Distance, Speed and Time problem involving quadratic equations... HELP!!?

Question

A salesman drives from Ajax to Barrington, a distance of 120 mi, at a steady speed. He then increases his speed by 10 mi/h to drive the 150 mi from Barrington to Collins. If the second leg of his trip took 6 min more time than the first leg, how fast was he driving between Ajax and Barrington?

Please help... I'm stuck with finding the time of both trips.. >_<

Answer 1

A salesman drives from Ajax to Barrington,

a distance of 120 mi, at a steady speed.

He then increases his speed by 10 mi/h

to drive the 150 mi from Barrington to Collins.

If the second leg of his trip took 6 min more time than the first leg,

how fast was he driving between Ajax and Barrington?

Assuming x to be the speed in miles per hour of the first leg of his trip

driving between Ajax and Barrington,

120/x - 150/(x + 10) = 0.1

120(x + 10) - 150x = 0.1x(x + 10)

0.1x^2 + x = -30x + 1200

0.1x^2 + 31x - 1200 = 0

x^2 + 310x - 12000 = 0

x = 34.8025

Answer 2

You are not given enough info to find the times first.



"How fast was he driving between Ajax and Barrington" means you are asked to find the speed during that trip, so let's focus on that and label it with a variable S.



Since D=RT, T=D/R. Since R is given in mph, time is measured in hours.

Ajax to Barrington: T=120/S

Barrington to Collins takes 6 minutes longer: T+6/60 = 150/(S+10)



Now replace T in the second equation by 120/S

120/S + 0.1 = 150/(S+10)



Multiply everything by S(S+10)

120(S+10) +0.1S(S+10) =150 S



120S +1200+ 0.1S^2 +S=150 S



0.1S^2 -29S +1200=0



S^2-290S+12000=0



(S-50)(S-240)=0



S=50 or s= 240



50 mph for the trip between Ajax and Barrington

(the 240 mph value actually works but is an unlikely driving speed)



That also means that he was driving at 60mph for the second trip.

You can now directlt find the times to double check these values, but notice they were not

specifically required

Answer 3

Do not be taken in by the difference in the units...



6 minutes = 1/10 hour



so,

let r = rate/speed



from Ajax to Barrington:



t = 120 / r



from Barrington to Collins:



t + (1/10) = 150 / (r + 10)



substitute and solve for r...



(120/r) + (1/10) = 150/(r + 10)



LCD: 10r(r + 10)



120(10)(r + 10) + r(r + 10) = 150(10r)



expand and solve...



r^2 - 290r + 12000 = 0



(r - 50)(r - 240) = 0



r = 50 mph or r = 240 mph



euclid

Answer 4

Let the speed from Ajax To Barrington be x mph



Time taken = 120 /x hours



Speed from Barrington to Collins = (x+10) mph



Time taken = 150/ ( x+10) hours



150/ (x+10) - 120/x = 6/60 = 1/10





MULTIPLY BOTH SIDES BY 10x(x+10)



1500x - 1200( x+ 10) = x ( x+10)



1500x -1200x - 12000 = x^2 +10x



300x - 12000 = x^2 +10x



x^2 - 290x = -12000



ADD 21025



x^2 -290x + 21025 = 9025



( x - 145)^2 = 9025



x = 145 + \/9025 = 240 mph ANSWER



CHECK



120/240 = 30 minutes



150/250 = 15/25 = 3/5 hours = 36 minutes



36-30 = 6 minutes

Answer 5

let speed from A to B be x mph



Time taken from A to B = 120/x hours

Time taken from B to C = 150/(x+10) hours



But time from B to C = 120/x + 1/10 hours



so



150/(x+10) = (1200 + x))/10x...............using 10/x as common denominator on RHS



150 *10x = (1200 +x)(x+10)



1500x = 1200x + 12000 + x^2 +10x



x^2 - 290x + 12000 = 0



x = [290 +/- 190]/2



x = 50mph or 240mph



suggest the second answer is too high#



Answer SPEED from A to B = 50mph and 60mph from B to C [

Answer 6

vt = 120, so v = 120/t

(v+10)(t+0.1) = 150 = vt + 10t + 0.1v + 1 = 120 + 10t + 0.1v + 1 = 121 + 10t + 0.1v

29 mi = 10t + 0.1v = 10t + 0.1(120)/t = 10t + 12/t = 29 multiply both sides by t

10t^2 + 12 = 29t

10t^2 - 29t + 12 = 0

t = ((29 +/- sqrt(841 -4(10)12))/20 = ((29 +/- sqrt(841-480))/20 = ((29 +/- sqrt(361))/20=

((29 +/- 19))/20 = 48/20, or 10/20

t = 2.4 hr and 0.5 hr

v = 50 mi/h and 240 mi/h.