SOLUTION: A freight tain leaves a station at 4:00 p.m. traveling ay 30 kilometers per hour. A passenger train leaves 1 hour later, traveling at 50 kilometers per hour. At what time will the

Question 21339: A freight tain leaves a station at 4:00 p.m. traveling ay 30 kilometers per hour. A passenger train leaves 1 hour later, traveling at 50 kilometers per hour. At what time will the passenger train overtake the frieght train.
Found 2 solutions by Photonjohn, venugopalramana:
Answer by Photonjohn(42) (Show Source): You can put this solution on YOUR website!
Freight - 30 mph and travels t hours
Passenger - 50 mph and travels t-1 hours
then 30t = 50(t-1)
30t = 50t -50
20t = 50
t = 2.5 hours of travel until overtaken.

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
A freight tain leaves a station at 4:00 p.m. traveling ay 30 kilometers per hour. A passenger train leaves 1 hour later, traveling at 50 kilometers per hour. At what time will the passenger train overtake the frieght train.
You may edit the question. Maybe convert formulae
SEE THE FOLLOWING EXAMPLES TO DO ..
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SHORT ANSWER TO YOUR QUESTION IS
INITIAL DISTNCE OF SEPERATION =30 KM
RELATIVE SPEED =50-30=20KMPH
TIME REQUIRED TO OVER TAKE =30/20=1.5 HRS=1HR30MTS.
SO OVER TAKING HAPPENS AT 4=00+1=00+1=30=6=30PM
SEE BELOW FOR EXPLANATIONS..COME BACK IF YOU DO NOT UNDERSTAND
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Travel_Word_Problems/19596: A freight train and an express train leave towns 390 km apart, traveling toward one another. The freight train travels 30 km per hour slower than the express train. They pass one another 3 hour later. What are their speeds? Thanks for your help.1 solutions Answer 9484 by venugopalramana(345) on 2005-11-15 00:20:54 (Show Source): Initial distance between 2 trains=390 kmFinal distance between 2 trains=0 km�.(since they pass one another)Hence distance traveled by both together=390-0=390 kmLet speed of express train =x kmph30 kmph slower than above=x-30 kmph=speed of goods trainsince both are traveling in opposite directions ,their relative speed is sum of their speeds =x+x-30=2x-30 kmph (note that with each passing hour,when the trains move in opposite directions,their distance of separation comes down by he sum of their speeds) time of travel =3 hrs. so we have�..relative speed =distance trvelled between final and initial stgs/time difference from beginning to end of travel2x-30=390/3=1302x=130+30=160x=160/2=80speed of express = 80 kmphspeed of goods = x-30 =80-30=50 kmph

Travel_Word_Problems/19282: If it takes 18 minutes to travel by bicycle from point A to B in a headwind and 6 minutes to travel from B to A with a tailwind, how long will it take to cover this distance with no wind?SEE THIS WHICH IS SIMILAR TO YOUR PROBLEM AND SOLVE YOUR PROBLEMA boat travels at the speed of 20 Miles per hour in still water. It travels 48 miles up stream, and then returns to the starting point in a total of five hours. The speed of the current is ____ miles per hour.distance =speed * time..or...time=distance/speedStill water speed of boat =20 mphlet the stream velocity be =x mphso its speed upstream =20-xits speed down stream =20+xdistance travelled either way =48 milestime taken to travel upstream=48/(20-x)time taken to travel down stream =48/(20+x)total time of travel =5 hours..hence[48/(20-x)]+[48/(20+x)]=548[(20+x+20-x)/((20-x)(20+x))]=548*40=5(20^2-x^2)=2000-5x^25x^2=2000-1920=8=x^2=80/5=16x=4 mph1 solutions Answer 9323 by venugopalramana(345) on 2005-11-12 06:55:46 (Show Source):

