Question

Fernell and Dabney shared the driving
Florida for spring break. Fernell averaged 50 mph, and
Dabney averaged 64 mph. If Fernell drove for 3 hours
longer than Dabney but covered 18 miles less than dabney, then for how many hours did Fernell drive?

Answer 1

Answer:

15 hours

Step-by-step explanation:

Given:

Fernell's speed $v_F= 50$ mph

Dabney's speed $v_D= 64$ mph

Denote:

$D_F$ - distance covered by Fernell

$D_D$ - distance covered by Dabney

$t_F$ - Fernell's time

$t_D$ - Dabney's time

1. If Fernell drove for 3 hours longer than Dabney, then his time is 3 hours more than Dabney's time and

$t_F=t_D+3$

2. If Fernell covered 18 miles less than Dabney, then

$D_D-D_F=18$

Use formula $D=t\cdot v$

$D_D=v_D\cdot t_D\Rightarrow D_D=64\cdot t_D\\ \\D_F=v_F\cdot t_F\Rightarrow D_F=50\cdot (t_D+3)$

Subtract from the first equation the second equation and equate it to 18:

$64t_D-50(t_D+3)=18\\ \\64t_D-50t_D-150=18\\ \\14t_D=168\\ \\t_D=12\ \text{hours}$

$t_F=t_D+3=12+3=15\ \text{hours}$