Question bank — Equations of motion (SUVAT) — derivations from calculus

False as one equation — acceleration changes at cut-out, so $a$ isn't constant over the whole trip. You must split into two constant-$a$ phases and apply SUVAT to each separately.

False. $s$ is displacement; a ball thrown straight up returns to your hand with $s=0$ even though it travelled a large distance up and back.

False. Negative $a$ means acceleration points in the negative direction; whether that speeds up or slows down depends on the sign of the velocity. A ball falling downward with down chosen negative has negative $a$ but is speeding up.

True — solving gives $v=\pm \sqrt{u^2+2as}$. Both roots are physical if the object can be moving either way at that displacement (e.g. a ball passing a height on the way up and again on the way down).

False. They are just consequences of the two definitions $v=\frac{ds}{dt}$, $a=\frac{dv}{dt}$ integrated under constant $a$ — not new physics, only calculus bookkeeping.

False. That midpoint rule holds only when $v$ rises linearly, i.e. constant $a$. For varying $a$ the velocity–time graph curves and the average is the true integral, not the endpoint midpoint.

False. The omitted variable is simply the one not appearing in that particular relation; it still exists and can be recovered from another equation. "Missing $t$" means you can solve without knowing $t$, not that time stopped mattering.

False — that's the whole point. It omits $a$, so you can find displacement from the two velocities and time even when the acceleration value is unknown (as long as it was constant).

The formula is fine but the direction is wrong: if down is positive then $s$ is a downward (positive) fall. "Upward" contradicts the chosen sign convention. Always fix a positive direction before reading off signs.

Error: at the peak velocity is momentarily zero but acceleration is still $g$ downward the entire time. $a$ is the rate of change of $v$, and $v$ is changing fastest through zero, so $a\ne 0$.

Error: braking opposes the motion, so with motion positive the acceleration must be $a=-5$. Using $+5$ describes speeding up, which is why the algebra gave a nonsensical negative $s$.

Error: SUVAT requires constant $a$, not an average value. Using a mean acceleration in $s=ut+\frac{1}{2}a{t}^2$ silently assumes the linear-$v$ profile, which air resistance violates — the answer is only a rough estimate, not exact. See Free fall and g for when the approximation is safe; the figure there shows how the real $v$–$t$ curve bends away from the straight SUVAT line.

Error: SUVAT applies per-axis (Vectors). Horizontally $a=0$ so you use only the horizontal velocity component; the vertical component belongs to a separate equation with $a=-g$. See Projectile motion, where the velocity arrow is split into its two component sketches.

Error: in Simple Harmonic Motion acceleration is $a=-{\omega }^{2}x$, which changes with position. Constant-$a$ formulas don't apply; you must integrate the actual $a$ — its $v$–$t$ graph is a sine wave, not a straight line.

Because it rewrites acceleration as a derivative with respect to position instead of time, so integrating over $s$ never introduces $t$ — time is eliminated by construction.

If $a$ is constant it factors out of $\int adt$, giving $at$; if $a=a(t)$ it stays inside and the integral becomes whatever antiderivative $a(t)$ has — generally not a neat SUVAT term.

$ut$ is the displacement you'd get travelling at the fixed initial speed $u$; $\frac{1}{2}a{t}^2$ is the extra displacement the acceleration piles on. Splitting them shows constant-speed motion plus an acceleration correction.

The limits encode the initial condition ($v=u$ at $t=0$) directly, so the constant is already baked in — definite integration and "constant + solve for C" give the same result. See Differentiation and Integration.

Because $ds=vdt$, so summing tiny strips of height $v$ and width $dt$ (their areas) accumulates total displacement — that sum is the integral $\int vdt$. See Velocity-time graphs and the shaded strip in the figure below.

The $v$–$t$ line goes straight from height $u$ to height $v$, enclosing a trapezium of parallel sides $u,v$ and width $t$, whose area $\frac{1}{2}(u+v)t$ is exactly $s$. The figure shows this trapezium shaded.

With $a=0$: $v=u$ (constant velocity), $s=ut$, and $v^2=u^2$. Motion reduces to steady speed — this is the horizontal axis of a projectile.

All give the starting state: $v=u$, $s=0$, and $v^2=u^2$ trivially. A sanity check that $t=0$ means "no motion yet".

$v=at$, $s=\frac{1}{2}a{t}^2$, $v^2=2as$, and $s=\frac{1}{2}vt$. This is the classic dropped-stone / free-fall setup in Free fall and g.

Velocity $v=0$ at the peak, but acceleration $a=-g$ is unchanged and displacement $s$ is at its maximum (non-zero). Only $v$ vanishes; the "instant of rest" is not an instant of no acceleration.

Displacement $s=0$, so $v^2=u^2+0=u^2$, giving $v=-u$: same speed, opposite direction on return. The equation cleanly predicts the symmetric landing speed.

The object passes that displacement twice — e.g. a ball reaches a height once going up and again coming down. Both times are real events; choose the one matching the phase you care about.

It means the object never reaches that displacement — e.g. a ball can't rise higher than its energy allows, so $u^2+2as<0$ signals "this position is unreachable" rather than a computational slip.