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

Answer 1

Answer:

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

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

Answer 2

Answer:

A taxi starts from Monument Circle and travels 5 kilometers to the east for 5 minutes. Then it travels 10 kilometers to the south for 10 minutes.

The taxi is moving with reference to

. For each leg of the trip, the taxi’s

stays the same, but its

changes