Johnny was able to drive an average of 27 miles per hour faster in his car after the traffic cleared. He drove 25 miles in traffic before it cleared and then drove another 104 miles. If the
total trip took 3 hours, then what was his average speed in traffic?

40 mph

Step-by-step explanation: let s = speed in traffic

(s+17) = out of traffic

:

Write a time equation; time = dist/speed

traffic time + normal speed time = 4 hrs

46%2Fs + 80%2F%28%28s%2B17%29%29 = 4

multiply by s(s+17) to cancel the denominators

46(s+17) + 80s = 4s(s+17)

46s + 782 + 80s = 4s^2 + 68s

126s + 782 = 4s^2 + 68s

Arrange as a quadratic equation on the right

2s^2 - 29s - 391 = 0

Use the quadratic formula a=2; b=-29; c=-391

I got a positive solution of

s = 23 mph in traffic

:

:

:

Check this by finding the actual times. Normal speed: 23 + 17 = 40 mph

46/23 = 2 hrs in traffic

80/40 = 2 hrs at normal speed

total time 4 hrs

0 = 4s^2 + 68s - 126s - 782

4s^2 - 58s - 782 = 0

simplify, divide by 2

so  the answer is 40mph

y=mx+b

step-by-step explanation:

m slope, b y-intercept

more detail

step-by-step explanation: