Question

Adam takes a bus on a school field trip. The bus route is split into the five legs listed in the table. Find the average velocity for each leg of thetrip. Then arrange the legs of the trip from highest velocity to lowest.Leg Distance (km) Time (min)A189B2515C 248D4812E157ghts reserved.earchleg Aleg Bleg Cleg Dleg E69°F Suna

Answer 1

Given the table:

Leg DIstance(km) time(min)

A 18 9

B 25 15

C 24 8

D 48 12

E 15 7

Let's find the average velocity for each leg of the trip.

To find the average velocity, apply the formula:

`v=(d)/(t)`

Where:

v is the velocity

d is the distance

t is the time.

Thus, we have the following:

• Average velocity for leg A:

`v_A=\frac{18\text{ km}}{9\text{ min}}=2\text{ km/min}`

The average velocity for leg A is 2 km/min.

• Average velocity for leg B:

`v_B=\frac{25\text{ km}}{15\text{ min}}=1.67\text{ km/min}`

The average velocity for leg B is 1.67 km/min

• Average velocity for leg C:

`v_c=\frac{24\text{ km}}{8\text{ min}}=3\text{ km/min}`

The average velocity for leg C is 3 km/min

• Average velocity for leg D:

`v_D=\frac{48\text{ km}}{12\text{ min}}=4\text{ km/min}`

The average velocity for leg D is 4 km/min.

• Average velocity for leg E:

`v_E=\frac{15\text{ km}}{7\text{ min}}=2.14\text{ km/min}`

The average velocity for leg E is 2.14 km/min.

Now, to arrange the legs of the trip from the highest velocity to the lowest, we have:

• Leg D ==> 4 km/min

,

• Leg C ==> 3 km/min

,

• Leg E ==> 2.14 km/min

,

• Leg A ==> 2 km/min

,

• Leg B ==> 1.67 km/min

ANSWER:

• The average velocity for leg A is , 2 km/min.

,

• The average velocity for leg B is , 1.67 km/min.

,

• The average velocity for leg C is , 3 km/min.

,

• The average velocity for leg D is , 4 km/min.

,

• The average velocity for leg E is , 2.14 km/min.

Arranging from the highest velocity to the lowest:

0. Leg D

,

1. Leg C

,

2. Leg E

,

3. Leg A

,

4. Leg B