Rectilinear motion under variable force

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2014, 2015, 2016)
• Priority tier: T3
• Marks (count): 13 (1), 15 (2)
• Average solve time: ~16 min
• Difficulty mix: hard 2, medium 1
• Section: B | Dominant type: computation

Why This Chapter Matters

Rectilinear motion under a variable force is the bridge between calculus and Newtonian mechanics: every question reduces to one integration. Three UPSC questions across three years confirm it is a reliable T3 source of 13–15 marks. The two core techniques — the energy first-integral (multiply both sides by $\dot x$ and integrate) and the $v\,dv/dx$ substitution — cover every variant the examiner has ever set. Master the substitution mechanics and the resulting integral evaluations and you will never lose marks here.

Minimum Theory

Equation of motion. A particle of unit mass constrained to a line with position $x$ satisfies $\ddot x = f(x)$ for a position-dependent acceleration. Two substitutions reduce this to a first-order problem.

Energy first integral. Multiply $\ddot x = f(x)$ by $\dot x$ and recognise the left side as $(d/dt)(\dot x^2/2)$. Integrating gives $\tfrac12\dot x^2 = \int f(x)\,dx + C$. Apply the initial condition $\dot x(0)=0$ at $x=x_0$ to determine $C$. Take the appropriate square-root sign (particle moves toward $x=0$, so $\dot x < 0$).

Arc-length form. Alternatively write $\ddot x = v\,dv/dx$ where $v = \dot x$. This directly gives $v\,dv = f(x)\,dx$. Once $v(x)$ is known, separate $dx/v(x) = dt$ and integrate for $T$.

Special integrals. Two evaluations appear repeatedly in UPSC: $\int_0^\infty e^{-u^2}\,du = \frac{\sqrt\pi}{2} \qquad\text{(Gaussian)}$ $\int_0^1 s^m(1-s)^{n-1}\,ds = B(m+1,n) = \frac{\Gamma(m+1)\Gamma(n)}{\Gamma(m+n+1)} \qquad\text{(Beta)}$ In particular $\int_0^1 s^2(1-s)^{-1/2}\,ds = B(3,1/2) = \Gamma(3)\Gamma(1/2)/\Gamma(7/2) = 2\cdot\sqrt\pi/\left(\tfrac{15}{8}\sqrt\pi\right) = 16/15$.

Question Archetypes

time-to-centre (2 question(s); 2015, 2016)

Recognition Cues

• “Starts from rest at distance $a$ from centre of force”; “force/acceleration directed toward $O$.”
• The acceleration magnitude is a power of $x$ (e.g. $k/x$, or $\mu(a^5/x^2)^{1/3}$).
• Asked to “find the time to arrive at $O$” — single-variable integration problem.

Solution Template

1. Write the equation of motion $\ddot x = -f(x)$ (negative because attraction is toward origin, $x>0$).
2. Substitute $v\,dv/dx = -f(x)$ and integrate with $v=0$ at $x=a$ to get $v^2(x)$.
3. Since the particle moves toward $O$, $\dot x < 0$. Write $dt = -dx/v(x)$ and set up $T = \int_0^a dx/|v(x)|$.
4. Choose a substitution that regularises both endpoints and reduces to a known integral (Gaussian or Beta).
5. Evaluate and state whether $T$ depends on $a$.

Worked Example

2015 Paper 1, 2015-P1-Q6d (13 marks)

A mass starts from rest at a distance $a$ from the centre of force which attracts inversely as the distance. Find the time of arriving at the centre.

The force attracts inversely as the distance, so the acceleration magnitude is $k/x$ (not $k/x^2$).

Step 1 — Energy first integral. $\ddot x = -k/x$. Multiply by $\dot x$:

$\frac{d}{dt}\!\left(\frac{\dot x^2}{2}\right) = -\frac{k\dot x}{x} = -k\frac{d}{dt}(\ln x).$

Integrate from $0$ to $t$ using $\dot x(0)=0$, $x(0)=a$:

$\frac{\dot x^2}{2} = k\ln\!\left(\frac{a}{x}\right).$

Step 2 — Time integral. Taking the negative root (particle falls toward origin):

$T = \int_0^a \frac{dx}{\sqrt{2k\ln(a/x)}}.$

Step 3 — Gaussian substitution. Let $x = ae^{-u^2}$, so $\ln(a/x) = u^2$ and $dx = -2uae^{-u^2}\,du$. When $x=a$, $u=0$; when $x\to 0$, $u\to\infty$.