Travel_Word_Problems/19151: OK...Brent rides a moped at 24 miles per hour. His wife Jan rides in the opposite direction at 8 miles per hour. How much time will pass when they are 24 miles apart? I know the answer is 3/4 hour, but I'm having a hard time getting that using the old rate x time = distance method. Can you bail me out? 1 solutions Answer 9226 by venugopalramana(345) on 2005-11-10 04:38:26 (Show Source): Brent rides a moped at 24 miles per hour. His wife Jan rides in the opposite direction at 8 miles per hour. How much time will pass when they are 24 miles apart? I know the answer is 3/4 hour, but I'm having a hard time getting that using the old rate x time = distance method. Can you bail me out? I'm having a hard time getting that using the................................ old rate x time = distance method. Can you bail me out? your formula is ok but you have to understand relative speed cocept and use it ..it is as followsBrent rides a moped at 24 miles per hour...so in one hour he goes 24 miles in one directionHis wife Jan rides in the opposite direction at 8 miles per hour...so in one hour she goes 8 miles in opposite direction.so initially if they were together ,after 1 hour their distance of seperation is 24+8=32 miles as they were going in opposite directions.(just for your idea if they were going in same direction their distance of seperation would have been 24-8=16 miles)so now we define relative speed as sum of the 2 speeds if they are in opposite directions (and difference between 2 speeds ifthey are in same direction)so relative speed in this case =24+8=32 mph now for distance again you have to consider 2 aspects ..one...what is the distance between them at start it is 0 here as they started at the same place.....next...we have to use the distance between them at the end ..here it is 24 miles ...so the additional distance of seperation between them from begining to end is 24-0=24(had the start been not 0 distance ,you will account for that here) now for time same way as above for distance..both started at same time ...so begining time =0end time = not known =t say...so difference in time =t-0=tnow you can use your formula with 3 new terms..relative speed ,difference in distance,difference in time,inplace of speed ,distance and time....so we get rate x time = distance....gives32*t=24t=24/32=3/4 hour =3*60/4 =45 mts....hope it is clear...

Quadratic_Equations/18997: Bobby traveled a distance of 50x + 150 miles at 50 miles per hour on the Gulf Freeway. Find a binomial that represents the time that he traveled.1 solutions Answer 9148 by venugopalramana(345) on 2005-11-09 11:23:24 (Show Source): Bobby traveled a distance of 50x + 150 miles at 50 miles per hour on the Gulf Freeway. Find a binomial that represents the time that he traveled.time*speed=distance...or...time=distance/speed=(50x+150)/50=(50x/50)+(150/50)=x+3time of travel =x+3...which is a binomial

Travel_Word_Problems/18170: I could really use some help on this, I am terrible at wordproblems. Tom Quig traveled 240 miles east of St. Louis. For most of the trip he averaed 60mph, but for one period of time he was slowed to 20 mph due to a major accident. If the total time of travel was 6 hours, how many miles did he drive at the reduced speed? Here is what I did : (60)* *(60)240 =3x80 =x Am I even close on this answer. Please help, thank you so much.lizzy1 solutions Answer 9063 by venugopalramana(345) on 2005-11-08 11:00:39 (Show Source): THIS WAS SOLVED BY ME IN PROBLEM NO.18707..MY[VENUGOPALRAMANA(268)] ANSWER NO.8986.PLEASE REFER

Travel_Word_Problems/18707: Please help me out with this one. I went thru the other questions and didn't see this type of question. Thanks Joe traveled 290 miles east of St. Louis. For most of the trip he averaged 70mph, but for one period of time he was slowed to 20mph due to a major accident. If the total time of travel was 7 hours, how many miles did he drive at the reduced speed?1 solutions Answer 8986 by venugopalramana(345) on 2005-11-07 10:57:09 (Show Source): Joe traveled 290 miles east of St. Louis. For most of the trip he averaged 70mph, but for one period of time he was slowed to 20mph due to a major accident. If the total time of travel was 7 hours, how many miles did he drive at the reduced speed?let the duration of travel at reduced speed of 20 mph be =x hrstotal duration of travel =7 hrs.so duration of travel at normal speed of 70 mph =7-x hrsdistance = speed *time distance travelled at reduced speed =20*x=20xdistance travelled at normal speed =70*(7-x)=70*7-70*x=490-70xtotal distance travelled=20x+490-70x=290490-290=70x-20x200=50xx=200/50=4 hrs.so distance travelled at reduced speed =4*20 =80 miles

