Extremely long Honors Pre-calc question

Question

jack travels from his house, point a, to point b, and then back to his house. the shape of the graph that models jack's distance, in meters, from his house given t hours after leaving his house is parabolic in shape as shown. it takes 1 hour to walk to point b which is 3 miles away from jack's house. A. Write a quadratic function that will model Jack's distance from his house given t hours after leaving his house B. How long after starting will Jack be 1 mile from his house (round to nearest minute) C. Jill meets jack on the path when jack has been traveling for 30 minutes. She will walk directly to Jack's house and arrives there at the same time he returns. Write a linear function that models Jill's distance from Jack's house

Answer 1

PART A:

 

(0,0)

(2,0)

(1,3)

 

y=At^2 + Bt + C

 

(0,0) ---> C=0

 

y = At^2 + Bt

 

3 = A + B  --> B = 3-A

 

0 = 4A + 2B

 

substituting first equation into the second

 

0 = 4A + 2(3-A)

 

0 = 4A + 6 - 2A

 

0 = 2A + 6

 

-6 = 2A

 

A = -3

 

B = 3 - A = 3 - (-3) = 6

 

y = -3t^2 + 6t

 

PART B:

 

At one mile:

1 = -3t^2 + 6t

 

3t^2 - 6t + 1 = 0

 

t = [ 6 +or- sqrt( 36 - 4(3)(1) ) ]/(2*3)

 

  = [ 6 +or- sqrt ( 36 - 12)] /6

 

 = [ 6 +or- sqrt ( 24)]/6

 

 = [ 6 +or- 2*Sqrt(6)]/6

 

 = 1 +or- sqrt(6)/3

 

 

1 + sqrt(6)/3 =  1.8165 hours or 1 hour 49 min

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PART C:

 

30 miniutes = 1/2 hour

 

y  = -3(1/2)^2 + 6(1/2) = -3/4 +3 = 9/4

 

Jack and Jill are 9/4 miles away when they meet.

They will meet again 1.5 hours later.

 

So (1/2, 9/4) and (1.5, 0) are on the line.

 

slope is 9/4 divided by -1 = -9/4

 

y = -9/4T + B

 

0 = -9/4*1.5 + B

 

B = 9/4*1.5 = 3.375

 

Her line equation is y = -9/4t+3.375