Question

A finch rides on the back of a galapagos tortoise, which walks at the stately pace of 0.060 m/s. after 1.0 minutes the finch tires of the tortoise's slow pace, and takes flight in the same direction for another 1.3 minutes at 12 m/s

Answer 1

The finch traveled a distance of 19.2 m by riding on the tortoise and then flying in the same direction.

Step-by-step explanation:

Based on the given information, the finch rides on the back of a galapagos tortoise, which walks at a pace of 0.060 m/s for 1.0 minute. During this time, the tortoise covers a distance of 0.060 m/s × 60 s = 3.6 m. After that, the finch takes flight in the same direction for 1.3 minutes at a speed of 12 m/s. During this time, the finch covers a distance of 12 m/s × 1.3 min = 15.6 m. So, in total, the finch traveled a distance of 3.6 m + 15.6 m = 19.2 m.

The subject of this question is Physics because it involves the motion of the finch and the tortoise and calculating distances and speeds.

Answer 2

The average speed is
`6.809\ m/sec`

Step-by-step explanation:

The question is

What was the average speed of the finch for this 2.3-minute interval?

we know that

To determine the average speed divide the total distance by the total time

Let

x ----> the total distance in meters

y ----> the total time in seconds

s ----> average speed in m/sec

so

`s=(x)/(y)`

step 1

Find the total distance x

To find the distance multiply the speed by the time

For the first 1.0 minutes

1 minute=60 sec

x=0.060(60)=3.6 m

For the other 1.3 minutes

1 minute=60 sec

x=12(60*1.3)=936 m

the total distance is

x=3.6+936=939.6 m

step 2

Find the total time y

y=(1+1.3)(60)=138 sec

step 3

Find the average speed

`s=(x)/(y)`

substitute the values

`s=(939.6)/(138)=6.809\ m/sec`