91Ó°ÊÓ

• Initial velocity$v_i=\text{30″¾i}/\text{h}$
• Final velocity $v_f=6\text{0″¾i}/\text{h}$
• Initial time$t_1=3\text{ h}$
• Final time $t_2=1\text{ h}$

Average speed is the ratio of distance covered in a particular period of time.

(a)

The average speed of an object is calculated by

$v_{avg}=\frac{d_t}{t}$

Here, $d_t$- Total distance, and t -Total time

Distance travel in first 3h

$\begin{array}{l}{d}_1=30×3\\ d_1=90\text{″¾}\end{array}$

Distance travel in 1h

$\begin{array}{l}{d}_2=60×1\\ d_2=60\text{″¾i}\end{array}$

Total distance

$\begin{array}{l}{d}_t=60+90\\ d_t=150\text{″¾i}\end{array}$

Now,

$\begin{array}{l}{v}_{avg}=\frac{150}{4}\\ v_{avg}=37.5\text{″¾i}/\text{s}\end{array}$

The value of average velocity is$37.5\text{″¾i}/\text{s}$

(b)

The given equation from the question

$\begin{array}{l}{v}_{avg,x}=\frac{v_{ix}+v_{fx}}{2}\\ v_{avg,x}=\frac{30+60}{2}\\ v_{avg,x}=45\text{ ″¾i}/\text{h}\end{array}$

The value of average velocity is$v_{avg,x}=45\text{ ″¾i}/\text{h}$

(c)

The given formula in the question is being only valid for a constant rate like baseballs, which means the result generated is wrong. It didn’t consider the different time intervals.