algebra 2-

Question

Two cars leave town A at the same time and travel 480 miles (each) to town B.  The speed of the slower ca is 20mph less than the speed of the faster one.  If the slower car arrives at town B 2 hours after the faster car. What is the sped of (in mph) of the faster car

Answer 1

Hi Billy,

 

This problem requires 2 equations to solve.

 

Remember distance = rate * time or d = rt

 

Car 1

Let d = 480

Let r = r

Let t = t

 

d = rt

480 = rt

 

Solve for t in terms of r since we need to know the speed

t = ⁴⁸⁰/_r

 

Car 2

Let d = 480

Let r = r - 20 since it is going 20mph slower than Car 1

Let t = t+2 since it takes 2 hours longer to get from town A to town B

 

d = rt

480 = (r - 20)(t + 2); multiply the binomials

480 = rt + 2r - 20t - 40; solve for t in terms of r, first subtract 2r from both sides

480 - 2r = rt - 20t - 40; add 40 to both sides

520 - 2r = rt - 20t; factor out a t from the right side

520 - 2r = t(r - 20); divide both sides by (r - 20)

^{(520 - 2r)}/_{(r - 20)} = t

 

Set the equations equal to each other

^{(520 - 2t)}/_{(r - 20)} = ⁴⁸⁰/_r; ; cross multiply

 

480(r - 20) = r(520 - 2r); distribute

480r - 9600 = 520r -2r²; add 2r² to both sides

2r² + 480r - 9600 = 520r; subtract 520r to both sides

2r² - 40r - 9600 = 0; factor out a 2

2(r² - 20r - 4800) = 0; divide both sides by 2

r² - 20r - 4800 = 0; factor

(r - 80)(r + 60) = 0; set each factor equal to 0

r - 80 = 0            r + 60 = 0

r = 80                 r = -60

r = 80 has to be used since we are talking about speed; so

 

Car 1 is traveling 80mph

 

Find t then check

Use Car 1 equation to keep things simple

 

480 = 80t; divide both sides by 80

t = 6hrs.

 

Check

 

Since Car 1 is traveling 80mph then Car 2 must be traveling 60mph thus satisfying condition 1

Since Car 1 travels from town A to town B in 6 hours then Car 2 must travel in 8 hours satisfying condition 2.

 

I hope this was helpful.

 

Sincerely, 

Patrick S.