Exercises — Average velocity vs instantaneous velocity

WHAT we do: apply $v_{avg}=\Delta x/\Delta t$ directly (1-D, so signed numbers). $\Delta x=10-4=6\text{m},\Delta t=4-1=3\text{s}.$ WHY divide: we want position change per second — dividing "how far" by "how long" is the universal recipe for a rate. $v_{avg}=\frac{6}{3}=2\text{m/s (in the }+x\text{ direction, i.e. }+2\hat{x}).$

• (A) tangent slope $=$ instantaneous velocity — one point, no smearing.
• (B) chord (secant) slope $=$ average velocity — two points apart over a while. WHY the tangent is instantaneous: a tangent touches the curve at a single instant, so its steepness reports the velocity at that one moment with no other time mixed in. WHY the chord is average: a chord's steepness is total rise $\Delta x$ over total run $\Delta t$ between its two endpoints — that is precisely $\Delta x/\Delta t$, the average. This is the picture in the parent note: slide the two chord-points together and the secant becomes the tangent, so the average becomes the instantaneous.

WHAT it looks like: at the top the position–time graph has a flat tangent (a horizontal peak). A flat tangent has slope $0$, so $v=\frac{dx}{dt}=0\text{m/s}.$ The ball is momentarily still in velocity, even though gravity is still pulling (acceleration $\ne 0$).

WHAT we do: read the position at both ends, subtract to get displacement, then convert that displacement into a rate. $x(1)=2(1)+3(1)=2+3=5\text{m},x(3)=2(9)+3(3)=18+9=27\text{m}.$ $\Delta x=27-5=22\text{m},\Delta t=3-1=2\text{s}.$ WHY divide by $\Delta t$: $22$ m is only how far the car moved; to turn it into a velocity (metres travelled per second) we must divide by how long it took. Without that division the answer would be a displacement, not a rate. $v_{avg}=\frac{22}{2}=11\text{m/s}.$

WHY the power rule: $x$ is a sum of powers of $t$, and $\frac{d}{dt}(t^n)=n{t}^{n-1}$ turns each power into its slope-formula directly — faster than the limit. $v(t)=\frac{dx}{dt}=2(2t)+3=4t+3.$ $v(2)=4(2)+3=11\text{m/s}.$ Notice: the midpoint of $[1,3]$ is $t=2$, and $v(2)=v_{avg}$ from 2.1. That is the constant-acceleration midpoint theorem, not luck (see L4).

WHAT the displacement is: north then east are perpendicular, so the straight-line displacement (red arrow) is the hypotenuse of a right triangle with legs $60$ and $30$. $|\Delta \vec{r}|=\sqrt{60^2+30^2}=\sqrt{3600+900}=\sqrt{4500}=30\sqrt{5}\approx 67.08\text{m}.$ Total time $\Delta t=20+20=40$ s. (a) WHY divide the displacement by $\Delta t$: the red arrow's length $30\sqrt{5}$ m is only how far the drone effectively moved; to turn that net displacement into a velocity (net metres per second) we apply the rate recipe $\Delta \vec{r}/\Delta t$ — dividing "how far net" by "how long." $|{\vec{v}}_{avg}|=\frac{30\sqrt{5}}{40}=\frac{3\sqrt{5}}{4}\approx 1.68\text{m/s}.$ (Direction: along the red arrow, north-of-east.) (b) WHY divide the path length by $\Delta t$: average speed is a rate too, but of distance travelled, not net displacement. So we take the total path length and again divide by "how long" — same recipe, different numerator. Path length (a scalar, no direction) $=60+30=90$ m: $\text{avg speed}=\frac{90}{40}=2.25\text{m/s}.$ As promised, $|{\vec{v}}_{avg}|\approx 1.68\le 2.25=$ avg speed. See Distance vs displacement.

