(a) $\langle v\rangle=\dfrac{200\text{ cm}}{20\text{ s}}=10\text{ cm/s}$. (b) $v_{max}\approx25\text{ cm/s}$. (c) $t_0\approx16\text{ s}$.
(ii) Problem Statement
"A point moves rectilinearly in one direction. Fig. 1.1 shows the distance $s$ traversed by the point as a function of the time $t$. Using the plot find: (a) the average velocity of the point during the time of motion; (b) the maximum velocity; (c) the time moment $t_0$ at which the instantaneous velocity is equal to the mean velocity averaged over the first $t_0$ seconds."
Reference Answer Verification (from the supplied solution image):
(a) $\langle v\rangle=\dfrac st=\dfrac{200\text{ cm}}{20\text{ s}}=10\text{ cm/s}$. (b) The slope $\dfrac{ds}{dt}$ is greatest along line $ac$, so $v_{max}=\dfrac{bc}{ab}=\dfrac{100\text{ cm}}{4\text{ s}}=25\text{ cm/s}$. (c) The tangent at point $d$ passes through the origin, so $t_0=16\text{ s}$. Our work matches.
Note: the sim draws the curve as a smooth S-shaped curve that matches the figure ($s=2.0$ m at about $t=20$ s, steepest near $t=12$ s). The live numbers come from this curve, so they differ slightly from graph readings.
(iii) The Big Idea in Plain Words
On a distance-time graph, slope means velocity. A steep line means fast. A flat line means stopped.
Symbols: $s$ is the distance moved. $t$ is the time. $\langle v\rangle$ is the mean (average) velocity, which is total distance over total time. $v_{max}$ is the greatest velocity. $t_0$ is the special time asked for in part (c).
Part (a): join the start to the end with a straight line and find its slope. Part (b): find the steepest part of the curve. Part (c): draw a straight line from the origin that just touches the curve. The touching point gives $t_0$.
Units: the graph is in metres, so $2.0\text{ m}=200\text{ cm}$ and $0.10\text{ m/s}=10\text{ cm/s}$.
(iv) The Solution Viewed in Each Reference Frame
1. Ground or Lab Frame (Stationary Observer)
Origin fixed at the startWhy: mean velocity is total distance over total time.
$$\langle v\rangle=\frac st=\frac{200\text{ cm}}{20\text{ s}}=10\text{ cm/s}$$
Why: the instantaneous velocity is the slope $\dfrac{ds}{dt}$, so the greatest velocity is the steepest slope.
$$v_{max}=\frac{bc}{ab}=\frac{100\text{ cm}}{4\text{ s}}=25\text{ cm/s}$$
Why: the mean velocity over the first $t_0$ seconds is $\dfrac{s}{t_0}$, the slope of the line from the origin to the curve.
$$\frac{ds}{dt}=\frac{s}{t_0}\;\Rightarrow\;\text{the tangent passes through the origin}\;\Rightarrow\;t_0=16\text{ s}$$
2. Medium Frame (Observer Moving at Steady Speed $u$)
Origin moves at $u$Why: in a frame moving at $u$, every distance changes to $s'=s-ut$ and every velocity to $v'=v-u$.
Times do not change.
Why: subtracting a steady speed lowers every velocity by the same amount.
(a) $\langle v\rangle'=\langle v\rangle-u$. (b) $v'_{max}=v_{max}-u$.
Why: the condition in (c) is $v'=\dfrac{s'}{t}$. Write it out.
$\dfrac{ds}{dt}-u=\dfrac{s-ut}{t}=\dfrac st-u$, so $\dfrac{ds}{dt}=\dfrac st$. The $u$ cancels, so $t_0=16\text{ s}$ in every frame. Add $u$ back to (a) and (b) and you recover the ground answers.
3. Centre-of-Mass (CM) Frame
This frame does not apply. The point is its own centre of mass, so its velocity and momentum are $0$ here. The point speeds up and slows down, so this frame accelerates and would need a pseudo-force (a "fake" force seen only by an accelerating observer). It shows nothing new, and the ground answers stand.
