Mathematics, 29.06.2019 23:00 july00

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.

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Answer from: DESIREEHANCOTTE10

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. $\frac{d}{x+10} =3.5$ ... 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. $3.5 = \frac{5d}{6x}$ ... ii

Divide i by ii to eliminate d.

$\frac{5}{6x} =\frac{1}{x+10} \\5x+50 = 6x\\x =50$

ie Speed of slower bus = 50mph and

speed of faster bus = 60 mph

Use ii equation

$\frac{5d}{6(50)} =3.5\\d=210$ miles

Answer from: cloudFF

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

$\large {\boxed {\bold {x = v \times \: t}}}$

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),

$\rm t=\dfrac{x}{v}$

in 3.5 hours :

time for B1 = time for B2

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

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)

$\rm 3.5\:h=\dfrac{x}{10+v}\\\\3.5=\dfrac{x}{60}\\\\x=210\:miles$

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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...