On the first part of her trip Natalie rode her bike 16 miles and on the second part of the trip she rode her bike 42 miles. Her average speed during the second part of the trip was 6 mph faster than her average speed on the first part of the trip. Find her average speed for the second part of the trip if the total time for the trip was 5 hours.

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