(a) $\langle v\rangle=\dfrac{\pi R}{T}\approx 50\ \text{cm/s}$; (b) $|\langle\vec v\rangle|=\dfrac{2R}{T}=32\ \text{cm/s}$; (c) $|\langle\vec w\rangle|=\dfrac{2\pi R}{T^2}\approx 10\ \text{cm/s}^2$.
(ii) Problem Statement
"A point traversed half a circle of radius $R=160$ cm during time interval $T=10.0$ s. Calculate the following quantities averaged over that time: (a) the mean velocity $\langle v\rangle$; (b) the modulus of the mean velocity vector $|\langle\vec v\rangle|$; (c) the modulus of the mean vector of the total acceleration $|\langle\vec w\rangle|$ if the point moved with constant tangent acceleration."
Key idea / reference check: (a) is distance over time, (b) is straight-line shift over time, and (c) is the change of the velocity vector over time. The reference gives 50 cm/s and 32 cm/s, which are rounded values; exactly $\pi R/T=50.27$ cm/s and $2R/T=32.0$ cm/s. It writes the last answer as $2\pi R/T^2$, which is $10.05\approx10$ cm/s$^2$. All three match.
(iii) The Big Idea in Plain Words
A bug walks round half of a circular track. It starts at point $i$ and ends at point $f$, on the other side of the circle.
Symbols: $R$ = radius of the circle. $T$ = time for the trip. $v_0$ = speed at $i$. $v$ = speed at $f$. $w_t$ = tangential acceleration, which means how fast the speed itself changes.
Mean speed counts the whole curved path. The mean velocity vector only counts the straight shift from $i$ to $f$, which is the diameter $2R$. The mean acceleration vector is the change of the velocity vector divided by $T$. At $i$ the bug moves up and at $f$ it moves down, so the two speeds add.
(iv) The Solution Viewed in Each Frame
1. Ground Frame (stationary observer)
Why: the figure is drawn in this frame, so we use it for all three answers.
Path length is half the circumference: $s=\pi R$. So $\langle v\rangle=s/T=\pi R/T=50.3$ cm/s.
Why: the mean velocity vector uses only the start and end points.
The shift is the diameter, $|\Delta\vec r|=2R$. So $|\langle\vec v\rangle|=2R/T=32.0$ cm/s.
Why: for constant $w_t$ the mean speed is the average of the end speeds.
$\langle v\rangle=\dfrac{v_0+v}{2}=\dfrac{\pi R}{T}$, so $v_0+v=\dfrac{2\pi R}{T}$. At $i$ the velocity is $+v_0\hat y$ and at $f$ it is $-v\hat y$, so $|\Delta\vec v|=v_0+v$. Then $|\langle\vec w\rangle|=\dfrac{v_0+v}{T}=\dfrac{2\pi R}{T^2}=10.05$ cm/s$^2$.
2. Mean-Velocity Frame (moves at $u=2R/T$ along $+x$)
Why: this frame makes the net shift zero, which shows what the mean velocity vector really means.
Here $\vec v\,'=\vec v-\vec u$ and $\Delta\vec r\,'=\Delta\vec r-\vec uT=0$. So $\langle\vec v\,'\rangle=0$.
Why: $\vec u$ is constant, so it cancels in a difference of velocities.
$\Delta\vec v\,'=\Delta\vec v$, so $|\langle\vec w\,'\rangle|=2\pi R/T^2$ again. The path length in this frame is different, so (a) is asked in the ground frame.
3. Centre-of-Mass Frame
Does not apply. There is a single point, so there is no system whose centre of mass can be used.
4. Medium Frame
Does not apply. There is no moving water, air or other medium in this problem.
- Changes: the path, the mean velocity vector (it becomes zero in the mean-velocity frame) and the path length.
- Same: the change of velocity $\Delta\vec v$, so $|\langle\vec w\rangle|=2\pi R/T^2$ in both frames.
(v) The Solution by Each Method
Method 1: Graph Method (fastest)
- Why: on a speed–time graph, the area is the distance covered.
With constant $w_t$ the graph is a straight sloping line from $v_0$ to $v$, and its area is $\pi R$. - Why: the mean speed is the height of a flat line with the same area.
That height is the mid-point of the line: $\langle v\rangle=\pi R/T=50.3$ cm/s. This is the dashed line in the bottom graph. - Why: the other two answers need vectors, not areas.
$|\langle\vec v\rangle|=2R/T=32.0$ cm/s and $|\langle\vec w\rangle|=(v_0+v)/T=2\pi R/T^2=10.05$ cm/s$^2$.
Method 2: Calculus Method
- Why: constant $w_t$ gives a simple speed law.
$\dfrac{dv}{dt}=w_t\Rightarrow v=v_0+w_tt$. - Why: a mean is an integral divided by the time.
$\langle v\rangle=\dfrac1T\int_0^T(v_0+w_tt)\,dt=\dfrac{v_0+v}{2}=\dfrac{\pi R}{T}$. - Why: the mean vector acceleration is the change of velocity over $T$.
$|\langle\vec w\rangle|=\dfrac{|\vec v-\vec v_0|}{T}=\dfrac{v_0+v}{T}=\dfrac{2\pi R}{T^2}$.
Method 3: Force Method
Does not suit this problem: no masses or forces are given, only the motion.
Method 4: Energy Method
Does not suit this problem: no mass, work or energy is involved.
Fastest: graph method. Most reliable in an exam: calculus. All give the same answer.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|
(vii) Common Exam Traps
- Mixing up speed and velocity: (a) uses the curved distance $\pi R$, but (b) uses the straight shift $2R$.
- Subtracting the speeds in (c). The directions at $i$ and $f$ are opposite, so $|\Delta\vec v|=v_0+v$, not $v-v_0$.
- Using only the tangential acceleration $w_t$ or the centripetal $v^2/R$. The question wants the mean of the total acceleration vector, found from $\Delta\vec v/T$.
- Using $\langle v\rangle=(v_0+v)/2$ when $w_t$ is not constant. It works here only because $w_t$ is constant.
Verified: the live simulation, every frame and every method give the same answer, in cm and seconds.