Question 66567: Two cars leave town going in the same direction. One travels 55 mph and the other travels 65 mph How long will it take before they are 180 miles apart?
Hint: This is different from problem #3, because we do not know the total distance. We do know that one car is 180 miles ahead of the other, so the slow car's distance can be represented by " d " and the distance of the fast car, which would be ahead, can be represented by "d + 180." The time is the same again, and can be represented by "t." This time we have two unknowns, so we need two equations, one for the fast car, and one for the slow car. Again we are using the basic formula d = rt , and substituting the values that are relevant to our situation. The equation for the slow car is: d = 55t. You write the equation for the fast car, and solve the system by substitution. Don't forget the units in your answer.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website!
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I SUGGEST THAT YOU LEARN THE CONCEPT OF RELATIVE SPEED TO
SOLVE PROBLEMS IN TIME AND DISTANCE .SEE 2 EXAMPLES BELOW FOR CONCEPT
AND WORKING.
SOLUTION OF YOUR PROBLEM
Two cars leave town going in the same direction. One travels 55 mph and the other travels 65 mph How long will it take before they are 180 miles apart?
RELATIVE SPEED =65-55=10 MPH.
DISTANCE OF SEPERATION = 180
TIME NEEDED =180/10=18 HRS.
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QUESTION
1.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.
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ANSWER
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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