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

Answer:

$a=1.33 m/s^{2}$

A. $F=2.67N$

B. $x=37.50m$

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:

$a=\frac{v_{f}-v_{o}  }{t_{f}-t_{o}  }=\frac{(10-0)m/s}{(7.5-0)s}=1.33m/s^{2}$

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

$t=\frac{v_{f} }{a}=\frac{10m/s}{1.33m/s^{2} }=7.51s$

Now, from the equation of position we know that

$x=x_{o}+v_{o}t+\frac{1}{2}at^{2}$

$x=\frac{1}{2}(1.33m/s^2)(7.51s)^2=37.50m$