$T = \int_0^\infty \frac{2uae^{-u^2}\,du}{\sqrt{2k}\cdot u} = \frac{2a}{\sqrt{2k}}\int_0^\infty e^{-u^2}\,du = \frac{2a}{\sqrt{2k}}\cdot\frac{\sqrt\pi}{2}.$

$\boxed{T = a\sqrt{\frac{\pi}{2k}}.}$

Common Traps

• “Attracts inversely as the distance” means $F = k/x$, not $k/x^2$. The inverse-square law would give a different integral.
• The sign of $\dot x$: since the particle moves toward the origin, always take $\dot x = -|v|$; the integral becomes $\int_0^a dx/|v|$.
• The substitution $x = ae^{-u^2}$ regularises both singular endpoints simultaneously: the integrand’s singularity at $x=a$ (where $\ln(a/x)\to 0$) is cured by the $u$ factor in the Jacobian, and the singularity at $x=0$ maps to the tail of the Gaussian.
• The answer $T\propto a/\sqrt{k}$ is proportional to $a$ — a useful sanity check.

2016 Paper 1, 2016-P1-Q8c (15 marks)

A particle moves in a straight line. Its acceleration is directed towards a fixed point $O$ in the line and is always equal to $\mu\!\left(\dfrac{a^5}{x^2}\right)^{\!1/3}$ when it is at a distance $x$ from $O$. If it starts from rest at a distance $a$ from $O$, find the time the particle will take to arrive at $O$.

The acceleration magnitude is $\mu(a^5/x^2)^{1/3} = \mu\,a^{5/3}x^{-2/3}$.

Step 1 — Speed. Use $v\,dv/dx = -\mu\,a^{5/3}x^{-2/3}$. Integrate with $v=0$ at $x=a$:

$\frac{v^2}{2} = \mu a^{5/3}\!\left(3x^{1/3}\right)\Big|_{x}^{a} = 3\mu a^{5/3}(a^{1/3}-x^{1/3}).$

$v^2 = 6\mu a^{5/3}(a^{1/3}-x^{1/3}).$

Step 2 — Time integral.

$T = \int_0^a \frac{dx}{\sqrt{6\mu a^{5/3}(a^{1/3}-x^{1/3})}}.$

Step 3 — Substitution $x = as^3$. Then $x^{1/3} = a^{1/3}s$, $dx = 3as^2\,ds$, limits $s:0\to 1$.

$6\mu a^{5/3}(a^{1/3}-a^{1/3}s) = 6\mu a^{5/3}\cdot a^{1/3}(1-s) = 6\mu a^2(1-s).$

$T = \int_0^1 \frac{3as^2\,ds}{\sqrt{6\mu a^2(1-s)}} = \frac{3}{\sqrt{6\mu}}\int_0^1 \frac{s^2}{\sqrt{1-s}}\,ds.$

Step 4 — Beta integral. With $u = 1-s$:

$\int_0^1 \frac{s^2}{\sqrt{1-s}}\,ds = \int_0^1 \frac{(1-u)^2}{\sqrt{u}}\,du = 2 - \frac{4}{3} + \frac{2}{5} = \frac{16}{15}.$

$\boxed{T = \frac{3}{\sqrt{6\mu}}\cdot\frac{16}{15} = \frac{8}{15}\sqrt{\frac{6}{\mu}}.}$

Notably, $T$ is independent of $a$.

Common Traps

• Simplify $(a^5/x^2)^{1/3} = a^{5/3}x^{-2/3}$ before integrating; then $\int x^{-2/3}\,dx = 3x^{1/3}$.
• Choose $x = as^3$ (cube substitution) because the exponent $1/3$ in the acceleration calls for it; $x = as^2$ does not simplify cleanly.
• The $a$ factor cancels completely, giving $T$ independent of starting distance — state this explicitly as a self-check.
• At $x=a$ the integrand has a $\sqrt{1-s}$ singularity (integrable); at $x=0$ the integrand vanishes ($s=0$). Both endpoints are integrable.

variable-acceleration-path (1 question(s); 2014)

Recognition Cues

• “Acceleration $\mu/y^2$ toward the $x$-axis”; particle is “projected” parallel to one axis with a given initial speed.
• Two separate equations of motion: $\ddot x = 0$ (no force in $x$) and $\ddot y = -\mu/y^2$ (force in $y$).
• Asked to find the “parametric equation of the path” — eliminate time to get $(x(t), y(t))$ in terms of a parameter.

