Question 1195978: A motorist leaves Caribou City at 5:03 am, drives the 163 miles to Mooseville , takes a 36 minute break and then drives a further 328 miles to Shellberg, arriving at 8:11 pm.
A.) Assuming the motorist drives at a constant speed throughout the journey, at what speed (to the nearest mile per hour) does he/she travel?
B.) If the motorist had gone 5 miles per hour slower than the speed you recorded in part (a), at what time (to the nearest minute) would he or she arrive?
C.) If the motorist had gone 1 mile per hour faster than the speed you recorded in part (a), at what time (to the nearest minute) would he or she arrive?
Found 3 solutions by math_tutor2020, Alan3354, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part A
The gap from 5:03 AM to 8:11 PM is 15 hours, 8 minutes
This is because...- The jump from 5:03 AM to 5:03 PM is 12 hours
- The jump from 5:03 PM to 8:03 PM is another 3 hours (12+3 = 15 hours so far)
- The jump from 8:03 PM to 8:11 PM is another 8 minutes (since 11-3 = 8)
That's how I got a total duration of 15 hrs, 8 min. This accounts for actual driving, and the break when the person isn't driving.
The driver took a 36 minute break.
Rewrite 36 as 8+28
I'm doing it like this so we can pull out the 8.
We'll subtract off those first 8 minutes to get
(15hr,8min) - (8min) = 15 hrs
Then subtract off the remaining 28 minutes of that break
15 hrs - 28 min = (14 hrs + 1 hr) - 28 min
15 hrs - 28 min = (14 hrs + 60 min) - 28 min
15 hrs - 28 min = 14 hrs + (60 min - 28 min)
15 hrs - 28 min = 14 hrs + 32 min
After ignoring the break, s/he actually spent 14 hrs & 32 min driving.
Let's convert that figure to hours only
14 hrs + 32 min = 14 hrs + (32/60 hr)
14 hrs + 32 min = 14 hrs + 0.533333 hr
14 hrs + 32 min = 14.533333 hrs
The decimal value is approximate.
The total distance traveled was:
163 mi + 328 mi = 491 mi
Now we can then say:
distance = rate*time
rate = distance/time
rate = (491 mi)/(14.533333 hrs)
rate = 33.784404 mph
This value is approximate.
Round to the nearest whole number to get 34 mph
Answer: 34 mph
This answer is approximate.
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Part B
In part A, we found his/her speed to be roughly 34 mph.
If they go 5 mph slower, then they now have a speed of 34-5 = 29 mph.
The total distance stays the same.
Let's compute the time duration spent driving (ignore the 36 min break for now)
distance = rate*time
time = distance/rate
time = (491 mi)/(29 mph)
time = 16.931034 approximately
This is the number of hours spent driving.
Let's convert to hours,minutes format
16.931034 hr = 16 hr + 0.931034 hr
16.931034 hr = 16 hr + (0.931034 hr)*(60 min/1 hr)
16.931034 hr = 16 hr + (0.931034*60) min
16.931034 hr = 16 hr + 55.86204 min
16.931034 hr = 16 hr + 56 min
I rounded to the nearest minute.
We've yet to include the 36 minute break.
Let's do so now.
(16 hr + 56 min) + 36 min
16 hr + (56 min + 36 min)
16 hr + (92 min)
16 hr + (60 min + 32 min)
16 hr + (1 hr + 32 min)
(16 hr + 1 hr) + 32 min
17 hr + 32 min
This is the total duration from when the motorist left Caribou City to when s/he arrived at Shellberg.
Rewind the clock back to 5:03 AM
- Add on 12 hours to get to 5:03 PM
- Add on 5 more hours (12+5 = 17 hrs so far) to get to 10:03 PM
- Then add on the final 32 minutes to get to 10:35 PM
Answer: 10:35 PM
This answer is approximate.
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Part C
The steps shown here are very similar to part B.
In fact, because of this similarity, I effectively copy/pasted the main outline of the steps. The only main changes are the numbers.
In part A, we found his/her speed to be roughly 34 mph.
If they go 1 mph faster, then they now have a speed of 34+1 = 35 mph.
The total distance stays the same.
Let's compute the time duration spent driving (ignore the 36 min break for now)
distance = rate*time
time = distance/rate
time = (491 mi)/(35 mph)
time = 14.028571 approximately
This is the number of hours spent driving.
Let's convert to hours,minutes format
14.028571 hr = 14 hr + 0.028571 hr
14.028571 hr = 14 hr + (0.028571 hr)*(60 min/1 hr)
14.028571 hr = 14 hr + (0.028571*60) min
14.028571 hr = 14 hr + 1.71426 min
14.028571 hr = 14 hr + 2 min
I rounded to the nearest minute.
We've yet to include the 36 minute break.
Let's do so now.
(14 hr + 2 min) + 36 min
14 hr + (2 min + 36 min)
14 hr + 38 min
This is the total duration from when the motorist left Caribou City to when s/he arrived at Shellberg.
Rewind the clock back to 5:03 AM
- Add on 12 hours to get to 5:03 PM
- Add on 2 more hours (12+2 = 14 hrs so far) to get to 7:03 PM
- Then add on the final 38 minutes to get to 7:41 PM
Answer: 7:41 PM
This answer is approximate.
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! A motorist leaves Caribou City at 5:03 am, drives the 163 miles to Mooseville , takes a 36 minute break and then drives a further 328 miles to Shellberg, arriving at 8:11 pm.
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Using the 24 hour clock makes finding the difference easier.
8:11PM = 2011
2011
-0503
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1508, 15 hours 8 minutes
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There is no advantage to the AM and PM.
There are disadvantages.
Answer by ikleyn(52879)  (Show Source):
You can put this solution on YOUR website! .
The numbers are very strange in this assignment.
What I want to say, is THIS:
There are tasks whose purpose is to develop conceptual understanding,
and such tasks are very valuable.
But in the current problem, the major task (as it seems to me) is
to torture a student with non-sensical arithmetic.
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