Question

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.

Answer 1

Answer:

the buses are at 50 mph  and 60 mph

the distance = 210 miles

Step-by-step explanation:

r1 = rate of bus 1

r2 = rate of bus 2

r1 = r2+10   (The speed of one of the buses was 10 mph greater than the speed of the other bus.)

distance = rate * time

for bus 1

distance = r1 * (3.5)

bus 2 was 1/6 short of making it to town b

so it made it 5/6 of the way

5/6 distance = r2 * (3.5)

multiply by 6/5 on each side

distance = 6/5 * r2 * 3.5

set the distances equal

r1 * (3.5)=6/5 * r2 * 3.5

r1 = 6/5  * r2

now set rate 1 equal  (r1 = r2+10)

6/5  * r2 = r2 + 10

solve for r2 by subtracting r2 on each side

6/5 r2 -r2 = 10

6/5 = 1.2

1.2 r2 - r2 =10

.2 r2 = 10

divide by .2 on each side

r2 = 10/.2

r2 = 50

r1 = 50 + 10

r1 =60

to find the distance

distance = r1 * (3.5)

distance = 60 * 3.5

distance = 210 miles

Answer 2

The speed of the buses are $V_1 = 50 mph, V_2=60 mph$ and the distance between A and B is $210 miles$

Explanation:

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.

Distance is the numerical measurement of how far apart objects are. In physics , distance refers to a physical length or an estimation based on other criteria. Whereas speed is  the rate at which someone or something moves or operates or is able to move or operate. A bus is a vehicle designed to carry many passengers as high as 300 passengers

$V_1 =$ speed of bus 1

$V_2 =$ speed of bus 2 $= V_1+10$ MPH

$distance = speed * time$

Bus 2 distance $=V_2 * (3.5) hours$

Bus 1 was 1/6 shorter from town b, therefore it made $\frac{5}{6}$ of the way

$\frac{5}{6} distance_1 =V_1 * (3.5)$

$distance_1= \frac{V_1 * (3.5)}{\frac{5}{6} }$

$distance_1= \frac{6}{5} * V_1 * 3.5$

Then we set the same distance for all buses

$V_2 * (3.5)= \frac{6}{5} * V_1 * 3.5\\(V_1+10) * (3.5)= \frac{6}{5} * V_1 * 3.5\\(V_1+10)= \frac{6}{5} * V_1\\(V_1+10)- \frac{6}{5} * V_1=0\\V_1(1-\frac{6}{5}) = -10 \\V_1= \frac{ -10}{1-\frac{6}{5}} \\ V_1=50 mph$

Therefore $V_2= V_1 + 10 = 50+10 = 60 mph$

The distance between A and B:

$distance = V_2 * (3.5)\\distance = 60 * 3.5\\distance = 210 miles$

Therefore the speed of the buses are $V_1 = 50 mph, V_2=60 mph$ and the distance between A and B is $210 miles$

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