Question

Jon recently drove to visit his parents who live 648 miles away. On his way there his average speed was 14 miles per hour faster than on his way home (he ran into some bad weather). If Jon spent a total of 27 hours driving, find the two rates (in mph). Round your answer to two decimal places, if needed.

Answer 1

Answer:

speed of the Jon visiting parents = 56 mph

           speed of the Jon when returning from home = 56 - 14 = 42 mph

Explanation:

given,

distance of Jon parent's house = 648 mile

avg speed when he was visiting his parent's house be 'x' mph

avg speed when he is returning from his parent's house be 'x-14' mph

total time taken = 27 hours

total distance = speed × time

648 = x × t₁

$t_1 = \dfrac{648}{x}$

648 = ( x - 14 ) × t₂

$t_2 = \dfrac{648}{x-14}$

t = t₁ + t₂

$27 = \dfrac{648}{x} + \dfrac{648}{x-14}$

$1 = 24 (\dfrac{1}{x} + \dfrac{1}{x-14})$

x² - 62 x + 336 = 0

x² - 56 x - 6 x + 336 = 0

(x - 56 )(x - 6)=0

on solving

x = 56 ,6

hence, speed of the Jon visiting parents = 56 mph

           speed of the Jon when returning from home = 56 - 14 = 42 mph

Answer 2

Answer:56 mph ,42 mph

Explanation:

Given

Distance traveled by john on his way to parents house=648 miles

Let x be the speed during his return trip so during arrival its speed is x+14

Total distance=648+648=1296 miles

Total time=27 hours

$distance =speed \times time$

$27=\frac{648}{x}+\frac{648}{x+14}$

27(x+14)(x)=648(2x+14)

$x^2-34x-336=0$

x=42,-8

-8 is not possible so x=42 mph

arriving rate=42+14=56 mph

Return rate=42 mph