Question 444160: My probelm is this: A truck enters a highway driving 60mph. A car enters the highway at the same place 12 minutes later and drives 67mph in the same direction. How long before the car passes the truck?
I have My Math Lab on coursecompass.com and it kind of explains the solving process but I seem to get lost at the end.
I know that the problem needs to be set up with d(t)=60(t + 10) and d(c)=67t but after that I get lost. Please help.
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! Truck 60 mph
Car 67 mph
Difference in time=00:12mins 0.2 hours
Truck will have covered 0.2 miles before Car enters the highway
catch up distance= 0.2 miles
catch up speed = 67-60 mph
catch up speed = 7 mph
Catchup time = catchup distance/catch up speed
catch up time= 0.2 / 7
catch up time= 0.03 hours
catch up time= 1.71 minutes
..
The other method
let the car meet after time = t hours
time truck takes = 12 minutes more (t+0.2)hours
Distance same
D= r*t
67*t = 60*(t+0.2)
67t=60t+12
7t= 12
t=12/7
1.71 hours
m.ananth@hotmail.ca
Answer by ikleyn(53427)  (Show Source):
You can put this solution on YOUR website! .
My problem is this: A truck enters a highway driving 60mph.
A car enters the highway at the same place 12 minutes later and drives 67mph in the same direction.
How long before the car passes the truck?
I have My Math Lab on coursecompass.com and it kind of explains the solving process but I seem to get lost at the end.
I know that the problem needs to be set up with d(t)=60(t + 10) and d(c)=67t but after that I get lost. Please help.
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Calculations in the post by @mananth, related to his first solution, are incorrect.
His answer in the first part is wrong.
I came to make a job accurately, as it should be done.
Truck 60 mph
Car 67 mph
Difference in time=00:12mins 0.2 hours
Truck will have covered 0.2*60 = 12 miles before Car enters the highway. <<<---=== It is where @mananth made hist 1st error
catch up distance= 0.2 * 60 = 12 miles. <<<---=== It is where @mananth made hist 1st error
catch up speed = 67-60 mph
catch up speed = 7 mph
Catchup time = catchup distance/catch up speed
catch up time= 12 / 7 = 1.7143 hours = 1 hour 43 minutes, approximately
..
The other method
let the car meet after time = t hours
time truck takes = 12 minutes more (t+0.2)hours
Distance same
D= r*t
67*t = 60*(t+0.2)
67t=60t+12
7t= 12
t=12/7
1.7143 hours = 1 hour 43 minutes, approximately.
---------------------------
Solved correctly.
As you can see, first version solution my @mananth was totally wrong, while his second version was correct,
but he even did not notice what his first version was wrong.
So, he completely relies on his computer solution, and NEVER even re-reads what the computer creates.
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