Exercises — Graphs — x-t, v-t, a-t; areas and slopes meaning

WHAT we do: velocity is the slope of the $x$–$t$ graph — "rise over run." WHY: slope of $x$–$t$ $=\frac{dx}{dt}=v$ (slope slides down the ladder $x\to v$). WHAT IT LOOKS LIKE: a single straight line — the steepness never changes. $v=\frac{\Delta x}{\Delta t}=\frac{14-2}{4-0}=\frac{12}{4}=3\text{m/s}$ A straight line means slope is the same everywhere, so velocity is constant at $3$ m/s (in the $+x$ direction, since the slope is positive).

Read the slope in each case (slope of $x$–$t$ = velocity).

• (a) flat: slope $=0\Rightarrow v=0$ ⇒ object at rest (NOT constant velocity — see the trap below).
• (b) up: slope positive and constant ⇒ constant velocity in $+x$.
• (c) down: slope negative and constant ⇒ constant velocity in $-x$.
• (d) bending up: slope increasing ⇒ velocity increasing ⇒ $a>0$, speeding up in $+x$.

WHAT: displacement is the area under the $v$–$t$ graph. WHY: each thin strip is $dx=vdt$, a rectangle; add them all → area (area adds up the ladder $v\to x$). WHAT IT LOOKS LIKE: one big rectangle, height $8$, width $5$. $\Delta x=\text{area}=8\times 5=40\text{m}$

Acceleration = slope of $v$–$t$. WHY: acceleration is defined as how fast velocity changes, i.e. the "rise over run" of the $v$–$t$ graph taken over a tiny run: $a=\frac{\Delta v}{\Delta t}\to \frac{dv}{dt}$. Because the line here is straight, the slope is the same everywhere, so we can use the whole rise over the whole run: $a=\frac{\Delta v}{\Delta t}=\frac{16-4}{3-0}=\frac{12}{3}=4{\text{m/s}}^2$ Displacement = area under $v$–$t$. Here $u$ is the initial velocity (the height where the graph starts, $u=4$ m/s) and $v$ is the final velocity (where it ends, $v=16$ m/s). The area is a trapezium (defined above): average of the two parallel heights times the width. $\Delta x=\frac{u+v}{2}t=\frac{4+16}{2}\times 3=10\times 3=30\text{m}$ (Same as rectangle $4\times 3=12$ plus triangle $\frac{1}{2}\times 3\times 12=18$; $12+18=30$ ✓.)

(a) Accelerations = slopes of each straight segment. (Slope of $v$–$t$ $=\frac{dv}{dt}=a$.)

• Phase 1: $a_1=\frac{20-0}{4}=5{\text{m/s}}^2$.
• Phase 2: flat line ⇒ $a_2=0$.
• Phase 3: $a_3=\frac{0-20}{5}=-4{\text{m/s}}^2$ (negative = slowing, since $v>0$ but $a<0$).

(b) Displacement = total area (all above the axis here, so all positive).

• Phase 1 (triangle): $\frac{1}{2}\times 4\times 20=40$ m.
• Phase 2 (rectangle): $20\times 6=120$ m.
• Phase 3 (triangle): $\frac{1}{2}\times 5\times 20=50$ m.
• Total $=40+120+50=210$ m.

(c) Average velocity. WHY this formula: average velocity is defined as total displacement divided by total time — "if you had moved at one steady speed, what speed would cover the same displacement in the same time?" On the graph that displacement is the total area under $v$–$t$, and the time is the full width, so: $\bar{v}=\frac{\text{total displacement}}{\text{total time}}=\frac{\text{area under whole }v\text{–}t}{\text{total width}}=\frac{210}{4+6+5}=\frac{210}{15}=14\text{m/s}.$ See Instantaneous vs Average Velocity: average velocity is total-displacement-over-total-time, not the average of the phase speeds.

Split into signed areas. By our sign convention, region above the time axis is $+$ (motion in $+x$), below is $-$ (motion in $-x$).

• $0$–$2$ s (rectangle above): $+6\times 2=+12$ m.
• $2$–$4$ s (triangle above, $v$ from $+6$ to $0$): $+\frac{1}{2}\times 2\times 6=+6$ m.
• $4$–$6$ s (triangle below, $v$ from $0$ to $-6$): $-\frac{1}{2}\times 2\times 6=-6$ m.
• $6$–$8$ s (rectangle below): $-6\times 2=-12$ m.

