Velocity Time Graph with Air Resistance

 

❓ Question

An object is dropped from a height $h$ above the ground. Apart from gravity, an additional drag force

$$F_{\text{drag}}=-kv$$

acts on the object, where $k$ is a positive constant.

Find the nature of the graph between velocity $v$ and time $t$.

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✍️ Short Explanation

The forces acting on the falling body are:

• Weight $mg$ (downward),
• Drag force $kv$ (upward, opposite to motion).

Taking downward direction as positive,

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

Hence,

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

The acceleration decreases as the velocity increases.

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🔷 Step 1 — Initial Condition

Initially, the object is released from rest:

$$v=0 \quad \text{at} \quad t=0.$$

So,

$$a=g-\frac{k}{m}(0)=g.$$

Thus, the $v$-$t$ graph starts from the origin with initial slope equal to $g$.

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🔷 Step 2 — As Time Increases

As the object falls,

• velocity $v$ increases,
• drag force $kv$ increases,
• net acceleration

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

gradually decreases.

Therefore, the slope of the $v$-$t$ graph continuously decreases.

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🔷 Step 3 — Terminal Velocity

Eventually,

$$mg=kv,$$

so the net force becomes zero.

Hence,

$$a=0$$

and the velocity attains a constant value called the terminal velocity:

$$v_T=\frac{mg}{k}.$$

After this point, the graph becomes horizontal.

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✅ Final Answer

The $v$-vs-$t$ graph:

• starts from the origin,
• initially has slope $g$,
• rises with decreasing slope (concave downward),
• finally approaches a horizontal line asymptotically at

$$\boxed{v=\frac{mg}{k}}.$$

Mathematically, the velocity is

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

which is an exponentially increasing curve approaching the terminal velocity.

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🔷 JEE Shortcut

For linear drag force $F=-kv$:

• $a=g-\dfrac{k}{m}v$
  
• Initial slope of $v$-$t$ graph $=g$
  
• Final velocity (terminal velocity)

$$\boxed{v_T=\frac{mg}{k}}$$

So always choose the graph that starts at the origin and asymptotically approaches a constant horizontal value.

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