Respuesta :
Answer:
The speed of the buses are 58.2 mph and 49.2 mph.
Step-by-step explanation:
Let the speed of bus 1 is x mph.
So, the other bus 2 has speed (x - 9) mph.
Then time-traveled to cover 585 miles by bus 1 will be [tex]\frac{585}{x}[/tex] hours.
Again time-traveled to cover 585 miles by bus 2 will be [tex]\frac{585}{x - 9}[/tex] hours.
Therefore, from the condition given
[tex]\frac{585}{x - 9} - \frac{585}{x} = 9[/tex]
⇒ [tex]65(\frac{1}{x-9} -\frac{1}{x}) = 1[/tex]
⇒ [tex]x^{2} - 9x - 585 = 0[/tex]
Applying the Sridhar Acharya formula
[tex]x = \frac{9+\sqrt{9^{2}- 4 \times (-585) } }{2}[/tex] {Neglecting the negative root}
⇒ x = 58.2 mph
So, x - 9 = 49.2 mph
Therefore, the speed of the buses are 58.2 mph and 49.2 mph. (Answer)
Answer:
The speed of bus 1 is 63 mile per hour And bus 2 is 54 mile per hour
Step-by-step explanation:
According to question
Let the speed of bus 1 = [tex]S_1[/tex] = S mph
and The sped of bus 2 = [tex]S_2[/tex] = S - 9 mph
Let The distance travel by bus 1 = [tex]D_1[/tex] = D mile
And The distance travel by bus 2 = [tex]D_2[/tex] = ( 585 - D ) mile
he time taken by both buses to cover distance = T hour
Now, Distance = Speed × Time
So , [tex]D_1[/tex] = [tex]S_1[/tex] × T
Or, D = S × T ......... 1
Again [tex]D_2[/tex] = [tex]S_2[/tex] × T
Or, ( 585 - D ) = ( S - 9 ) × T ..........2
From eq 1 an eq 2
585 - S ×T = S ×T - 9 T
Or, 585 + 9 T = 2 × S ×T
Or, 585 + 9 × 5 = 2 × S × 5
Or, 585 + 45 = 10 × S
So , 630 = 10 × S
∴ S = [tex]\frac{630}{10}[/tex] = 63 mph
And S -9 = 63 - 9 = 54 mph
Hence The speed of bus 1 is 63 mile per hour And bus 2 is 54 mile per hour Answer
