Quadratic Equations (Speed)

Question

During the first part of a trip, Cheryl drove 150 mi at a certain speed. She then drove another 300 mi at a speed that was 10 mph faster. If the total time of Cheryl's trip was 8 hr, what was her speed on each part of the trip?

Answer 1

First define variables:

 

S1 = speed 1 (for the 150 mi)

S2 = speed 2 (for the 300 mi)

 

T1 = time driven at S1 

T2 = time driven at S2

 

distance/speed = time 

therefore the time driven each leg is:

 

 T1 = 150mi/S1

 T2 = 300mi/S2

 

we know the combined time = 8 hrs, so

T1 + T2 = 8  hrs

 

or,  

150/S1 + 300/S2 = 8    and we also know that S2 = S1 + 10   (2nd speed is 10mph faster)

 

150/S1 + 300/(S1+10) = 8    ok, now we have one equation, one unknown (S1) 

 

getting a common denom. 

 

[150*(S1+10) + 300*S1] / [S1*(S1+10)]  = 8       (now mult both sides by denom on left)

 

[150*(S1+10) + 300*S1] = 8*[S1*(S1+10)]         (now, do all the multiplying)

 

150*S1 + 1500 + 300*S1 = 8*(S1² + 10*S1)  =   8*S1² + 80*S1    (now, combine like terms and put on rt side)

 

0 = 8*S1² - 370*S1 - 1500      (OK, here is our quadratic, to solve for S1)

I'm not going to try to factor, so I'll use the quadratic formula)

[-b +/ sqrt(b2 - 4ac) ] / (2a)      a=8, b=-370, c= -1500

 

S1 = [370 +/- sqrt(3702 - 4*8*(-1500)] / (2*8)

 

S1 = (370 + 430)/16     (note: the 370 - 430 is not a real soln because results in a negative speed S1)

 

S1 = 50 MPH   

S2 = 60 MPH     (since S2 = S1+10) 

 

Using our earlier  T1 = 150/S1 so 

 

T1 = 3 HRS 

T2 = 5 HRS      (since T2 = 8 - T1)

 

 

now, let's check to see if these are correct

 

Leg 1 = 150 mi = 50mph * 3 hrs   checks

Leg 2 = 300 mi = 60mph * 5 hrs   checks 

 

hope this helps !!!

 

 

 

Answer 2

Dear Haley,

 

Question:  What was her speed on each part of the trip?  You are looking for the value of "X."  You also want (X+10).

 

 

1.  Draw a diagram and then a table to represent data.

 

.__150 mi________>_____300mi__________________.     Total time=8 hours

 

X mi/h                                        (X+10) mi /h      

 

 

First part of trip                    Rate   *          t                =            =Distance

 

                                            X mi/h         t                            t(X)  =  150 mi

 

Second part of trip                (X+10) mi/h   8-t         =    (8-t)(X+10=300 mi

 

                                                                 Total distance=               450 mi     

 

 

Looking at the table, you can see that tx=150, so  solve for "t."   t=150/X   

 

 

2.  Add the two equations in the table together, and keep in mind that you need to solve for "X."  You don't want "t."

      

tx + (8-t)(X+10) =450 mi              Substitute tx=150  and t=150/X

 

150+ (8-150/X)(X+10)=450  You want to get rid of the denominator of "X" by multiplying every term

 

by "X."

150X+-450X + 8X² -150X +80X-1500=0               Simplify by collecting terms.

 

8X² -370X-1500=0                    Simplify further by dividing every term by 2.

 

  4X²  -185X -750=0              I suggest that you use the quadratic equation to find "X."          

 

a=4, b=-185 and c=-750           I will let you finish the work from here.  If your answer is correct, you should be able to substitute it into the time formula of t=150/x.  When you substitute the correct time and rate into the equations in the table, you should get out 150 miles for one answer and 300 for the other.

 

Your first rate is "X=_____mi/h, and your second rate or speed is (X+10) mi/h   Answers  

 

If I have helped you, please give me a thumbs up.

 

                                                            Susan C.