Travel_Word_Problems/18444: Joe traveled against the wind in a small plane for 3 hr. The return trip with the wind took 2.8 hr. Find the speed of the wind if the speed of the plane in still air is 180mph.1 solutions Answer 8840 by venugopalramana(345) on 2005-11-04 06:38:04 (Show Source): Joe traveled against the wind in a small plane for 3 hr. The return trip with the wind took 2.8 hr. Find the speed of the wind if the speed of the plane in still air is 180mph.let the wind speed be = x mphspeed of the plane in still air is 180mph...hence speed against the wind =180-x time of travel against wind =3 hrswe have distance travelled=d =speed*time=3*(180-x)=540-3x speed pro wind =180+xtime of travel pro wind=2.8 hrshence distance travelled=d =speed*time=2.8*(180+x)=504+2.8xso 504+2.8x=540-3x2.8x+3x=540-504=365.8x=36x=36/5.8=360/58=180/29 mph

Complex_Numbers/17943: A boat travels at the speed of 20 Miles per hour in still water. It travels 48 miles up stream, and then returns to the starting point in a total of five hours. The speed of the current is ____ miles per hour. 1 solutions Answer 8659 by venugopalramana(345) on 2005-11-01 08:19:31 (Show Source): A boat travels at the speed of 20 Miles per hour in still water. It travels 48 miles up stream, and then returns to the starting point in a total of five hours. The speed of the current is ____ miles per hour. distance =speed * time..or...time=distance/speedStill water speed of boat =20 mphlet the stream velocity be =x mphso its speed upstream =20-xits speed down stream =20+xdistance travelled either way =48 milestime taken to travel upstream=48/(20-x)time taken to travel down stream =48/(20+x)total time of travel =5 hours..hence [48/(20-x)]+[48/(20+x)]=548[(20+x+20-x)/((20-x)(20+x))]=548*40=5(20^2-x^2)=2000-5x^25x^2=2000-1920=8=x^2=80/5=16x=4 mph

