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Question 17341: Martina leaves home at 9 A.M., bicycling at a rate of 24mi/h. Two hours later, John leaves, driving at a rate of 48mi/h. At what time will JOhn catch up with Martina?
I have been trying to figure this out for several hours now. Is this a trick question, because I cannot figure it out. I know that distance= rate times time (d=r x t), but it doesn't give the distance.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Martina leaves home at 9 A.M., bicycling at a rate of 24mi/h. Two hours later, John leaves, driving at a rate of 48mi/h. At what time will JOhn catch up with Martina?
I have been trying to figure this out for several hours now. Is this a trick question, because I cannot figure it out. I know that distance= rate times time (d=r x t), but it doesn't give the distance. this is OK ..but you have to learn the concept of RELATIVE SPEED HERE to do the problem.let me explain the principle first before we solve your problem
let us say there are 2 persons ..you and your friend martina..let us say you are at one place to start with at your home and both of you start at the same time.you walk at 4 mph(miles per hour)say and martina walks at 3 mph.now see what happens after 1 hr.using your formula d=r*t,you will be 4 miles away from your home and martina will be 3 miles away...after 2 hours the distances will be 8 and 6 miles respectively...like this though you started at the same time and at the same place that is zero distance between you and martina,your distance of seperation increased to 4-3=1 mile in 1 hour ,8-6=2 miles in 2 hrs...and so on..so now you are ready to take this concept of relative speed ...relative speed between you and martina is 4 mph - 3 mph = 1mph.in this case as you are travelling in the same direction.so you shoud use this relative speed for r in your formula of d=r*t...ok..the d in this case is the distance of seperation between you and martina at the start .it may be zero as we assumed in this case or may not be zero as given in your problem.similarly you mave to adjust for difference in time of start...but the principle is same so procedure is
1.use relative speed.. this will be difference in speeds if you are travelling in same direction.BUT IT WILL BE SUM OF THE SPEEDS IF YOU ARE TRAVELLING IN OPPOSITE DIRECTIONS.CAN YOU GUESS WHY?BECAUSE AS WE ANALYSED ABOVE EVERY HOUR YOU TWO ARE APPROACHING EACH OTHER (HERE OFCOURSE AT THE BEGINING YOU TWO WILL HAVE TO BE SEPERATED BY A CERTAIN DISTANCE)AND THE DISTANCE OF SEPERATION REDUCES BY THE SUM OF YOUR SPEEDS EVERY HOUR.
2.find the difference in time of start if any and its effect on distance of seperation at the start.
3find the distance of seperation at start using given data and time gap mentioned in 2.
4.note the end criteria..and use your standard formula d=r*t to meet the end criteria using relative speed for r and difference in distance of seperation between start and end positions.
now let us see your example...
1.relative speed =48-24=24 mph=r
2.difference in time of start =2 hrs.
3.distance of seperation at the start = the head start martina got in 2 hrs cycling at 24 mph=2*24=48 miles..
4.end criteria is that they meet at the end ..that is distance of seperation at the end is zero.so difference in distance of seperation from start to end ..... =48-0=48=d
so using d=r*t,we get 48=24*t..or..t=48/24=2 hrs. that is john will catch up with martina in 2 hrs from his time of start that is 11=00+2....that is 13=00 hrs.
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