Mean velocity is total distance over total time, so $\langle v\rangle=\dfrac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}$.
(ii) Problem Statement
"A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion."
Reference Answer Verification (from the supplied solution image):
The reference writes $\tfrac s2=v_0t_1$ and $\tfrac s2=(v_1+v_2)t$, then $\langle v\rangle=\dfrac{s}{t_1+2t}=\dfrac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}$. Our work gives exactly this result.
(iii) The Big Idea in Plain Words
Mean velocity is the total distance divided by the total time. You cannot just average the speeds. The point spends different times at each speed.
Symbols: $s$ is the whole distance. $t_1$ is the time for the first half. $2t$ is the time for the second half. $T=t_1+2t$ is the whole time. The point never turns back, so distance and displacement are equal.
The trick: in the second half each speed lasts the same time $t$. So we find $t_1$ and $2t$ separately and add them.
(iv) The Solution Viewed in Each Reference Frame
1. Ground or Lab Frame (Stationary Observer)
Origin fixed at the startWhy: the first half has one speed, so time equals distance divided by speed.
$$\frac s2=v_0t_1\;\Rightarrow\;t_1=\frac{s}{2v_0}$$
Why: in the second half, each speed lasts the same time $t$, so the two distances add up to $s/2$.
$$\frac s2=(v_1+v_2)\,t\;\Rightarrow\;2t=\frac{s}{v_1+v_2}$$
Why: mean velocity is total distance over total time.
$$\langle v\rangle=\frac{s}{t_1+2t}=\frac{s}{\frac{s}{2v_0}+\frac{s}{v_1+v_2}}=\frac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}$$
2. Medium Frame (Observer Moving at Steady Speed $u$)
Origin moves at $u$Why: a steady medium (like a river) moving at $u$ shifts every velocity by the same amount.
Each velocity becomes $v'=v-u$. Times do not change.
Why: the mean velocity is just distance moved in this frame over the same time $T$.
$$\langle v\rangle'=\frac{s-uT}{T}=\langle v\rangle-u$$
Why: adding $u$ back must recover the ground answer.
$\langle v\rangle=\langle v\rangle'+u=\dfrac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}$. The sidebar slider $u$ shows this live.
3. Centre-of-Mass (CM) Frame
This frame does not apply. The centre of mass of a single point is the point itself, so its velocity is always $0$ and its momentum is $0$. The speed jumps from $v_0$ to $v_1$ to $v_2$, so this frame accelerates and would need a pseudo-force (a "fake" force $-ma$ seen only in accelerating frames). It adds nothing, and the answer stays $\langle v\rangle$ once you return to the ground frame.
What Changes and What Stays the Same Across Frames
- What changes: every measured velocity (by $-u$) and the distance moved.
- What stays the same: all times $t_1$, $2t$, $T$, the speed jumps' timing, and the ground answer $\langle v\rangle$.
(v) The Solution by Each Method
Method 1: Equivalent-Speed Shortcut
Fastest- Why: equal times mean a plain average of speed.
The second half acts like one speed $w=\dfrac{v_1+v_2}{2}$. - Why: equal distances at two speeds give the harmonic mean.
$\langle v\rangle=\dfrac{2v_0w}{v_0+w}=\dfrac{2v_0(v_1+v_2)}{2v_0+v_1+v_2}$.
Method 2: Graph Method ($v$-$t$ and $x$-$t$)
- Why: area under a $v$-$t$ graph is distance.
Draw three flat steps: height $v_0$ for $t_1$, then $v_1$ for $t$, then $v_2$ for $t$. Areas: $v_0t_1=\tfrac s2$ and $(v_1+v_2)t=\tfrac s2$. - Why: on the $x$-$t$ graph, the straight line from start to end has slope equal to the mean velocity.
The dashed chord in the simulation's graph rises $s$ over time $T$: $\langle v\rangle=\dfrac sT=\dfrac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}$.
Method 3: Calculus Method
- Why: the mean value of a changing quantity is its integral divided by the interval.
$\langle v\rangle=\dfrac{\int v\,dt}{\int dt}$. - Why: split the integral at each speed change.
$\int v\,dt=v_0t_1+v_1t+v_2t=\tfrac s2+\tfrac s2=s$ and $\int dt=t_1+2t$. - Why: put in $t_1$ and $2t$ from above.
$\langle v\rangle=\dfrac{s}{\frac{s}{2v_0}+\frac{s}{v_1+v_2}}=\dfrac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}$.
Method 4 & 5: Force, Energy and Relative-Motion Methods
Force Method: not suited. No forces or accelerations are given, only speeds.
Energy Method: not suited. No mass or work is given.
Relative-Motion Method: only a change of frame by $u$, which shifts the answer by $u$ (see frame 2). It does not shorten the work, so the shortcut in Method 1 is used instead.
Fastest: Method 1. Most reliable in an exam: Method 3 (total distance ÷ total time). All give $\langle v\rangle=\dfrac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}$.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|---|---|---|
| $s$ | total distance | m | |
| $v_0$ | speed over the first half of the distance | m/s | |
| $v_1,\,v_2$ | speeds for the two equal time halves | m/s | |
| $t_1,\,2t,\,T$ | first-half, second-half, total time | s | |
| $u$ | medium speed (Medium frame only) | m/s | |
| $\langle v\rangle$ | mean velocity (answer) | m/s |
(vii) Common Exam Traps
- Using $\frac{v_0+v_1+v_2}{3}$. Wrong: the times are not equal.
- Using $\frac{v_0+(v_1+v_2)/2}{2}$. This is right only if both halves take equal time.
- Mixing "half the distance" (first part) with "half the time" (second part).
- Writing $t$ instead of $2t$ for the whole second-half time.
- Changing frames but forgetting to subtract $u$ from every velocity.
Verified: live numbers, all frames and all methods give the same $\langle v\rangle$, in m/s.