Carcass wrote:
John and Karen begin running at opposite ends of a trail until they meet somewhere in between their starting points. They each run at their respective constant rates until John gets a cramp and stops. If Karen runs 50% faster than John, who is only able to cover 25% of the distance before he stops, what percent longer would Karen have run than she would have had John been able to maintain his constant rate until they met.
Let the distance to be travelled is 100m
Speed of John be 10m/s
So, speed to Karen is 15m/s
Relative Velocity (towards each other) = 15 + 10 = 25m/s
Time taken to meet in between = \(\frac{distance}{Velocity_{rel}} = 100/25 = 4\)sec
Distance travelled by John in 4 sec (Meeting point) = 10(4) = 40m
Distance travelled by Karen in 4 sec (Meeting point) = 15(4) = 60m
John was able to cover only 25% of the distance = 0.25(40) = 10m
Time required to do so = \(\frac{10}{10} = 1\)sec
Karen now has to run till 90m mark as John stopped at 10m mark (100m - 10m).
In that 1 sec, Karen already ran 15m. So she will run 90m - 15m = 75m more at 15m/s
Time required by Karen to do so = \(\frac{75}{15} = 5\) sec
Initially Karen would take 4 sec but not it will take her 5 sec to new Meeting point. Percent change = \(\frac{(5-4)}{4}(100) = 25\)%