Question

A fly undergoes a displacement of -5.80 m while accelerating at -1.33m/s^2 for 4.22 s what was the initial velocity of the fly?

Answer 1

The fly's average velocity over the 4.22 second interval is

$\bar v=\dfrac{-5.80\,\mathrm m}{4.22\,\mathrm s}=-1.37\,\dfrac{\mathrm m}{\mathrm s}$

If the fly's acceleration is uniform over the interval, then we also have

$\bar v=\dfrac{v_i+v_f}2\implies v_i+v_f=-2.74\,\dfrac{\mathrm m}{\mathrm s}$

With the given acceleration we can get another relation involving the initial and final velocities:

$-1.33\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{v_f-v_i}{4.22\,\mathrm s}\implies -v_i+v_f=-5.61\,\dfrac{\mathrm m}{\mathrm s}$

Solving this system, in particular for $v_i$, we find

$v_i=1.44\,\dfrac{\mathrm m}{\mathrm s}$