Answer:
14 mph ( average speed during the second part of the trip )
Step-by-step explanation:
Let´s call "x" the average speed during the first part then
t = 5 hours
t = t₁ + t₂ t₁ and t₂ times during part 1 and 2 respectively
l = t*v ( distance is speed by time ) t = l/v
First part
t₁ = 16/x and t₂ = 42 / ( x + 6)
Then
t = 5 = 16/x + 42 /(x + 6)
5 = [ 16 * ( x + 6 ) + 42 * x ] / x* ( x + 6 )
5 *x * ( x + 6 ) = 16*x + 96 + 42 x
5*x² + 30*x - 58*x - 96 = 0
5*x² - 28*x - 96 = 0
We obtained a second degree equation, we will solve for x and dismiss any negative root since negative time has not sense
x₁,₂ = [28 ± √ (28)² + 1920 ] / 10
x₁,₂ = ( 28 ± √2704 )/ 10
x₁ = 28 - 52 /10 we dismiss that root
x₂ = 80/10
x₂ = 8 mph average speed during the first part, and the average speed in the second part was 6 more miles than in the firsst part. then the average spedd dring the scond part was 8 + 6 = 14 mph