Instantaneous velocity $=$ slope of the graph (see Position-time graphs, Slope and tangent in calculus). WHY slope = velocity: slope is rise-over-run $=\Delta x/\Delta t$, and on a position–time graph "rise" is change in position and "run" is change in time — exactly the velocity recipe. So a steeper tilt means a faster velocity, and the direction of the tilt sets the sign.

• A rises ⇒ positive slope ⇒ $v>0$ (position increasing, so moving in $+x$).
• B flat ⇒ zero slope ⇒ $v=0$ (position not changing, so at rest).
• C falls ⇒ negative slope ⇒ $v<0$ (position decreasing, so moving back toward smaller $x$). Ranking: $v_A>v_B=0>v_C$.

Average velocity ignores the wiggles; it only sees start and end. $\Delta x=x(9)-x(0)=10-0=10\text{m},\Delta t=9-0=9\text{s}.$ WHY divide: the $10$ m is the net position change; the rate recipe $\Delta x/\Delta t$ turns it into net metres per second. $v_{avg}=\frac{10}{9}\approx 1.11\text{m/s}.$ Geometrically this is the slope of the single chord from the first point to the last — flatter than segment A, steeper (less negative) than segment C.

"At rest" means $v=0$. Differentiate: $v(t)=\frac{dx}{dt}=3t^2-12t+9=3(t^2-4t+3)=3(t-1)(t-3).$ Set to zero: $t=1$ s and $t=3$ s. WHAT it looks like: at both instants the position graph has a flat tangent (a local peak at $t=1$, a local valley at $t=3$) — the object stops and reverses direction. Between them $v<0$ (check: $v(2)=3(1)(-1)=-3<0$), so it moves backward there.

Instantaneous velocity: $v(t)=\frac{dx}{dt}=v_0+at$, so at the midpoint $v(t_m)=v_0+a\cdot \frac{t_1+t_2}{2}.$ Average velocity: $v_{avg}=\frac{x(t_2)-x(t_1)}{t_2-t_1}.$ Compute the numerator. Subtracting the two positions, the constant $x_0$ cancels (same at both times) and the linear and quadratic pieces group: $x(t_2)-x(t_1)=v_0(t_2-t_1)+\frac{1}{2}a(t_2^2-t_1^2).$ WHY factor next: we are about to divide by $(t_2-t_1)$, so we want that exact bracket to appear as a factor and cancel cleanly. The difference of squares gives it to us: $t_2^2-t_1^2=(t_2-t_1)(t_2+t_1).$ WHY this identity: $(t_2-t_1)(t_2+t_1)=t_2^2-t_1^2$ when you expand — the cross terms $-t_1{t}_2$ and $+t_2{t}_1$ cancel, leaving only the squares. Now divide every term by $(t_2-t_1)$; the bracket cancels wherever it appears: $v_{avg}=v_0+\frac{1}{2}a(t_2+t_1)=v_0+a\cdot \frac{t_1+t_2}{2}=v(t_m).■$ WHY it works intuitively: the velocity $v_0+at$ is a straight line in time, and the average of a straight line over an interval always lands on its midpoint value. See Equations of motion under uniform acceleration and Acceleration as derivative of velocity.

