# If a second car leaves later at a higher speed, how long will it take to catch up?

If a Skoda sets off at an average speed of 20 miles per hour at 7 am and a Porsche sets off at an average speed of 40 miles per hour at 4 pm in the same direction, at what time will the Porsche catch the Skoda?

Could you please explain how it works?

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I assume they start from the same point, albeit at different times? What have you tried? In what sense is 'average' meant here? – Raskolnikov Dec 12 '10 at 19:56
The two cars start at the same point. – sammville Dec 12 '10 at 19:57
I believe that Raskolnikov's point is that the question is ambiguous if you only know the average speeds. (And "average" would have to be defined relative to some period of time.) My guess is that the intended question assumes that both cars actually travel at constant speed. – Jonas Meyer Dec 12 '10 at 20:01
In real life, if a Corvette ZR-1 doing 200mph passes a stationary top fuel dragster at which point it starts a 1/4 mile run, then by the finish line the dragster would have caught up with the vette and passed it. Note the dragster hits 200mph about 1/3 down the track. – ja72 May 23 '11 at 17:15

Hint: Distance covered is speed$\times$ time traveled. If you assume the time at which the Skoda leaves to be $t=0$, distance covered by Skoda after $t$ hours is $20t$ and distance covered by Porsche after 4pm is $40(t-9)$.

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Maybe part of the exercise is to explain why these simple linear formulas work? You know, the theorem about the integral of a continuous function over an interval $[a,b]$. – Raskolnikov Dec 12 '10 at 19:58
@Raskolnikov: It's tagged as precalculus. So I did not think of this. I will be happy to modify my post if the OP gives such inputs. – Timothy Wagner Dec 12 '10 at 20:01
Please can you explain the formulars? – sammville Dec 12 '10 at 20:03
Sorry Timothy, I didn't pay attention to the tags. – Raskolnikov Dec 12 '10 at 20:10
@sammville: The average speed of an object is the ratio distance traveled by the object in a given time to the time traveled. So distance covered in a given span of time is the product of the average speed over that time interval with the time interval. Of course, here you are assuming the average speed is the same regardless of distance traveled, so you are in fact assuming constant speed. Once you have this formula, you just need to account for the fact that both cars travel at different times, so the time for which they have traveled is different. Can you work with this? – Timothy Wagner Dec 12 '10 at 20:10

If you are geometrically inclined, like I am, you can visualize the path of the Skoda as a line on a graph where the X axis is the time of day (in military time) and the Y axis is distance traveled in miles.

Here are some points for the Skoda --

• (7,0), because at 7 am, the Skoda has traveled 0 miles.
• (8,20), because at 8 am, the Skoda has traveled 20 miles.
• (12,100), because at 12 pm, the Skoda has traveled 100 miles.

If you plot these points on your graph, you'll see that it makes a straight line with slope 20 and X-intercept of 7. From that you could figure out that the Y intercept is -140.

Similarly, the Porsche would have a slope of 40 and an X-intercept of 16 (we're using military time). From there you could find out that the Y intercept is -640.

So you have

• Skoda's traveled distance = $20*t - 140$
• Porsche traveled distance = $40*t - 640$

We can graph these on a chart --

So you want to know when their traveled distance is equal. So --

$Skoda \, distance = Porsche \, distance$

$20*t - 140 = 40*t - 640$

$20*t = 40*t - 500$

$20*t = 500$

$t = 25$

So, at 25 o' clock, or 1 am the next day, their traveled distances will be the same.

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From $7$ am to $4$ pm the Skoda would have travelled $180$ miles. Thus the Porsche has to cover $180$ miles extra to overtake the Skoda. In each hour the Porsche can only cover $20$ miles extra, so the Porsche would need another extra $9$ hours to do that, thus finishing at $1$ am.

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HINT $\:$ The Porsche travels $\rm 20\ t$ more miles than the Skoda during the time $\rm\:t\:$ they're both driving. This must be equal to the $\ 20\cdot 9$ miles of the Skoda's head start. $\$ Hence $\rm\ t\ =\ \ldots$

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Let's say Ds = distance traveled by the Skoda and Dp = distance traveled by the Porsche. If we set t to be the time after 4pm that the cars are traveling, then we set up the formulas as follows:

Dp = 40t Ds = 20t + 20(9) *the Skoda has been travelling from 7am to 4pm, or 9 hours.

When the cars meet, the distance traveled should be the same.

40t = 20t + 180

Subtract 20t from both sides: 20t = 180

Divide both sides by 20: t = 9

So, after 4pm, the Porsche should catch up with the Skoda 9 hours later, or at 1 am.

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