I'm learning for a physics exam which is still a bit away but I started today and I'm studying with books. One of it has got questions and answers though the answers prove to be unhelpful at times as they only show the answer without a solution.

The question I ran into now is:

"800m in front of a car going 80 km/h is a second car going 60 km/h. After how much time and distance has the first car reached the second car?"

Now, what I did to get to the solution was transforming it to m/s first getting:

$$ v_1 = 22.22 \frac{m}{s} \\ v_2 = 16.67 \frac{m}{s} $$

Then I checked how much time the first car would need to travel 800m

$$ \frac{800m}{22.22\frac{m}{s}} = 36s $$

Next I did the math how far the second car would have moved during 36 seconds, leading me to:

$$ v_2 * 36 s = 600m $$

I added that distance to the original distance and got $1400m$ and did the math to see how much time the first car would need to travel that:

$$ \frac{1400m}{22.22\frac{m}{s}} = 63s $$

So I came to the conclusion that it would take the first car 63 seconds to reach the second car and 1400m to do so.

The solution according to my book would be 144 seconds and 3200m. It doesn't say how to arrive at that conclusion. Where did I go wrong?


2 Answers 2


Velocity is a relative term. You always need look at velocity as "compared to what?" With that in mind, in the reference frame of the first car, the second car is approaching him with the velocity of $20km/h.$
To understand this, imagine that you are sitting in your [first] car. From your point of view, the second car is approaching you (since you are going faster than he is compared to the ground below you, and from your point of view, you are sitting still).

Now all you need to do now is divide $800m$ with $5.556m/s$ to get 144 seconds, and then multiply that 144 seconds with the velocity of the first car ($22.22m/s$). Do you understand? The basic concepts used here are reference frames.

  • $\begingroup$ Shortcut special - At 20 km/hr the car covers 800m gap. At 80 km/hr (4 times 20) the car would travel 4 times the 800m which is 3200m. This does not require computation of time for computing total distance. $\endgroup$
    – kpv
    Oct 14, 2017 at 22:35

Initially the 1st car is at A and the 2nd car is at B.

After 36s the 1st car reaches B. In the meantime, the 2nd car has moved to C which is 1400m from A. It takes the 1st car 63s to reach C from A.

However, when the 1st car reaches C the 2nd car is no longer at C, it has now moved to another point D which is 1850m from A. So now you must calculate how long it takes the 1st car to reach D from A.

When the 1st car reaches D, the 2nd car has moved to another point E.... Following your method, you need to go though many cycles of calculation until the 1st car gets close enough to the 2nd car. You get there eventually, but it takes a long long time if you want to get the correct answer.

Alternatively, think about the relative velocity of the 1st car with respect to the 2nd. How long does it take the 1st car to cover the initial distance between the cars at the relative velocity?


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