The energetic dog

I used a slightly different approach here:

The dog ran uphill $100m$ using time $(100/3)s$

During the $(100/3)s$ , the boy walked up $(100/3)m$

To find the time they met, distance the both of them will cover is equal to $100m$ minus distance covered by the boy

$(1)(t)+(5)(t)=100-(100/3)$

$t=(100/9)s$

Thus, distance covered by dog downhill is:

$s=(5)t$

$s=(500/9)m$

And the boy covered distance of:

$s=(100/9)s$

So the dog runs uphill $(500/9)m$ again using time $((500/9)/3)s$ or $(500/27)s$

During the $(500/27)s$ , the boy walked up $(500/27)m$

Then, to find the time they met

$(1)(t)+(5)(t)=100-(100/3)-(100/9)-(500/27)$

$t=(500/81)s$

And, distance covered by dog will be

$s=(5)t$

$s=(2500/81)m$

Until here, I was confirm that distance covered by the dog will now be:

$S=100+(500/9)+(500/9)+(2500/81)+(2500/81)+... ...$

So at this point, I suspect that it might be a geometric progression with

$a=100, T2=(500/9), T3=(2500/81)$

$r=((500/9)/100)=((2500/81)/(500/9))=(5/9)$

So the total distance covered by the dog would be

$S=a+T2+T2+T3+T3+... ...$

$S=2[a/(1-r)]-100$

$S=350m$

If looked carefully, this question actually resemble the question about the total vertical distance traveled by a bouncing ball.

Let t1 and t2 be the time the dog takes to run uphill and downhill between 2 consecutive meetings. The distance from the first "place of meeting" to the summit is equal to the sum of the distance from the second "place of meeting" to the summit and the distance the boy has moved, so we have 3t1 = 5t2 + 1(t1 + t2) or t1 = 3t2. This ratio is true in any two consecutive meetings, so by adding all of those ratios in every consecutive meetings together we still have t1 = 3t2. But we know the dog takes 100 seconds to finish its journey, so it takes 75 seconds running uphill and 25 seconds running downhill. The total journey of the dog is 75 x 3 + 25 x 5 = 350m.

Dog uphill: dog=100m, he=100/3 dog downhill: dog= 5/6(100-100/3) he=100/3 + 1/6 (100-100/3) total: dog 1400/9 human= 400/9 dog "walks" 3.5 more than the human, as the boy walks 100m the dog runs 350

I solved the solution using ratios.

I realised that the dog would run for 100s. I then worked out the ratio of the time spent running at 5m/s and 3m/s. In the first stage, the dog runs uphill 100m at 3m/s. 100/3 = 33.3 . The Dog runs downhill for 55.5m at 5m/s, and 55.5/5 = 11.1.

Therefore, the ratios of the times spent running at different speeds are 1:3. Then, we simply multiply the total time the dog runs, which is 100 seconds, by these ratios and speeds, to obtain the total distance.

[(100 * 3/4) * 3] + [(100*1/4) * 5)] = 350m.

I did it in a pretty easy way. (I apologize if it's not explained very well) So if the distance between the boy and the summit in the beginning is 100m and his speed is 1m/s, it means until they both reach the summit, they have travelled for 100 seconds. The dog takes 100/3 seconds to reach the summit for the first time. Then he runs down to the boy and up to the summit again. The distance of the dog running down to the boy and the distance of the dog running up to the summit every time is the same (excluding the first time).

We know that after the first time to the summit the dog runs for (100 - 100/3) seconds in total, which is 200/3 seconds. We also know that the speed of the dog is 3 and 5. This means 5/8 of the 200/3 seconds is spent on going up hill, and 3/8 of the 200/3 seconds is spent on going down hill. 5/8 x 200/3 = 125/3 s 3/8 x 200/3 = 25 s

125/3 s x 3 m/s = 125m and this is the total distance of the dog running up hill excluding the first time. 25 s x 5 m/s = 125m and this is the total distance of the dog running down hill. We can see that the distance is the same as I said before.

Now we just need to add everything together. The first time when the dog runs up to the summit, he runs 100 m. 100m + 125m x 2 = 350m

And 350m is the total distance.

The total trip takes 100 seconds. If the dog were only running uphill, he would cover a distance of 300 m.

Let $t$ be the time during which the dog was running downhill.

When it comes to the displacement of the dog, every second running downhill brings it 5 meters down instead of 3 meters up--a difference of 8 meters. Thus the displacement of the dog is $\Delta = 300 - 8t$ .

When it comes to the distance ran by the dog, every second running downhill makes it go 5 meters instead of 3 meters--a difference of 2 meters. Thus the distance traveled by the dog is $s = 300 + 2t$ .

We know that the dog's overall displacement is 100 meters uphill. Therefore $300 - 8t = 100 \ \ \ \ \therefore\ \ \ \ \ t = 25\ \ \ \ \therefore\ \ \ \ s = 300 + 2t = \boxed{350}.$

I also used Seraph Yang approach. After the first 100m, the dog goes down and up by equal distance. After first up hill let the dog go total Xm down and Xm up.The total time is 100sec,
since the boys speed is 1m/sec, and dog has spent 100/3 sec for moving up, the dog has only 200/3 sec left to move these Xm + Xm. the the dog takes X/5 + X/3 sec. = 200/3.
This gives X = 125m. Total for dog = 100 + 125 + 125 = 350.................................................................

This is only a different approach. Some of the others are better.