$w_x$ is the slope of the $v_x$ graph: $+1,\,0,\,-1,\,+2,\,0$. The coordinate $x$ is the signed area: it peaks at $x=3$ when $t=4$, then ends at $x=0$ at $t=7$. The distance $s$ is the unsigned area: it ends at $s=6$ units.
(ii) Problem Statement
"A point travels along the $x$ axis with a velocity whose projection $v_x$ is presented as a function of time by the plot in the figure. Assuming the coordinate of the point $x=0$ at the moment $t=0$, draw the approximate time dependence plots for the acceleration $w_x$, the $x$ coordinate, and the distance covered $s$."
What this problem teaches: one velocity graph hides three more graphs. Slopes give acceleration. Areas give position. Areas counted as positive give distance.
(iii) The Big Idea in Plain Words
A point slides along a line. We are told how fast it goes at each moment. We must find how its speed changes, where it is, and how far it has walked in total.
Symbols: $t$ is time. $v_x$ is velocity along $x$ (negative means moving left). $w_x$ is acceleration, the rate at which $v_x$ changes. $x$ is the position. $s$ is the total path length.
The graph has five straight pieces, so each piece is easy:
| Piece | Time | $v_x$ | $w_x$ (slope) | $x$ at end | $s$ at end |
|---|---|---|---|---|---|
| oa | 0 to 1 | 0 → 1 | +1 | 1/2 | 1/2 |
| ab | 1 to 3 | stays 1 | 0 | 5/2 | 5/2 |
| b4 | 3 to 4 | 1 → 0 | −1 | 3 | 3 |
| 4d | 4 to 6 | 0 → −2 | −1 | 1 | 5 |
| d7 | 6 to 7 | −2 → 0 | +2 | 0 | 6 |
Shape of the plots: $w_x$ is flat steps with jumps. $x$ is smooth, with a peak at $t=4$. $s$ is smooth and never falls. After $t=7$, $w_x=0$, $x=0$ and $s=6$ stay constant.
(iv) The Solution Viewed in Each Frame
1. Ground (Lab) Frame (stationary observer)
Why: the graph was recorded by a still observer, so this is the natural frame.
Read the slopes: $w_x=+1,\,0,\,-1,\,+2,\,0$ on the five pieces.
Why: position is the signed area under the $v_x$ graph.
$x$: $\tfrac12+2+\tfrac12=3$ at $t=4$. Then $-2$ gives $1$ at $t=6$. Then $-1$ gives $0$ at $t=7$.
Why: distance ignores direction, so every area counts as positive.
$s=\tfrac12+2+\tfrac12+2+1=6$ units.
2. Centre-of-Mass Frame (origin rides on the point)
Why: for a single point, the centre of mass is the point itself.
In this frame $x'=0$ and $v'=0$ at all times, so the momentum about the centre of mass is zero trivially. The ground slides past with velocity $-v_x$.
Why: this frame accelerates, so a pseudo-force appears.
Any small object resting in it would feel $-m\,w_x$, pointing opposite to $w_x$. No such object is in the problem.
Why: undo the shift to compare with the ground.
The ground's displacement is $-x$, so the point's $x$ is its negative: $x(7)=0$. The ground's path length is the same: $s=6$.
3. Medium Frame
Does not apply. There is no water, air or other fluid in this problem. The view equals the ground frame.
- Changes: the picture. In the centre-of-mass frame the point is fixed, the ground moves with $-v_x$, and a pseudo-force $-m\,w_x$ appears.
- Same: the final answers. $x(7)=0$, $s(7)=6$, and the peak $x=3$ at $t=4$, in every frame.
(v) The Solution by Each Method
Method 1: Graph Method (areas and slopes)
- Why: the graph is given, so use its slopes and areas directly.
Slopes: $w_x=+1,0,-1,+2,0$. - Why: split the area into triangles and a rectangle.
Triangle 0–1: $\tfrac12$. Rectangle 1–3: $2$. Triangle 3–4: $\tfrac12$. Triangle 4–6 (below axis): $2$. Triangle 6–7 (below axis): $1$. - Why: signed sum gives $x$; unsigned sum gives $s$.
$x=\tfrac12+2+\tfrac12-2-1=0$. $s=\tfrac12+2+\tfrac12+2+1=6$.
Method 2: Calculus Method (the book's route)
- Why: acceleration is the derivative of velocity.
$w_x=\dfrac{dv_x}{dt}$ gives $+1,0,-1,+2,0$ on the five pieces. - Why: integrate $v_x$ piece by piece, starting each piece where the last ended.
0–1: $x=\tfrac{t^2}{2}$. 1–3: $x=t-\tfrac12$. 3–6: $v_x=4-t$, so $x=4t-\tfrac{t^2}{2}-5$ (gives $3$ at $t=4$, $1$ at $t=6$). 6–7: $v_x=2(t-7)$, so $x=(t-7)^2$, which is $1$ at $t=6$ and $0$ at $t=7$. - Why: for $s$, integrate $|v_x|$ so it only grows.
After $t=4$ use $|v_x|=t-4$ then $2(7-t)$. This gives $s(6)=5$ and $s(7)=6$.
Method 3: Force Method
Only partly suits this problem. No mass or forces are given, so no free-body diagram has numbers. We can still say $F=m\,w_x$, so the force is proportional to the slope. It cannot give $x$ or $s$.
Method 4: Energy Method
Does not suit this problem. Nothing is given about forces, heights or masses, and energy does not tell direction or position.
Method 5: Relative-Motion Method
The ground frame already gives the graph, so it is the shortest frame. One line: signed area $=0$, unsigned area $=6$. A moving frame would only add steps.
Fastest: Method 1 (just areas). Most reliable in an exam: Method 2 (every number comes from a formula). Both give $x(7)=0$ and $s(7)=6$.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|
(vii) Common Exam Traps
- Mixing up $x$ and $s$. The distance $s$ never decreases. The coordinate $x$ can.
- Counting area below the axis as positive in $x$. It must be negative. Only $s$ counts it as positive.
- Putting the peak of $x$ where $v_x$ is largest. The peak is where $v_x=0$, at $t=4$.
- Drawing $w_x$ as slanted lines. The $v_x$ pieces are straight, so $w_x$ is flat steps.
- Putting jumps in $x$ or $s$. Both are smooth. Only $w_x$ jumps, at $t=1,3,6,7$.
Verified: the live plots, both frames and all methods agree — $x(7)=0$, $s(7)=6$, peak $x=3$ at $t=4$ for the book values, with $v$ in units per time unit, $w$ in units per time unit squared.