(a) Displacement = net signed area: $\Delta x=+12+6-6-12=0\text{m}.$ The particle ends where it started. (b) Distance = sum of magnitudes (below-axis counted as positive): $\text{distance}=|12|+|6|+|6|+|12|=36\text{m}.$ See Distance vs Displacement: they differ exactly because the velocity changed sign.

Climb the ladder $a\to v$ using area under $a$–$t$ ($\Delta v=\int adt$).

• By $t=4$: $\Delta v=3\times 4=12\Rightarrow v(4)=2+12=14$ m/s.
• $4$–$7$: $a=0\Rightarrow$ no change $\Rightarrow v(7)=14$ m/s.
• $7$–$10$: $\Delta v=-2\times 3=-6\Rightarrow v(10)=14-6=8$ m/s.

Now climb $v\to x$ using area under $v$–$t$ (each phase is a trapezium; average of start and end heights times width).

• Phase 1 ($v:2\to 14$ over $4$ s): $\frac{2+14}{2}\times 4=8\times 4=32$ m.
• Phase 2 ($v=14$ constant, $3$ s): $14\times 3=42$ m.
• Phase 3 ($v:14\to 8$ over $3$ s): $\frac{14+8}{2}\times 3=11\times 3=33$ m.
• Total displacement $=32+42+33=107$ m.

All velocities stayed positive, so distance $=$ displacement $=107$ m. Uses Equations of Uniformly Accelerated Motion and Differentiation and Integration in Physics.

Acceleration = slope ($a=\frac{dv}{dt}$): $a=\frac{25-5}{8}=\frac{20}{8}=2.5$ m/s². Displacement = trapezium area: $\Delta x=\frac{u+v}{2}\times t=\frac{5+25}{2}\times 8=15\times 8=120$ m. Check with $v^2=u^2+2ax$ (itself derivable from the graph — see parent): $v^2=5^2+2(2.5)(120)=25+600=625=25^2✓$ The third kinematic equation is just the trapezium area written without $t$.

(a) Shape: $a=-10$ is constant, so $v$–$t$ is a straight line with slope $-10$, starting at $+20$ m/s. It crosses zero (the top) and continues into negative $v$ (falling in $-x$, i.e. downward). $v(t)=20-10t$ (b) Time to top = where $v=0$: $0=20-10t\Rightarrow t=2$ s. (At the top the ball is momentarily at rest — $v=0$, but $a$ is still $-10$; it never pauses in the sign of $a$.) (c) Total flight time: by symmetry the down-trip mirrors the up-trip, so total $=4$ s. Check: return to start means net displacement $0$, so solving $\Delta x=20t-\frac{1}{2}(10)t^2=0$ gives $t(20-5t)=0\Rightarrow t=4$ s ✓. (d) Max height = area under $v$–$t$ from $0$ to $2$ s (triangle above axis): $\frac{1}{2}\times 2\times 20=20$ m. (e) Net displacement = net signed area from $0$ to $4$ s: triangle above ($+20$ m) plus equal triangle below ($-20$ m) $=0$ m — back to the hand. Total distance $=|+20|+|-20|=40$ m $=2\times$ max height. See Free Fall and Projectile Motion and Vectors — Components and Signs for why "up $=+$" fixes every sign automatically.

(a) Displacement = area under the curved $v$–$t$ graph = integral: $\Delta x={\int }_0^3{t}^{2}dt={\left[\frac{t^3}{3}\right]}_0^3=\frac{27}{3}=9\text{m}.$ (b) WHY not a triangle/trapezium: those formulas assume the top edge is a straight line (constant acceleration). Here $a=\frac{dv}{dt}=2t$ changes with time, so the top edge curves. The only reliable "area" for a curved edge is the integral — the limit of adding infinitely many thin rectangles $vdt$ (this is exactly the area idea from the parent note, taken to its limit). See Differentiation and Integration in Physics. Sanity check by bounding: replace the curve by the straight chord from $(0,0)$ to $(3,9)$, giving a triangle of area $\frac{1}{2}\times 3\times 9=13.5$ m. Because the true curve $v=t^2$ lies below that chord across the interval, this straight-chord triangle over-estimates the real area. So the exact answer must be smaller than $13.5$ m — and indeed $9<13.5$ ✓. (There is no under-estimate here; the chord bounds the area from above.)