Complex_Numbers/17942: The speed of a commuter plane is 150 miles per hour slower than that of a passenger jet. The commuter plane travels 450 miles in the same time the jet travels 1150 miles. The speed of the commuter plane is ____ miles per hour and that of the jet is ____ miles per hour. 1 solutions Answer 8645 by venugopalramana(345) on 2005-10-31 23:28:24 (Show Source): SEE THE FOLLOWING EXAMPLE TO KNOW THE METHOD*******************************************************************************Susan's communte to work in the morning takes 45 minutes and coming home she is able to drive 10 miles per hour faster, so it only takes 30 minutes, how far is her communte (one-way)?let her speed of travel going to work be =xon return she drives 10 mph faster.hence her return speed =x+10let the distance of home to work =d=distance of work to home using the formula d=rt...or t=d/r...time of travel to work =d/x=45/60..or d=x*45/60=3x/4...(1)time of travel to home =d/(x+10)=30/60..or d=(x+10)30/60=(x+10)/2..(2)equating d from the 2 eqns.d=3x/4=(x+10)/2..crossmultiplying2*3x=4*(x+10)6x=4x+406x-4x=402x=40x=20 mphd=3x/4=3*20/4=15 miles. ********************************************************************************If Mr. X leaves now and drives 66 km/hr, he will reach Tuscaloosa in time for an appointment. On the other hand, if he has lunch first and leaves in 40 minutes, he will have to drive 90 km/hr to make the appointment. How far away is Tuscaloosa? (thanks for the assistance).let distance =dspeed of travel in case 1 =66...time of travel in case 1 ..=d/r=d/66speed of travel in case 2 =90...time of travel in case 2 ..=d/r=d/90difference in time of travel =d/66-d/90=40 minutes or 40/60 hrs=2/3d[(1/66)-(1/90)]=2/3d[(90-66)/(90*66)]=2/3d=2*90*66/(24*3)d=165
Travel_Word_Problems/17944: Mr. X biked 2.5 hrs up a mountain. Rested for 30 minutes. Biked down in 1.5 hrs. How far did he bike, if his rate of ascent was 3 km/hr less than his rate of decent? 1 solutions Answer 8644 by venugopalramana(345) on 2005-10-31 23:27:13 (Show Source): SEE THE FOLLOWING EXAMPLE TO KNOW THE METHOD*******************************************************************************Susan's communte to work in the morning takes 45 minutes and coming home she is able to drive 10 miles per hour faster, so it only takes 30 minutes, how far is her communte (one-way)?let her speed of travel going to work be =xon return she drives 10 mph faster.hence her return speed =x+10let the distance of home to work =d=distance of work to home using the formula d=rt...or t=d/r...time of travel to work =d/x=45/60..or d=x*45/60=3x/4...(1)time of travel to home =d/(x+10)=30/60..or d=(x+10)30/60=(x+10)/2..(2)equating d from the 2 eqns.d=3x/4=(x+10)/2..crossmultiplying2*3x=4*(x+10)6x=4x+406x-4x=402x=40x=20 mphd=3x/4=3*20/4=15 miles. ********************************************************************************If Mr. X leaves now and drives 66 km/hr, he will reach Tuscaloosa in time for an appointment. On the other hand, if he has lunch first and leaves in 40 minutes, he will have to drive 90 km/hr to make the appointment. How far away is Tuscaloosa? (thanks for the assistance).let distance =dspeed of travel in case 1 =66...time of travel in case 1 ..=d/r=d/66speed of travel in case 2 =90...time of travel in case 2 ..=d/r=d/90difference in time of travel =d/66-d/90=40 minutes or 40/60 hrs=2/3d[(1/66)-(1/90)]=2/3d[(90-66)/(90*66)]=2/3d=2*90*66/(24*3)d=165
Travel_Word_Problems/17945: If Mr. X leaves now and drives 66 km/hr, he will reach Tuscaloosa in time for an appointment. On the other hand, if he has lunch first and leaves in 40 minutes, he will have to drive 90 km/hr to make the appointment. How far away is Tuscaloosa? (thanks for the assistance).1 solutions Answer 8643 by venugopalramana(345) on 2005-10-31 23:18:12 (Show Source): SEE THE FOLLOWING EXAMPLE TO KNOW THE METHOD*******************************************************************************Susan's communte to work in the morning takes 45 minutes and coming home she is able to drive 10 miles per hour faster, so it only takes 30 minutes, how far is her communte (one-way)?let her speed of travel going to work be =xon return she drives 10 mph faster.hence her return speed =x+10let the distance of home to work =d=distance of work to home using the formula d=rt...or t=d/r...time of travel to work =d/x=45/60..or d=x*45/60=3x/4...(1)time of travel to home =d/(x+10)=30/60..or d=(x+10)30/60=(x+10)/2..(2)equating d from the 2 eqns.d=3x/4=(x+10)/2..crossmultiplying2*3x=4*(x+10)6x=4x+406x-4x=402x=40x=20 mphd=3x/4=3*20/4=15 miles. ********************************************************************************If Mr. X leaves now and drives 66 km/hr, he will reach Tuscaloosa in time for an appointment. On the other hand, if he has lunch first and leaves in 40 minutes, he will have to drive 90 km/hr to make the appointment. How far away is Tuscaloosa? (thanks for the assistance).let distance =dspeed of travel in case 1 =66...time of travel in case 1 ..=d/r=d/66speed of travel in case 2 =90...time of travel in case 2 ..=d/r=d/90difference in time of travel =d/66-d/90=40 minutes or 40/60 hrs=2/3d[(1/66)-(1/90)]=2/3d[(90-66)/(90*66)]=2/3d=2*90*66/(24*3)d=165

Human-and-algebraic-language/17843: Marcella leaves home at 9:00 A.M. and drives to school, arriving at 9:45 A.M. Ifthe distance between home and school 27 mi, what is Marcella"s average rate of speed?1 solutions Answer 8597 by venugopalramana(345) on 2005-10-31 11:01:03 (Show Source): Marcella leaves home at 9:00 A.M. and drives to school, arriving at 9:45 A.M. Ifthe distance between home and school 27 mi, what is Marcella"s average rate of speed?time of travel =9:45-9:00=45 mts=45/60 hrs=3/4 hrs.distance=27 miles speed = distance/time =27/(3/4)=27*4/3=36 miles per hour

