ddleigh115
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Two buses left town A for town B at the same time. The speed of one of the buses was 10 mph greater than the speed of the other bus. In 3 1/2 hours one bus reached town B, while the other bus was away from town B at a distance equal 1/6 of the distance between A and B. Find the speed of the buses and the distance between A and B.

Respuesta :

Answer:

Speed of slower bus = 50mph and

speed of faster bus = 60 mph

Step-by-step explanation:

Let speed of first bus be x mph

Then speed of second bus = x+10 mph.

Let A and B have a distance d.

Then given that faster bus reached in 3.5 hours.

i.e. [tex]\frac{d}{x+10} =3.5[/tex] ... i

THe other bus was away from B by 1/6 times d.

Distance travelled by slower bus = 1-1/6 = 5d/6  

Time taken = 3.5 hours

i.e. [tex]3.5 = \frac{5d}{6x}[/tex] ... ii

Divide i by ii to eliminate d.

[tex]\frac{5}{6x} =\frac{1}{x+10} \\5x+50 = 6x\\x =50[/tex]

ie Speed of slower bus = 50mph and

speed of faster bus = 60 mph

Use ii equation

[tex]\frac{5d}{6(50)} =3.5\\d=210[/tex] miles

ardni313

The speed of the bus = 50 mph and 60 mph

The distance between town A and town B : 210 miles

Further explanation

Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration

• At constant velocity motion:

the speed of vo = v = constant

acceleration = a = 0

Δx = vt or x = xo + vt

An equation of constant velocity motion

[tex] \large {\boxed {\bold {x = v \times \: t}}} [/tex]

x = distance = m

v = speed = m / s

t = time = seconds

The velocity graph (v) versus time (t) (v on the y axis and t on the x axis) shows a straight line parallel to the x axis (= time) because it has a constant speed

While the displacement of objects (Δx) is the area under the v-t line

Two buses left town A for town B

The speed for B1(first bus) = v

The speed for B2(second bus) = 10+v⇒10 mph greater

In 3.5 hours :

B1 reached town B, while B2 at the distance a distance equal 1/6 of the distance between A and B, then :

x = displacement of B1

x-1/6x=5/6x =  displacement of B2

Because time = t = displacement(x) : speed(v),

[tex]\rm t=\dfrac{x}{v}[/tex]

in 3.5 hours :

time for B1 = time for B2

[tex]\rm t1=t2\\\\\dfrac{x}{10+v}=\dfrac{\frac{5}{6}x }{v}\\\\xv=\frac{5}{6}x(10+v)\Rightarrow divide\:both\:side\:by\:x\\\\v=\dfrac{5}{6}(10+v)\\\\\dfrac{50}{6} =\dfrac{1}{6}v\\\\v=50[/tex]

So speed the first bus ,B1 = 50 mph and speed the second bus, B2 = 10+50 = 60 mph

The distance from town A and town B (use t1=time for B1)

[tex]\rm 3.5\:h=\dfrac{x}{10+v}\\\\3.5=\dfrac{x}{60}\\\\x=210\:miles[/tex]

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