Question

If a 1kg English swallow is carrying a 1kg coconut and flies northward from rest to a final velocity of 10m/s in 7.5 seconds, what is its acceleration? A. What force was exerted by the bird during that acceleration? B. How far did the bird travel during that time?

Answer 1

`a=1.33 m/s^(2)`

A.
`F=2.67N`

B.
`x=37.50m`

Step-by-step explanation:

From the exercise we know that the swallow:

`m_(total) =2kg`

`v_(o)=0\\v_(f)=10m/s\\t=7.5s`

To find its acceleration we must calculate:

`a=(v_(f)-v_(o)  )/(t_(f)-t_(o)  )=((10-0)m/s)/((7.5-0)s)=1.33m/s^(2)`

A. Now, from Newton's second law we know that force is:

`F=m*a`

`F=(2kg)*(1.33m/s^(2))=2.67N`

B. To find the distance which the bird travels we need to find how long does it take

`v_(f)=v_(o)+at`

Solving for t

`t=(v_(f) )/(a)=(10m/s)/(1.33m/s^(2) )=7.51s`

Now, from the equation of position we know that

`x=x_(o)+v_(o)t+(1)/(2)at^(2)`

`x=(1)/(2)(1.33m/s^2)(7.51s)^2=37.50m`