) Hence C

[Redhorsep]

The second car catches up with first car when they travel the same distance or distance traveled by both equal each other (think of it as if they start from one side to the other side, when the late starter car catches up with the early starter car, they arrive at the same point, which is the same distance from the point where they started).

Let first car=F
Second car=S

Distance traveled by first car: F X (T+1) (since first car gets to travel 1 more hour than second car)
Distance traveled by second car: S X T
Set both distance equal:

F X (T+1) = S X T
distribute: FT + F = ST
Now that's try to understand what the equation created as above means:

The F means the gains in distance that first car has over second car and it is the distance for second car to catch up, it is the same as F X 1, which is the speed of F times 1 hour, given F's speed, this is the distance that F can travel in hour, which is exactly the same as its hourly speed.

Now in order to find out that gain in distance, we must do: ST - FT, distance traveled by second car minus the distance travelved by the first car WITHOUT the gains in distance had the first car not started one hour earlier.

So: rewrite the equation, you get:
F= ST-FT
F = T X (S-F)

Now what does S-F mean? it is the difference between the speed of first car and second car, or you can understand it as relative speed (relative difference between two speeds).

Time is what we are looking for, and according to this equation,
T= F / (S-F)
according to the rephrased (and final equation), in order to find Time to catch up, we need:

difference between second car's speed and first car's speed
AND first car's speed, which as I explained earlier, is exactly the same as first car's gain over distance (extra distance traveled by first car because it started one hour earlier).

Which, as you can see, we need both statements to fulfill both requirements as the statement one gives us the relative speed (S-F) and using second statement with the first statement, we can also find first car's speed or its gains over distance. Hence, C.


"catch up" problem is the hardest out of all the speed problems, to me at least. It takes several problems to get a hang of it. But I derived an conceptual framework, hope it applies to as many problems that might come up in the future and can help you.

time to catch up=(relative distance or difference in distance or distance needs to catch up) / (relative speed or difference between the two car's speeds)