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 the trip. then arrange the legs of the trip from highest velocity to lowest

leg distance (km) time (min)
a 18 9
b 25 15
c 24 8
d 48 12
e 15 7

The answer is:

VA= 33 m/S

VB= 27.8 m/s

VC= 50 m/s

VD=66.7 m/s

VE=35.7 m/s

So, the Legs from the highest velocity to the lower are:

D - C - E - A - B

The explanation:

we are going to use the velocity formula:

Velocity = distance (m) / time (s)

So,

A) when D = 18 Km = 18000 m

and t = 9 min = 9*60 =540 s

So V = 18000m/ 540 S

       = 33 m/S

B) when D = 25 Km = 25000 m

and t = 15 min = 15 * 60 = 900 S

So V = 25000m / 900S

      = 27.8 m/s

C) when D = 24 Km = 24000 m

and t = 8 min = 8 * 60 = 480 s

So V = 24000m / 480s

      = 50 m/s

D) when D = 48 Km = 48000 m

and t = 12 min = 12 * 60 = 720 s

So V = 48000 m / 720s

        =66.7 m/s

E) when D = 15 km = 15000m

and t = 7 min = 7 * 60 = 420 s

So V = 15000m / 420s

       = 35.7 m/s

b

explanation: