Question

The correct match is:

List - I	List - II
a) A body covers first half of distance with a speed $V_1$ and second half of distance with a speed $V_2$	e) Average velocity is $\sqrt{\frac{gh}{2}}$
b) A body covers first half of the time with a speed $V_1$ and second half of the time with a speed $V_2$	f) Average speed is $\frac{v_1+v_2}{2}$
c) A body is projected vertically up from ground with certain velocity.Considering its total motion,	g) Average speed is $\frac{2V_1{V}_2}{V_1+V_2}$
d) A body freely is released from a height h	h) Average velocity is zero

• A. $a\to f,b\to g,c\to e,d\to h$
• B. $a\to g,b\to f,c\to h,d\to e$
• C. $a\to h,b\to g,c\to h,d\to e$
• D. $a\to e,b\to f,c\to h,d\to g$

Answer 1

The correct option is B $a\to g,b\to f,c\to h,d\to e$

Case A:
Let total distance be $s$.

So, body covers $\frac{s}{2}$ each with velocities $V_1$ and $V_2$.
Now, average speed $=\frac{totaldistance}{totaltimetaken}$

$\Rightarrow V_{avg}=\frac{s}{\frac{\frac{s}{2}}{V_1}+\frac{\frac{s}{2}}{V_2}}=\frac{2V_1{V}_2}{V_1+V_2}$

Case B:

Total distance covered $=v_1\frac{t}{2}+v_2\frac{t}{2}$ (where, total time $=t$)
So, $v_{avg}=\frac{v_1\frac{t}{2}+v_2\frac{t}{2}}{t}=\frac{v_1+v_2}{2}$

Case C:

During total motion, net displacement $=0$
Average velocity $=\frac{Totaldisplacement}{Totaltime}$ (not distance **)
Hence, $\overrightarrow{v_{avg}=0}$

Case D:

For free fall, acceleration due to gravity $=g$
Also acceleration is rate of change of velocity.
$\Rightarrow g=\frac{dv}{dt}=\frac{d}{dt}\left(\frac{dh}{dt}\right)=\frac{d^{2}h}{d{t}^2}$

On integrating, we have
$\Rightarrow \frac{g{t}^2}{2}=h$ (for 0 initial condition)

$\Rightarrow v_{avg}=\frac{h}{\sqrt{\frac{2h}{g}}}=\sqrt{\frac{gh}{2}}$