This a summary of the question (because it was really long)
- Everyday A leaves home before 5pm to pick B up from the train station at 5pm and drive home
- One day B catches an earlier train that arrives at the station at 4pm.
- B decides to walk home until he is intercepted by A and then they will drive home together
- They end up arriving home 10 minutes earlier than usual, how many minutes did B walk?
I have absolutely no clue where to start because it doesn't say anything about the distance. in the solutions manual it says that you're supposed to draw a distance-time graph and you can get the answer of 55mins but I don't understand how you would draw one either
This is the full question:
Pat works in the city and lives in the suburbs with Sal. Every afternoon, Pat gets on a train that arrives at the suburban station at exactly 5pm. Sal leaves the house before 5 and drives at a constant speed so as to arrive at the train station at exactly 5pm to pick up Pat. The route that Sal drives never changes.
One day, this routine is interrupted, because there is a power failure at work. Pat get to leave early, and catches a train which arrives at the suburban station at 4pm. Instead of phoning Sal to ask for an earlier pickup, Pat decides to get a little exercise, and begins walking home along the route that Sal drives, knowing that eventually Sal will intercept Pat, and then will make a U-turn, and they will head home together in the car. This is indeed that happens, and Pat ends up arriving at home 10 minutes earlier than on a normal day. Assuming that Pat's walking speed is constant, that the U-turn takes no time, and that Sal's driving speed in constant, for how many minutes did Pat walk?
Suppose Sal meets Pat $t$ minutes before reaching the station, i.e. $t$ minutes before $5$ pm. Sal will then have saved $t$ minutes on his outbound trip and will also save $t$ minutes on his homebound trip, in all $2t$ minutes. We know that $2t = 10$ minutes, so $t = 5$ minutes. Since Pat started walking at $4$ pm and met Sal at $5$ minutes before $5$ pm, Pat must have been walking for $55$ minutes.