1D Kinematics by Calculus

Most JEE motion problems are not constant-acceleration cookbook exercises. When $a$ depends on $t$, on $v$ or on $x$, you must use calculus: differentiation gives instantaneous rates, integration gives accumulated change.

Concept

For a particle on a line at position $x(t)$:

• Velocity $v = \dfrac{\mathrm{d}x}{\mathrm{d}t}$.
• Acceleration $a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}$.
• Chain rule trick: $a = v\dfrac{\mathrm{d}v}{\mathrm{d}x}$, useful when $a$ depends on $x$.

For constant acceleration, integrating gives the standard kinematics: $v = u + at,\qquad x = ut + \tfrac{1}{2}at^{2},\qquad v^{2} = u^{2} + 2ax.$

Displacement = $\int v\,\mathrm{d}t$; change of velocity = $\int a\,\mathrm{d}t$; area under $v$-$t$ graph = displacement; area under $a$-$t$ graph = velocity change.

Derivation

Suppose $a = -k v$ (linear drag). Then $\dfrac{\mathrm{d}v}{\mathrm{d}t} = -k v$ separates as $\int_{u}^{v} \frac{\mathrm{d}v}{v} = -k\int_{0}^{t}\mathrm{d}t \Rightarrow v = u e^{-k t}.$ Position follows from another integration: $x(t) = \dfrac{u}{k}(1 - e^{-k t})$. As $t\to\infty$, $x\to u/k$ (asymptotic stop).

If $a$ depends on $x$, e.g. $a = -\omega^{2} x$ (SHM), use $v\,\mathrm{d}v = a\,\mathrm{d}x$: $\int v\,\mathrm{d}v = \int -\omega^{2} x\,\mathrm{d}x \Rightarrow v^{2} = \omega^{2}(A^{2} - x^{2}).$

JEE Worked Example

Q. A particle starts from rest with $a(t) = 6t\,\text{m s}^{-2}$. Find its velocity and position at $t = 4\,\text{s}$.

Solution. $v = \int_{0}^{t} 6t'\,\mathrm{d}t' = 3 t^{2}$. So $v(4) = 48\,\text{m s}^{-1}$.

$x = \int_{0}^{t} 3 t'^{2}\,\mathrm{d}t' = t^{3}$. So $x(4) = 64\,\text{m}$.

Traps

• Using $v = u + at$ when $a$ is variable.
• Confusing average velocity ($\Delta x/\Delta t$) with average speed (total distance/$\Delta t$).
• Forgetting initial conditions when integrating.
• Sign errors during deceleration; choose a fixed positive direction.

Key Takeaways

• $v = \mathrm{d}x/\mathrm{d}t$, $a = \mathrm{d}v/\mathrm{d}t = v\,\mathrm{d}v/\mathrm{d}x$.
• Area under $v$-$t$ graph $=$ displacement (signed).
• Pick the integration variable that matches the dependence of $a$.