Solution Template

1. Solve the horizontal equation: $\ddot x = 0 \Rightarrow \dot x = u_0$ (constant), $x = u_0 t$.
2. For the vertical equation, use the energy integral: multiply $\ddot y = f(y)$ by $\dot y$ and integrate with $\dot y(0) = 0$ at $y = a$.
3. Separate $dt = dy/\dot y$ and write $x = u_0\int dy/\dot y$ to express $x$ as an integral over $y$.
4. Evaluate the integral by a trigonometric substitution.
5. Introduce a standard parameter (angle $\eta$) to write $x, y$ in recognisable parametric form.

Worked Example

2014 Paper 1, 2014-P1-Q8b (15 marks)

A particle is acted on by a force parallel to the axis of $y$ whose acceleration (always toward the axis of $x$) is $\mu y^{-2}$, and when $y=a$ it is projected parallel to the axis of $x$ with velocity $\sqrt{2\mu/a}$. Find the parametric equations of the path.

Step 1 — Horizontal. $\ddot x = 0$, $\dot x = \sqrt{2\mu/a}$ constant, so $x = \sqrt{2\mu/a}\cdot t$.

Step 2 — Vertical energy integral. $\ddot y = -\mu/y^2$. Multiply by $\dot y$:

$\frac{1}{2}\dot y^2 = \frac{\mu}{y} + C.$

At $t=0$: $\dot y=0$, $y=a$ gives $C = -\mu/a$. So

$\dot y^2 = \frac{2\mu(a-y)}{ay}.$

Step 3 — Express $x$ via $y$. Since $\dot y < 0$:

$x = \sqrt{\frac{2\mu}{a}}\int_a^y \frac{dy}{\dot y} = \int_y^a \sqrt{\frac{y'}{a-y'}}\,dy'.$

Step 4 — Trigonometric substitution. Let $y' = a\sin^2\phi$, so $dy' = 2a\sin\phi\cos\phi\,d\phi$ and $\sqrt{y'/(a-y')} = \tan\phi$. Then

$x = \int_{\phi_y}^{\pi/2} 2a\sin^2\phi\,d\phi = a\left[\phi - \frac{\sin 2\phi}{2}\right]_{\phi_y}^{\pi/2}.$

Step 5 — Cycloid parameter. Let $\eta = \pi - 2\phi_y$, so $\phi_y = (\pi-\eta)/2$. Then $\sin 2\phi_y = \sin\eta$ and $y = a\sin^2\phi_y = \tfrac{a}{2}(1+\cos\eta)$.

$x = a\!\left(\frac{\eta}{2} + \frac{\sin\eta}{2}\right) = \frac{a}{2}(\eta + \sin\eta).$

$\boxed{x = \frac{a}{2}(\eta+\sin\eta),\qquad y = \frac{a}{2}(1+\cos\eta).}$

This is the parametric equation of a cycloid (circle of radius $a/2$ rolling along $y=a$).

Common Traps

• The force “toward the $x$-axis” means $\ddot y = -\mu/y^2$ for $y>0$ — negative sign is mandatory.
• The integral $\int\sqrt{y/(a-y)}\,dy$ looks like a conic but it is not — the substitution $y = a\sin^2\phi$ transforms it into $\int(1-\cos 2\phi)d\phi$, giving a cycloid.
• Recognise the final form: standard cycloid $x = R(\eta+\sin\eta),\;y = R(1+\cos\eta)$ with $R = a/2$. State “cycloid” explicitly for full credit.
• At $\eta = 0$: $(x,y) = (0,a)$ — the initial point. Verify.

Marks-Aware Writing

A 13-mark answer must: set up the ODE correctly with proper sign; carry out the energy first integral and apply the initial condition; execute the substitution; evaluate the resulting standard integral; box the final answer. Shortcut or skip the integral evaluation and you lose 4–5 marks.

A 15-mark answer must additionally: justify the substitution choice explicitly; verify at least one limit; state any notable feature (e.g., $T$ independent of $a$, or path is a cycloid). Five marks are typically reserved for the integral evaluation — show all steps.

Practice Set

• 2019-P1-Q5d (10 m) — Inverse-square gravity work; use $W = \int_h^{4h} k/r^2\,dr$ with $k = Wh^2$. ()
• 2019-P1-Q7c (20 m) — Extended dynamics problem; energy methods. ()
• 2018-P1-Q6b (12 m) — Variable force in rectilinear motion. ()