Completeing the square problem dealing with time

[tklopfstein]

I worked on this problem for over an hour this morning and still couldn't get it to come out right:

Here is the problem and how I worked on it.
Seanna is competing in the bicycle Race Across America.
She rode 62 miles before lunch and 69 miles after lunch. She rode for one hour more after lunch than before lunch, but her speed after lunch was 2 mph slower. What were her speeds before and after lunch?

62= 69 + 2 Solve for t.
t t+1


t(62 = 69 +2)
t t +1

t+1(62=69t + 2t)
t+1

62t + 62 = 69t² + 69t + 2t² + 2t
-62t -62t

62= 71t² + 9t

71(t² + 9t)=62
71
1/71(71(t² + 9t)=62)
71
t² + 9t = 62
71 71
t² +( 9t *1/2)² = 62 + ( 9t *1/2)²
71 71 71
(t + 9_)² = 62 + 81____
142 71 20164
17608 + 81___
20164 20164

t + 9 = √17689
142 √20, 164

t + 9__ = .93648
142

-9 -9
142 142

t=.873 book says answer should be 14.5 miles and 16.5 miles
Can you tell me what I did wrong?

 

[tkhunny]

She rode 62 miles before lunch and 69 miles after lunch. She rode for one hour more after lunch than before lunch, but her speed after lunch was 2 mph slower. What were her speeds before and after lunch?

t=.873 book says answer should be 14.5 miles and 16.5 miles

We need a very serious chat about units.

"What were her speeds" Your answer has no units. The book's answers have the wrong units. Someone should make up someone's mind.

Slowly, deliberately, one step at a time.

Distance = Rate*Time

DBefore = RBefore*TBefore
DAfter = RAfter*TAfter

"She rode 62 miles before lunch and 69 miles after lunch."

DBefore = 62 miles
DAfter = 69 miles
62 miles = RBefore*TBefore
69 miles = RAfter*TAfter

"She rode for one hour more after lunch than before lunch"

TBefore + 1 h = TAfter
62 miles = RBefore*TBefore
69 miles = RAfter*(TBefore + 1 h)

"her speed after lunch was 2 mph slower"

RAfter = RBefore - 2 mph
62 miles = RBefore*TBefore
69 miles = (RBefore - 2 mph)*(TBefore + 1 h)

That's all the information that is given.

62 miles/RBefore = TBefore

69 miles = (RBefore - 2 mph)*((62 miles/RBefore) + 1 h)
69 miles = (62 miles) + [RBefore*(1 h)] - [(2 mph)*(62 miles/RBefore)] - [(2 mph)*(1 h)]

7 miles = [RBefore*(1 h)] - [(2 mph)*(62 miles/RBefore)] - (2 miles)
9 miles = [RBefore*(1 h)] - [(2 mph)*(62 miles/RBefore)]

(9 miles)*RBefore = RBefore²*(1 h) - 124 m²/h

RBefore²*(1 h) - (9 miles)*RBefore - 124 m²/h = 0

RBefore² - (9 mph)*RBefore - 124 m²/h² = 0

RBefore from the quadratic formula.

RBefore = [9 mph + sqrt((9 mph)² - 4*1*(-124 m²/h²)]/2*1
RBefore = [9 mph + 24.021 mph]/2 = 16.51 mph
Also
RBefore = [9 mph - 24.021 mph]/2, which makes no sense.

RAfter = RBefore - 2 mph = 16.51 mph - 2 mph = 14.51 mph

Don't lose your units. They WILL save you.

 

[tklopfstein]

thank you

Thank for your help. Can you tell me how your answer relates to the completing the square format? This is the format I was suppose to use? I'm sorry if that wasn't clear.

 

[Denis]

Tammy, I don't follow what you're doing...

Formula: speed = distance divided by time

Before lunch: @ x mph, 62 miles, h hours
x = 62/h

After lunch: @ x-2 mph, 69 miles, h+1 hours
x-2 = 69/(h+1)

Those are the 2 equations you need.
Simplify both in terms of h, then solve for x.

 

[tklopfstein]

Well you see this problem was under a whole series of problems dealing with completing the square so I understood that they wanted us to somehow take the distance formula and use it in the completing the square formula. I'm so mixed up. I don't know which was is up or down anymore. I don't understand how you got the 7 & the 9 for miles?

 

[Denis]

Ok; let's try again...

x=speed after lunch; so speed before lunch = x+2
time before lunch = h; so time after lunch = h+1

before lunch: ...@ (x+2) mph.....62 miles.......h hours
after lunch: .....@ x mph...........69 miles.......(h+1) hours

distance before lunch = speed * time = (x + 2)h [1]
distance after lunch = speed * time = x(h + 1) [2]

The difference between [2] and [1] is 69-62 = 7; so:
x(h + 1) - h(x + 2) = 7
hx + x - hx - 2h = 7
x - 2h = 7
2h = x - 7
h = (x - 7)/2 [3]

Substitute [3] in [1]:
(x + 2)(x - 7)/2 = 62
x^2 - 5x - 14 = 124
x^2 - 5x - 138 = 0
Using quadratic equation:
x = [5 +- sqrt(5^2 - 4(1)(-138))] / 2
x = [5 +- sqrt(577)] / 2
x = (5 +- 24.02~) / 2
x = 14.51~ or x = -9.51 (not possible)
So x = 14.51~

Looks like the 14.5 answer in your book is 14.51~ rounded.

A suggestion; when you get a problem like that one, set up a similar one
yourself: this will help you "SEE" what's going on.
As example:
before lunch: 20 mph for 5 hours = 100 miles
after lunch: 18 mph for 6 hours = 108 miles
Now pretend that's to be solved; problem would be worded:
100 miles before lunch, 108 miles after with speed reduced by 2 and
time increased by 1.
Now, by going through your solving steps, you KNOW if you're correct,
because you know the details....capish?

 

[tklopfstein]

thanks so much

I'm going to go over your solution tomorrow. It's making a little more sense now. Really appreciate the help.

Take care.