Question 644295: The word problem I am stuck on is: A man travels 72 miles to his destination in 1.5 hours. During the first leg of his trip he encountered construction during which he traveled 30mph. The second leg of his trip he was able to travel 60mph. How much time was spent in the construction zone?
I am not sure where to begin with this problem. I know that t will = the time the man spent in the construction zone because it is the unknown. I converted hours to minutes, so 90 minutes I assume is a little easier to handle. But constructing a formula is where I feel like a deer in the headlights. Any help you can offer would be greatly appreciated! Thank you- Andrea
Found 2 solutions by ankor@dixie-net.com, DrBeeee:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A man travels 72 miles to his destination in 1.5 hours.
During the first leg of his trip he encountered construction during which
he traveled 30 mph.
The second leg of his trip he was able to travel 60 mph.
How much time was spent in the construction zone?
:
Let t = time spent in construction zone, going 30 mph
then
(1.5-t) = time spent driving 60 mph
:
Write a distance equation: dist = speed * time
:
30t + 60(1.5-t) = 72
:
Since you are dealing in mph, it's best to leave the time in hrs
:
Is this enough to get you started on this?
Answer by DrBeeee(684)  (Show Source):
You can put this solution on YOUR website! This is called a distance problem, based on the use of the formula (1) d = r*t, where d is the distance travelled while moving at the rate r, for t amount of time. Rate r is usually miles per hour (abbreviated as mph) and time is given or measured in hours. When you multiply (miles/hour)*(hours) you get miles; the distance d. Other units are also used depending on the problem; all we need to is check that the denominator of the rate units cancels with the time units. For example above the units of "hours" cancelled. If you have a rate, for example, in centimeters per second, then time must be in seconds and distance will be in centimeters. Now back to the problem.
As with all algebra/math problems we define some variables to represent the physical properties of the problem. For this problem we have two distinct areas of travel (called legs), two different rates and we know the total distance is 72 miles and it takes 1.5 hours the travel over the two legs. In a sense we know the sum of the parts, we just don't know the parts!
Let r1 = rate of travel the first leg, in mph
Let t1 = the time to travel the first leg, in hours
Let d1 = distance of first leg, in miles
And likewise for the second leg
Let r2 = rate during second leg, mph
Let t2 = time to travel the second leg, hours
Let d2 = distance of second leg, in miles
We have two unknowns - the time and distance of legs one and two, therefore we need to set up two algebraic equations in terms of these unknowns.
What do we know?
a) the total distance is 72 miles,
b) the rate on first leg is 30mph
c) the rate on the second leg is 60mph
d) the total time is 1.5 hours
Now the tough part, Andrea, is to set up some equations, based on the formula and givens.
Firstly we know the total distance is given by
(1) d1 + d2 = 72
And the total time is given by
(2) t1 + t2 = 1.5
We also know that
(3) d1 = 30*t1
and
(4) d2 = 60*t2
All we need to do is solve these four equations for the unknowns. Think about it before proceeding. Pause.
There's two basic ways to solve them, here's my choice:
Substitute (3) and (4) into (1) and get
(5) 30*t1 + 60*t2 = 72
Now simultaneously solve (5) and (2)
Solve (2) for t1 yielding
(6) t1 = 1.5 - t2
Now sbstitute (6) into (5) and get
(7) 30*(1.5 - t2) + 60*t2 = 72 or
(8) 45 - 30*t2 + 60*t2 = 72 or
(9) 30*t2 = 27 or
(10) t2 = 0.9
Then from (6) we get
(11) t1 = 0.6
Using (3) we get
(12) d1 = 30*0.6 or
(13) d1 = 18
Using (4) we get
(14) d2 = 60*0.9 or
(15) d2 = 54
Let's check these values.
Is (2) correct?
Is (t1 + t2 = 1.5)?
Is (0.6 + 0.9 = 1.5)
Is (1.5 = 1.5)? Yes
Is (1) correct?
Is (d1 + d2 = 72)?
Is (18 + 54 = 72)?
Is (72 = 72)? Yes
Answer: The man spent 0.6hr or 36min in the construction zone.
Comment: Note in (1) - (15) I omit all units, such as miles, hours, etc. This is standard practice - all algbraic equations are written without units. It makes life simpler.
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