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

True — a straight line has constant slope, and slope of $x$–$t$ is velocity, so velocity is unchanging (even if the line is flat, giving $v=0$).

False — a flat $x$–$t$ line has slope $0$, so $v=0$; the object is at rest, not cruising at constant speed.

True — flat $v$–$t$ means $v$ is unchanging in value, i.e. constant velocity (and its slope $0$ means $a=0$).

False — integrating position over time gives units of $\text{m}\cdot \text{s}$, which is physically meaningless; for $x$–$t$ you take the slope, not the area.

False — negative area means net displacement is in the $-x$ direction; the object certainly moved, just backward overall.

True — equal positive and negative areas cancel to give displacement $0$, but the sum of their magnitudes (the distance) can be large.

True — if $v\ge 0$ throughout, velocity never reverses, so no area is subtracted and net area equals total area.

False — the graph fixes only changes in position (areas); different starting positions shift the whole $x$–$t$ curve up or down.

True — speed is the magnitude of slope, so a steep line (up or down) means fast motion; only the direction differs.

False — $a=0$ means velocity is constant; that constant can be any nonzero value, so the object may be moving steadily.

The sign of $v$ is direction, not deceleration; whether it slows depends on the slope ($a$) versus $v$'s sign, not on $v$ being negative.

Wrong operation — velocity is the slope of $x$–$t$; area under $x$–$t$ has no physical meaning (units $\text{m}\cdot \text{s}$).

Height is the instantaneous acceleration; $\Delta v$ is the area under $a$–$t$, i.e. accumulated acceleration over time.

At the top $v=0$ but $a=-g$ throughout; the $v$–$t$ slope is unbroken through zero, so acceleration never vanishes.

Upward curvature means $a>0$; that speeds up motion only if $v$ is also $+$. If the object moves in $-x$ ($v<0$), $a>0$ actually slows it down.

For displacement you subtract (signed), but for distance you add its magnitude — distance ignores direction.

The slope gives acceleration $a$; the area of that trapezium gives displacement — slopes and areas are different operations.

Its slope gives $a$ and its area gives $\Delta x$ — both operations yield physically meaningful quantities, unlike $x$–$t$ (only slope) or $a$–$t$ (only area). See Differentiation and Integration in Physics.

Taking a slope moves you down the chain $x\to v\to a$ (differentiation), while accumulating area moves you back up $a\to v\to x$ (integration) — opposite operations, opposite directions.

Each thin strip is $vdt$, and $v$ carries a sign, so backward motion contributes negative strips; the signed sum is net displacement. See Distance vs Displacement.

With constant acceleration the $v$–$t$ graph is a straight sloped line, so the region under it between two times is a trapezium (average height $\times$ width). See Equations of Uniformly Accelerated Motion.

$v$'s sign points the motion; whether $|v|$ grows depends on whether $a$ pushes with ($v,a$ same sign) or against ($v,a$ opposite) the motion.

Curvature means the slope (velocity) is changing, and a changing velocity is exactly what acceleration measures.

Velocity is momentarily zero there (e.g. top of a throw), but if the slope is nonzero the object immediately reverses — it does not rest. See Free Fall and Projectile Motion.

Zero — the positive $v$–$t$ area (going up) exactly cancels the equal negative area (coming down back to the start point).

Twice the maximum height — you sum the magnitudes of the equal up and down areas.

Zero area means zero $\Delta v$; a genuine velocity jump needs finite area, so an infinitely thin spike changes nothing.

Undefined at that instant — the slope jumps, meaning velocity changes discontinuously, which requires infinite (idealised) acceleration. See Instantaneous vs Average Velocity.

It implies infinite slope, i.e. infinite velocity — physically impossible, so real $x$–$t$ graphs never go vertical.

They become equal in magnitude, because no area is subtracted; sign changes in $v$ are the only thing that makes them differ. See Vectors — Components and Signs.