From the limit: we build the average velocity over a tiny window $[t,t+\Delta t]$ and then shrink the window. $x(t+\Delta t)=3(t+\Delta t{)}^2-2(t+\Delta t)=3t^2+6t\Delta t+3\Delta t^2-2t-2\Delta t.$ WHY subtract $x(t)$: the numerator of average velocity is the change in position, so we subtract the starting position $x(t)=3t^2-2t$; the constant-in-$\Delta t$ pieces ($3t^2$ and $-2t$) cancel: $\text{numerator}=6t\Delta t+3\Delta t^2-2\Delta t.$ WHY we can divide by $\Delta t$: every term contains a factor $\Delta t$, and while $\Delta t\ne 0$ (before the limit) dividing is legal: $\frac{6t\Delta t+3\Delta t^2-2\Delta t}{\Delta t}=6t+3\Delta t-2.$ WHY the limit finishes it: now let $\Delta t\to 0$ — the only surviving $\Delta t$ term, $3\Delta t$, vanishes: $v(t)=6t-2.✓$ Now set $v_{avg}[1,T]=v(T)$. $v_{avg}=\frac{x(T)-x(1)}{T-1}=\frac{(3T^2-2T)-(3-2)}{T-1}=\frac{3T^2-2T-1}{T-1}.$ WHY factor the top: we want to cancel the $(T-1)$ in the denominator. Factor $3T^2-2T-1=(T-1)(3T+1)$ (check by expanding: $3T^2+T-3T-1=3T^2-2T-1$ ✓): $v_{avg}=\frac{(T-1)(3T+1)}{T-1}=3T+1(T\ne 1).$ $v(T)=6T-2.$ Equate: $3T+1=6T-2\Rightarrow 3=3T\Rightarrow T=1.$ Comment: the algebra forces $T=1$, but that is the excluded endpoint (it makes $v_{avg}$ the indeterminate $0/0$), so there is no valid $T\ne 1$ where the average over $[1,T]$ equals the endpoint velocity $v(T)$. The clean, always-true match is instead the midpoint theorem of Exercise 4.1: for this quadratic $v_{avg}[1,T]=3T+1$ and $v$ at the midpoint $t=\frac{1+T}{2}$ is $6\cdot \frac{1+T}{2}-2=3T+1$ — an exact equality for every $T$. For example over $[1,3]$: $v_{avg}=3(3)+1=10$ and the midpoint $t=2$ gives $v(2)=6(2)-2=10$. ✓ Moral: average velocity matches instantaneous velocity at the midpoint, not the endpoint.

(a) $\frac{d}{dt}\sin (kt)=k\cos (kt)$, so with $k=\frac{\pi }{2}$: $v(t)=4\cdot \frac{\pi }{2}\cos \left(\frac{\pi }{2}t\right)=2\pi \cos \left(\frac{\pi }{2}t\right)\text{m/s}.$ (b) Over one period the particle returns to start: $x(4)=4\sin (2\pi )=0=x(0)$. So, applying the rate recipe with a zero net displacement, $v_{avg}=\frac{x(4)-x(0)}{4-0}=\frac{0-0}{4}=0\text{m/s}.$ (c) Track the path length. Starting at $x=0$, the particle swings out to $+4$ (a quarter period, $t:0\to 1$), back to $0$ ($t:1\to 2$), out to $-4$ ($t:2\to 3$), and back to $0$ ($t:3\to 4$). Each leg covers $4$ m of ground, so the total distance is $4+4+4+4=16$ m. WHY divide by $\Delta t$: average speed is distance travelled per second, so we take the path length and divide by the elapsed time: $\text{avg speed}=\frac{16}{4}=4\text{m/s}.$ Explaining the mismatch: average velocity is $0$ because it only sees the net displacement, and the round trip nets zero — the particle finishes exactly where it began. Average speed is $4$ m/s because it counts every metre actually travelled, which is far from zero. This is the round-trip lesson of the parent note, now with smooth (sinusoidal) motion instead of a straight walk. Consistent with the general rule, $|{\vec{v}}_{avg}|=0\le 4=$ avg speed, with equality only if the motion never reverses — which here it clearly does.

$v(t)=\frac{dx}{dt}=10t\text{m/s}\Rightarrow v(0)=0,v(2)=20\text{m/s}.$ $v_{avg}=\frac{x(2)-x(0)}{2-0}=\frac{5(4)-0}{2}=\frac{20}{2}=10\text{m/s}.$ (a) $v_{avg}=10$ sits exactly halfway between $0$ and $20$ — neither closer to start nor end. WHY: the motion is constant-acceleration, so velocity grows linearly; the average of a straight line is its midpoint value. (b) Set $v(t)=v_{avg}$: $10t=10\Rightarrow t=1$ s — the midpoint of $[0,2]$, confirming the theorem of Exercise 4.1.