What Changes and What Stays the Same Across Frames
- What changes: distances, velocities, the mean velocity (a) and the maximum velocity (b). Each shifts by $-u$.
- What stays the same: all times, and the answer to (c): $t_0\approx16\text{ s}$.
(v) The Solution by Each Method
Method 1: Graph Method (Slopes)
Fastest- Why: the slope of the line from the start to the end is the mean velocity.
The curve ends at $s=200\text{ cm}$ when $t=20\text{ s}$. So $\langle v\rangle=\dfrac{200}{20}=10\text{ cm/s}$. - Why: the steepest part of the curve has the biggest slope.
Pick two points $a$ and $c$ on that part. The rise $bc$ is $100\text{ cm}$ and the run $ab$ is $4\text{ s}$. So $v_{max}=25\text{ cm/s}$. - Why: a line from the origin to the curve has slope $\dfrac{s}{t_0}$. It equals the curve's own slope only if the line just touches the curve.
Rotate a ruler about the origin until it touches the curve at $d$. Read $t_0=16\text{ s}$.
Method 2: Calculus Method
- Why: velocity is the rate of change of distance.
$v=\dfrac{ds}{dt}$ and $\langle v\rangle=\dfrac{s(t)-s(0)}{t}$. With $s(0)=0$ and $s=200\text{ cm}$ at $t=20\text{ s}$: $\langle v\rangle=10\text{ cm/s}$. - Why: a maximum has zero slope. So set the slope of $v$ to zero.
$\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}=0$. This is the bend (inflection point) of the curve, where $v=v_{max}=25\text{ cm/s}$. - Why: the mean velocity over the first $t$ seconds is $\dfrac st$. When it is largest, its derivative is zero.
$\dfrac{d}{dt}\left(\dfrac st\right)=\dfrac{1}{t}\left(\dfrac{ds}{dt}-\dfrac st\right)=0\Rightarrow\dfrac{ds}{dt}=\dfrac{s}{t_0}$. This holds at $t_0\approx16\text{ s}$.
Methods 3, 4 and 5: Force, Energy and Relative-Motion Methods
Force Method: not suited. No mass or force is given, only a distance-time plot.
Energy Method: not suited. No mass or work is given.
Relative-Motion Method: not suited. Another frame only shifts velocities by $u$ (see frame 2). It does not shorten the work.
Fastest: Method 1 (graph). Most reliable in an exam: Method 2 (calculus), which gives exact conditions. Both give $10\text{ cm/s}$, $25\text{ cm/s}$ and $16\text{ s}$.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|---|---|---|
| $s$ | total distance moved | m | |
| $t$ | time of motion (curve reaches $s$) | s | |
| $\langle v\rangle$ | mean velocity, $s/t$ | m/s (1 m/s = 100 cm/s) | |
| $v_{max}$ | greatest slope of the curve | m/s | |
| $t_0$ | time where $ds/dt=s/t_0$ | s | |
| $u$ | medium speed (Medium frame only) | cm/s |
(vii) Common Exam Traps
- Reading the graph in m but answering in cm/s. $2.0\text{ m}=200\text{ cm}$, so $\langle v\rangle=10\text{ cm/s}$, not $0.1\text{ cm/s}$.
- Using the whole $25\text{ s}$ on the axis for part (a). The motion ends at $20\text{ s}$.
- Finding $v_{max}$ from the highest point of the curve. The highest point is the greatest distance, not the greatest velocity.
- Drawing any line from the origin in part (c). It must just touch the curve at one point.
- Using the tangent at the steepest point for $t_0$. That tangent does not pass through the origin.
Verified: the live numbers, all frames and all methods agree on $10\text{ cm/s}$, $25\text{ cm/s}$ and $16\text{ s}$ (to graph-reading accuracy), with units in m, s and cm/s.