Question

A cheetah can accelerate from rest to 25.0 m/s in 6.22 s. What is the cheetah's average speed for the first 3.11 s of its sprint

Answer 1

18.75m/s

Step-by-step explanation:

Given data

initial speed ,u= o m/s

final speed v= 25 m/s

time t= 3.11 seconds

Applying the first equation of motion we have

`v= u+at\\\\v-u=at\\\\a= (v-u)/(t) \\\\a= (25-0)/(6.22) \\\\a= 4.02 m/s^2`

Also applying
`v=u+at` with initial speed set at 0, we can also find speed at 3.11 seconds

`v=0+4.02*3.11\\v= 12.50 m/s`

Hence the average speed in the first 3.11 seconds is

`\bar v= (v)/(2) \\\\\bar v=( 12.50)/(2) \\\\ \bar v= 6.25 m/s`

Applying the expression
`\bar v=\bar u+at` we can now fing the average speed in the second 3.11 second

`\bar v=\bar u+at\\\\ \bar v=6.25+4.02*3.11\\\\ \bar v=18.75m/s`