Answer
(a) $(-9.81m/s^2)\hat y$
(b) $(-9.81m/s^2)\hat y$
(c) $(-9.81m/s^2)\hat y$
Work Step by Step
(a) We know that
$\vec a_{avg}=\frac{\vec \Delta v}{\Delta t}$
$\vec a_{avg}=\frac{\vec {v_{t+\Delta t}}-\vec v_t}{\Delta t}$
We plug in the known values to obtain:
$\vec{a_{avg}}=\frac{(16.6m/s)\hat x-[(9.81m/s^2)(t+\Delta t)]\hat y-[(16.6m/s)\hat x-(9.81m/s^2)t\hat y]}{\Delta t}$
$\vec a_{avg}=\frac{[(-9.81m/s^2)\Delta t]\hat y}{\Delta t}$......eq(1)
Now, the average acceleration for $\Delta t=1.00s$
$\vec a_{avg}=(-9.81m/s^2)\hat y$
(b) We can find the average acceleration for $\Delta t=2.50s$ from eq(1) as
$\vec a_{avg}=\frac{[(-9.81m/s^2)\Delta t]\hat y}{\Delta t}$
$\vec a_{avg}=(-9.81m/s^2)\hat y$
(c) We can find the average acceleration for $\Delta t=5.00s$ from eq(1) as
$\vec a_{avg}=\frac{[(-9.81m/s^2)\Delta t]\hat y}{\Delta t}$
$\vec a_{avg}=(-9.81m/s^2)\hat y$
