(a) $\vec v=\vec a(1-2\alpha t)$, $\vec w=-2\alpha\vec a$; (b) $\Delta t=\dfrac1\alpha$ and distance $s=\dfrac{a}{2\alpha}$.
(ii) Problem Statement
“A radius vector of a particle varies with time $t$ as $\vec r=\vec a\,t\,(1-\alpha t)$, where $\vec a$ is a constant vector and $\alpha$ is a positive factor. Find: (a) the velocity $\vec v$ and the acceleration $\vec w$ of the particle as functions of time; (b) the time interval $\Delta t$ taken by the particle to return to the initial point, and the distance covered during that time.”
No diagram is given, so the scene shows a straight line along $\vec a$. The tilt on screen is only for looks.
(iii) The Big Idea in Plain Words
A ball is thrown along a line. It slows down, stops, turns round and comes back to where it started.
Symbols: $\vec r$ = where the particle is, measured from its start. $\vec a$ = a fixed vector (its length is a speed, in m/s). $\alpha$ = a positive number with units 1/s. $t$ = time. $\vec v$ = velocity, how fast $\vec r$ changes. $\vec w$ = acceleration, how fast $\vec v$ changes.
The velocity falls steadily and passes through zero at $t=1/(2\alpha)$. That is the turning point. After it the particle returns. Distance counts the path both ways, so it is twice the outward trip.
(iv) The Solution Viewed in Each Frame
1. Ground Frame (stationary observer)
Why: velocity is the rate of change of position.
$\vec v=\dfrac{d\vec r}{dt}=\vec a(1-2\alpha t)$.
Why: acceleration is the rate of change of velocity.
$\vec w=\dfrac{d\vec v}{dt}=-2\alpha\vec a$. It is constant and points opposite to $\vec a$.
Why: “back at the start” means $\vec r=0$.
$\vec a\,t(1-\alpha t)=0$ gives $t=0$ or $t=1/\alpha$. So $\Delta t=1/\alpha$.
Why: the particle turns round when $v=0$, and distance must count both legs.
Turning time: $t_1=1/(2\alpha)$. Outward distance: $a\,t_1(1-\alpha t_1)=\dfrac{a}{4\alpha}$. The way back is equal. So $s=2\cdot\dfrac{a}{4\alpha}=\dfrac{a}{2\alpha}$.
2. Moving Frame (travels at constant $\vec u=\vec a$, the starting velocity)
Why: subtracting $\vec a t$ removes the steady part of the motion.
$\vec r\,'=\vec r-\vec a t=-\alpha\vec a\,t^2$ and $\vec v\,'=\vec v-\vec a=-2\alpha\vec a\,t$.
Why: $\vec u$ is constant, so it cannot change the acceleration.
$\vec w\,'=-2\alpha\vec a=\vec w$. Same as before.
Why: the particle is back at the start when it meets the frame origin again.
The ground start point sits at $-\vec a t$ in this frame. Set $-\alpha a t^2=-at$, so $t=1/\alpha$. Same $\Delta t$. The path here never turns back, so its length is $a/\alpha$. Distance is frame-dependent, and the question asks it in the ground frame: $s=a/(2\alpha)$.
3. Centre-of-Mass Frame
Does not apply. There is one particle, so its centre of mass is the particle itself.
4. Medium Frame
Does not apply. No water, air or other medium is mentioned.
- Changes: the velocity (it is $\vec v-\vec a$ in the moving frame) and the path length ($a/\alpha$ instead of $a/(2\alpha)$).
- Same: the acceleration $-2\alpha\vec a$ and the return time $1/\alpha$. The answers asked for are the ground-frame ones.
(v) The Solution by Each Method
Method 1: Calculus Method (most reliable)
- Why: differentiate once for velocity.
$\vec v=\vec a(1-2\alpha t)$. - Why: differentiate again for acceleration.
$\vec w=-2\alpha\vec a$. - Why: solve $\vec r=0$ for the return time.
$\Delta t=1/\alpha$. - Why: the speed changes sign, so split the integral at $v=0$.
$s=\int_0^{1/2\alpha}a(1-2\alpha t)\,dt+\int_{1/2\alpha}^{1/\alpha}a(2\alpha t-1)\,dt=\dfrac{a}{4\alpha}+\dfrac{a}{4\alpha}=\dfrac{a}{2\alpha}$.
Method 2: Graph Method (fastest for distance)
- Why: on a velocity–time graph, area is distance.
$v(t)$ is a straight line from $a$ down to $-a$. It crosses zero at $t=1/(2\alpha)$. - Why: area below the axis is still distance travelled.
Each triangle has area $\tfrac12\cdot\dfrac{1}{2\alpha}\cdot a=\dfrac{a}{4\alpha}$. Their sum is $s=\dfrac{a}{2\alpha}$. The two areas cancel for displacement, which matches $\vec r=0$ at $t=1/\alpha$. - Why: the slope of the line is the acceleration.
Slope $=\dfrac{-a-a}{1/\alpha}=-2\alpha a$.
Method 3: Relative-Motion Method
- Why: a frame moving at $\vec a$ removes the steady part.
There $\vec r\,'=-\alpha\vec a t^2$, which is plain constant acceleration from rest, with $\vec w=-2\alpha\vec a$. - Why: the start point moves at $-\vec a$ here.
Meeting it again gives $\alpha t^2=t$, so $\Delta t=1/\alpha$. Distance still needs the ground frame: $s=a/(2\alpha)$.
Method 4: Force Method
Does not suit this problem: no mass or force is given, only the motion.
Method 5: Energy Method
Does not suit this problem: no mass, work or energy is given.
Fastest: graph method (two triangles). Most reliable in an exam: calculus. All give the same answer.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|
(vii) Common Exam Traps
- Giving the distance as $0$ because $\vec r=0$ at the end. That is displacement. Distance counts both legs.
- Integrating $\vec v$ from $0$ to $1/\alpha$ without splitting at $t=1/(2\alpha)$. The signed integral cancels to zero.
- Writing $v=a(1-2\alpha t)$ as the speed for all times. After the turn it is negative, so the speed is $a(2\alpha t-1)$.
- Forgetting that $\vec w$ is constant and opposite to $\vec a$. It does not depend on $t$.
- Dropping the factor 2: the turning point is $1/(2\alpha)$, not $1/\alpha$.
Consistency check: the live readouts, both frames and the methods all give $\Delta t=1/\alpha$, $s=a/(2\alpha)$ and $\vec w=-2\alpha\vec a$. Units: $a$ in m/s, $\alpha$ in 1/s, $\Delta t$ in s, $s$ in m, $w$ in m/s².