Since this has been my second day of tirelessly browsing and roasting my brains out to find an answer (but in vain), I am posting a question to the following problem here:
Two pedestrians simultaneously head towards each other from two different locations. They meet 6 hours later. Together, they walked 76km in total. How much time does individual pedestrian need in order to walk 1km of path, if the first pedestrians walks this path of 1km one minute less than the other pedestrian?
I've tried solving it using a table. It has three columns (distance, rate and time) and two rows (pedestrian1, pedestrian2). The values for
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pedestrian1: $d$ (distance), $v$ (rate), $6$ (time)
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pedestrian2: $76 – d$ (distance), $v – 1/60$ (rate), $6$ (time)
Then, I've set two equations: $d = v * 6$ and $76-d = (v-1/60) * 6$ and added them together to produce $76 + 1/10 = 12v$ –> $v = 6.34$, which is wrong. The answers in the solutions are that the first pedestrians need 9 minutes per kilometer, while the second needs 10 minutes per kilometer.
Thank you for the time taken.
Best Answer
Using the formula $t=d/v$, you can write down two equations from the statements in the problem. The first says that by combining their speeds, the two pedestrians cover the 76 km in six hours: $${76 \text{ km}\over v_1+v_2} = 6\text{ hr}\cdot60\text{ min/hr}.$$ I’m measuring the speeds in km/min, so the time is converted into minutes. The second fact is that it takes the second pedestrian one more minute than the first to cover 1 km, so you have $$\frac1{v_1}+1=\frac1{v_2}.$$ Solve the two equations for $v_1$ and $v_2$ and then compute $1/v_1$ and $1/v_2=1/v_1+1$, or substitute $t_1=1/v_1$ and $t_2=1/v_2$ into the two equations and solve for the times directly. There will be two solutions to this system of equations, but one of them doesn’t make physical sense for this problem, so that one will be rejected.