Travel_Word_Problems/17806: Please help!!! Susan's communte to work in the morning takes 45 minutes and coming home she is able to drive 10 miles per hour faster, so it only takes 30 minutes, how far is her communte (one-way)? thanks in advance1 solutions Answer 8593 by venugopalramana(345) on 2005-10-31 04:54:43 (Show Source): Susan's communte to work in the morning takes 45 minutes and coming home she is able to drive 10 miles per hour faster, so it only takes 30 minutes, how far is her communte (one-way)?let her speed of travel going to work be =xon return she drives 10 mph faster.hence her return speed =x+10let the distance of home to work =d=distance of work to home using the formula d=rt...or t=d/r...time of travel to work =d/x=45/60..or d=x*45/60=3x/4...(1)time of travel to home =d/(x+10)=30/60..or d=(x+10)30/60=(x+10)/2..(2)equating d from the 2 eqns.d=3x/4=(x+10)/2..crossmultiplying2*3x=4*(x+10)6x=4x+406x-4x=402x=40x=20 mphd=3x/4=3*20/4=15 miles.

Travel_Word_Problems/17825: Greetings!Okay, who can help me understand word problems? I really need to know how to break down this problem and solve it. Could someone please help understand!Problem: Alonzo travels 230 miles in 3 hours, how many miles miles traveled in 7 hours? Thanks to all.1 solutions Answer 8588 by venugopalramana(345) on 2005-10-31 04:07:10 (Show Source): Alonzo travels 230 miles in 3 hours, how many miles miles traveled in 7 hours? it is assumed that the rate of travel is uniform ..that is if he goes 100 miles in 1 hour ,then he will go 200 miles in 2 hours,300 miles in 3 hrs...etc....so we get the eqn.d=r*t..where d=distance travelled,r=rate or speed of travel and t=time..let us first find his rate of travel.we have d=230..t=3...r=?230=3r...or r=230/3now to the second case we have r=230/3...t=7...d=?d=7*230/3=1610/3=536.66666

Linear-equations/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. 1 solutions Answer 8367 by venugopalramana(345) on 2005-10-27 01:53:11 (Show Source): 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 problemlet 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=r2.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=dso 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.

Travel_Word_Problems/16947: A student leaves the university at noon, bicycling south at a constant rate. At 12:30pm a second student leaves the same point and heads west, bicycling 7mph faster than the first student. At 2:00pm, they are 30 miles apart. How fast is each one going? Also, how would I properly set-up this equation? Below is what I thought is the answer, please let me know where I went wrong? r=d/t 30/2 = 15mph (first student)30/1.5 = 20 + 7 = 27 mph (second student)1 solutions Answer 8239 by venugopalramana(345) on 2005-10-25 06:33:24 (Show Source): let the speed of the 1st.student=x mph.time of start of 1st.student = 12=00 noon....time at end is 2=00 pm.duration of travel =2 hrs.distance travelled by 1st.student from start in south direction ........... =speed * duration = 2*x miles speed of 2nd.student is 7 mph more.hence it is x+7 mph.time of start of 2nd.student is 12=30 noon....time at end is 2=00 pm.duration of travel =1.5 hrs.distance travelled by 2nd.student from start in west direction ................= speed * duration = 1.5*(x+7) miles if we call their 2 end positions as A and B and start point as S,we can see that ASB is a right angled triangle at S with SA=2x and SB=1.5(x+7).and AB=30 as given in the problem.hence using pythogarus theorem in right angled triangle ASB , we have SA^2+SB^2=AB^2(2x)^2+[1.5(x+7)]^2=30^24x^2+2.25(x^2+14x+49)=9006.25x^2+31.5x+110.25=9006.25x^2+31.5x-789.75=0 using the standard formula for quadratic equation ..we get x=9mph.

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