1. Forces on the falling body

Take downward direction as positive. Then the forces are:

• Weight: $mg$ downward
• Viscous force: $kv$ upward, so in our sign convention it is $-kv$

Hence, the equation of motion is

$m\frac{dv}{dt}=mg-kv.$

2. Acceleration as a function of speed

From the above,

$a=\frac{dv}{dt}=g-\frac{k}{m}v.$

So as the body falls and $v$ increases, acceleration decreases linearly with $v$.

• Initially, when $v=0$: $a=g$
• As $v$ increases, $a$ decreases.
• At terminal speed $v_t$: $0=g-\frac{k}{m}v_t$ $v_t=\frac{mg}{k}$ and then $a=0.$

Thus:

• $v$ starts from $0$ and increases gradually toward the finite limit $\dfrac{mg}{k}$.
• $a$ starts from $g$ and decreases gradually toward $0$.

3. Time dependence of speed

Solving

$m\frac{dv}{dt}=mg-kv,$

we get

$v(t)=\frac{mg}{k}\left(1-e^{-kt/m}\right).$

This shows that $v$ rises from $0$ and asymptotically approaches the terminal velocity $\dfrac{mg}{k}$.

Its graph is an increasing curve with decreasing slope.

4. Time dependence of acceleration

Using $a=g-\dfrac{k}{m}v$ and the expression for $v(t)$:

$a(t)=g\,e^{-kt/m}.$

So acceleration decreases exponentially from $g$ to $0$.

Its graph is a decreasing curve approaching zero asymptotically.

5. Conclusion about the correct schematic graph

The correct option must show:

• $v$ vs $t$: starts at $0$, increases, and approaches a constant terminal value.
• $a$ vs $t$: starts at $g$, decreases, and approaches $0$.

This corresponds to Option B.

6. Comparison with stored answer

Stored correct answer: B

Derived answer: B

Hence, they agree.