JEE Physics 2026 Question with Solution

Given: A particle of mass $m$ falls from rest in a resistive medium with resistive force $F=-kv$.

Find: Which graph correctly represents $v$ versus $t$.

For downward motion, take downward direction as positive. Then the forces are weight $mg$ downward and resistive force $kv$ upward.

So, the equation of motion is

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

Rearranging,

$$\frac{dv}{dt}+\frac{k}{m}v=g$$

This is a first-order linear differential equation.

Using the integrating factor,

$$\text{I.F.}=e^{\frac{k}{m}t}$$

we get the solution

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

Now analyze the result:

• At $t=0$, $v(0)=0$, so the particle starts from rest.
• As $t\to\infty$, $v\to\frac{mg}{k}$, which is the terminal velocity.

Therefore, velocity increases from zero, rises rapidly at first, and then gradually levels off toward a horizontal asymptote.

So the correct graph is the one that starts at the origin and approaches a constant value asymptotically. Graph 1 matches this behavior.

Therefore, the correct option is A.

Given: Resistive force is proportional to velocity.

Find: The qualitative shape of the $v$-$t$ graph.

Whenever the resistive force is proportional to speed, the net downward force decreases as $v$ increases. Hence the acceleration is not constant.

Initially, when $v=0$,

$$\frac{dv}{dt}=g$$

so the particle begins with maximum acceleration.

As time increases, $kv$ increases, so

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

becomes smaller and smaller.

Thus the slope of the $v$-$t$ graph decreases continuously, and eventually becomes nearly zero when terminal velocity is approached.

This rules out:

• Graph 2, because it shows the slope increasing.
• Graph 3, because it shows constant slope, which would mean constant acceleration.
• Graph 4, because it shows decreasing velocity.

Hence the only correct graph is Graph 1, so the correct option is A.