Exercises — Average acceleration vs instantaneous acceleration

You have two endpoints and a clock — nothing about the middle. That is exactly the ingredients of the average. $a_{avg}=\frac{v_f-v_i}{\Delta t}=\frac{32-8}{6}=\frac{24}{6}=4{\text{m/s}}^2$ Instantaneous acceleration needs the slope at a point, which requires knowing $v$ as a function of time (a curve, not two dots) — so it is not computable here.

False. Acceleration is $\frac{d\vec{v}}{dt}$, and $\vec{v}$ is a vector — it carries direction. The speed (length of $\vec{v}$) is fixed, but the direction keeps turning, so $\vec{v}$ is constantly changing. A changing vector has a nonzero rate of change. See Uniform circular motion. The acceleration points toward the centre (centripetal); it is not zero.

Instantaneous acceleration is the derivative of velocity (Step 3 of the parent derivation). $a(t)=\frac{dv}{dt}=\frac{d}{dt}(4t^2-3t+1)=8t-3$ Why term by term: $\frac{d}{dt}(4t^2)=8t$, $\frac{d}{dt}(-3t)=-3$, $\frac{d}{dt}(1)=0$ (a constant does not change, so its rate is zero). $a(3)=8(3)-3=24-3=21{\text{m/s}}^2$

Average needs the two endpoint velocities, not derivatives. $v(1)=4(1)-3(1)+1=2\text{m/s}v(3)=4(9)-9+1=28\text{m/s}$ $a_{avg}=\frac{v(3)-v(1)}{3-1}=\frac{28-2}{2}=\frac{26}{2}=13{\text{m/s}}^2$ Compare: $a_{avg}=13\ne a(3)=21$. They differ because acceleration here is not constant ($a=8t-3$ grows with time), so the overall chord slope ($13$) sits below the steep tangent slope at the right end ($21$).

Use signed velocities — direction lives in the sign. $v_i=+12,v_f=-8\text{(west)}$ $a_{avg}=\frac{v_f-v_i}{\Delta t}=\frac{-8-(+12)}{5}=\frac{-20}{5}=-4{\text{m/s}}^2$ Magnitude $4{\text{m/s}}^2$, direction west (the minus sign = the $-$/east-negative direction). See Vectors: addition and subtraction for why $\Delta v$ uses final minus initial.

On a $v$–$t$ graph, acceleration = slope = (rise in velocity)/(run in time).

• Segment A, $0\to 2$ s: $v$ goes $0\to 8$. Slope $=\frac{8-0}{2-0}=4{\text{m/s}}^2$.
• Segment B, $2\to 5$ s: $v$ flat at $8$. Slope $=\frac{8-8}{5-2}=0{\text{m/s}}^2$ (cruising).
• Segment C, $5\to 8$ s: $v$ goes $8\to 2$. Slope $=\frac{2-8}{8-5}=\frac{-6}{3}=-2{\text{m/s}}^2$ (slowing). Whole-trip average uses only the two endpoints of the trip ($v(0)=0$, $v(8)=2$): $a_{avg}=\frac{2-0}{8-0}=0.25{\text{m/s}}^2$ The average ($0.25$) is a single chord slope from start-point to end-point — it ignores the bumps in between, exactly as the definition promises.

Instantaneous $a$ = tangent slope; on straight segments the tangent slope equals the segment slope.

• Magnitudes: $|+4|=4$ (A), $|0|=0$ (B), $|-2|=2$ (C).
• Largest magnitude: Segment A, $4{\text{m/s}}^2$.
• Zero: everywhere on Segment B ($2$–$5$ s), where the graph is flat.

Differentiate each component once to get velocity, again to get acceleration (parent Step 3, applied per-axis). $\vec{v}=\frac{d\vec{r}}{dt}=(2t)\hat{x}+(2)\hat{y}$ $\vec{a}=\frac{d\vec{v}}{dt}=(2)\hat{x}+(0)\hat{y}=2\hat{x}{\text{m/s}}^2$ Magnitude: $|\vec{a}|=\sqrt{2^2+0^2}=2{\text{m/s}}^2$, pointing purely east. Notice the acceleration is constant even though the speed keeps growing — the two ideas are independent.

Speed is unchanged ($10$ m/s both times), yet the direction flipped from north to east — so $\Delta \vec{v}$ is nonzero. Write the vectors with $\hat{x}$ = east, $\hat{y}$ = north: ${\vec{v}}_i=10\hat{y},{\vec{v}}_f=10\hat{x}$ $\Delta \vec{v}={\vec{v}}_f-{\vec{v}}_i=10\hat{x}-10\hat{y}$ This is the red arrow in the figure: to subtract, reverse ${\vec{v}}_i$ and add it to ${\vec{v}}_f$. $|\Delta \vec{v}|=\sqrt{10^2+(-10{)}^2}=\sqrt{200}=10\sqrt{2}\approx 14.14\text{m/s}$ $|{\vec{a}}_{avg}|=\frac{|\Delta \vec{v}|}{\Delta t}=\frac{10\sqrt{2}}{2}=5\sqrt{2}\approx 7.07{\text{m/s}}^2$ Direction: south-east (pointing along $\hat{x}-\hat{y}$, i.e. $45^{\circ }$ below east). Constant speed, real acceleration — the vector nature is doing all the work.

(a) Differentiate twice. $v=\frac{dx}{dt}=3t^2-18t+24,a=\frac{dv}{dt}=6t-18$ Set $a=0$: $6t-18=0\Rightarrow t=3$ s. (b) At $t=3$: $v(3)=3(9)-18(3)+24=27-54+24=-3$ m/s (moving in $-x$). A body speeds up when $v$ and $a$ have the same sign. Just before $t=3$, $a=6t-18<0$; and $v(3)=-3<0$. Same sign ($-,-$) → speeding up as it approaches $t=3$; exactly at $t=3$, $a=0$ is the instant the acceleration switches sign, the turning point of the speed trend.

Let $a$ be constant. Then velocity grows linearly: $v(t)=v_i+at$ (this is what "constant rate" means — see Derivatives as rates of change). Average over $[t_1,t_2]$: $a_{avg}=\frac{v(t_2)-v(t_1)}{t_2-t_1}=\frac{(v_i+a{t}_2)-(v_i+a{t}_1)}{t_2-t_1}=\frac{a(t_2-t_1)}{t_2-t_1}=a$ Instantaneous at any $t$: $\frac{dv}{dt}=\frac{d}{dt}(v_i+at)=a$. Both equal the same constant $a$, independent of which interval or point you pick. $■$ This is precisely why the equation $v=v_i+at$ (from the equations of motion) is allowed to use a single $a$ for the whole interval — there the average and instantaneous values coincide.

Ex 1.1 average acceleration ::: $4{\text{m/s}}^2$ Ex 2.1 $a(3)$ for $v=4t^2-3t+1$ ::: $21{\text{m/s}}^2$ Ex 2.2 $a_{avg}$ on $[1,3]$ ::: $13{\text{m/s}}^2$ Ex 2.3 average acceleration ::: $-4{\text{m/s}}^2$ (west) Ex 3.1 whole-trip average ::: $0.25{\text{m/s}}^2$ Ex 4.1 acceleration magnitude ::: $2{\text{m/s}}^2$ (east) Ex 4.2 average acceleration magnitude ::: $5\sqrt{2}\approx 7.07{\text{m/s}}^2$ (south-east) Ex 5.1 time acceleration is zero ::: $t=3$ s