Real-life math problem , how much faster did i run?

Question

So I was running today and I tried to run faster than usual. I always workout for $30$ minutes, which contains $25$ minutes of running and $5$ minutes of walking. But today, I ran $20$ minutes and walked for $10$. The distance in both workouts was the same, but I'm not sure of the absolute value of it. I'm assuming the same walking speed for both workouts. So I was wondering, is it possible to know how much faster was I running today in percentage or in absolute value? Meaning get an answer in form of $V_1=aV_2$ or $V_1=V_2+a$

I figured there are not enough data, so I tried to assume $6$ km/hour, walking speed, but still couldn't get a result.

Bonus: This really happened to me

Answer 1

Your walk speed is $v_0$ your normal run speed is $v_1$ but today it was $v_2$.

Normally you exersice the distance $s=5v_0+25v_1$.

Today you exersiced the same distance $s=10v_0+20v_2$.

So, what you have is $5v_0+25v_1=10v_0+20v_2\Leftrightarrow v_2=\frac{25v_1-5v_0}{20}$.

So, if for example $v_0=100$ (that is 100m per minute or 6km per hour) then $v_2=\frac{25v_1-500}{20}$.

You need $v_1$ (or $v_2$) in meters per minute. Or then you need to know the total distance $s$ in meters.

Answer 2

If your two running speeds were $r_1$ and $r_2$ and your walking speed was $w$ then $$25r_1+5w=20r_2+10w$$

That could be rewritten as something like $$r_2=\tfrac54r_1-w$$ or $$r_2-r_1 =\tfrac14(r_1-w)$$ or $$\frac{r_2}{r_1}=\frac54-\frac{w}{r_1}$$or something similar but not much simpler than that

Answer 3

Let us assume constant speed both during walking and running separately, that is, no acceleration. And, as you say, let us use walking speed as $6km/hr=\frac{5}{3}$ meter/sec. Now, in your usual routine, we have $(25\times 60)v_1+300\times\frac{5}{3}$ as the total distance, where $v_1$ is the initial running distance. In the second case, that is today, we have $(20v_2\times 60)+600\times\frac{5}{3}$ as the total distance. Thus, equating the distances as you said they are equal, we obtain $300(5v_1+\frac{5}{3})=600(2v_2+\frac{5}{3})\implies 5v_1=4v_2+\frac{5}{3}$ in meter/sec