Displacement vs. Total Distance

1. Displacement: $30 + (-50) = -20$ miles

   The car ends up 20 miles south of where it started.

2. Total distance: $30 + 50 = 80$ miles

   The car traveled 80 miles total.

1. $$\text{Displacement} = \int_0^4 (3t + 2)\,dt = \left[\frac{3t^2}{2} + 2t\right]_0^4 = (24 + 8) - 0 = 32 \text{ m}$$
2. Since $v(t) = 3t + 2 > 0$ for all $t \in [0, 4]$, we have $\vert v(t)\vert = v(t)$. $$\text{Distance} = \int_0^4 \vert v(t)\vert \,dt = \int_0^4 (3t + 2)\,dt = 32 \text{ m}$$
3. The answers are the same because the particle never reverses direction. When $v(t) \geq 0$ throughout the interval, every bit of motion is in the positive direction. Displacement equals total distance.

Step 1: Factor: $v(t) = (t-1)(t-4)$, so $v(t) = 0$ at $t = 1, 4$.

Step 2: Sign analysis:

• $[0, 1]$: $v(0.5) = 1.75 > 0$ ✓
• $[1, 4]$: $v(2) = -2 < 0$ ✓
• $[4, 5]$: $v(4.5) = 1.75 > 0$ ✓

1. $$\text{Displacement} = \int_0^5 (t^2 - 5t + 4)\,dt = \left[\frac{t^3}{3} - \frac{5t^2}{2} + 4t\right]_0^5$$ $$= \frac{125}{3} - \frac{125}{2} + 20 = \frac{250 - 375 + 120}{6} = \frac{-5}{6} \text{ m}$$

   The particle ends up $\frac{5}{6}$ m to the left of its starting position.

2. $$\text{Distance} = \int_0^1 v(t)\,dt + \int_1^4 \vert v(t)\vert \,dt + \int_4^5 v(t)\,dt$$ $$= \int_0^1 v(t)\,dt - \int_1^4 v(t)\,dt + \int_4^5 v(t)\,dt$$

   Let $F(t) = \frac{t^3}{3} - \frac{5t^2}{2} + 4t$. Then:

   • $F(0) = 0$
   • $F(1) = \frac{1}{3} - \frac{5}{2} + 4 = \frac{2 - 15 + 24}{6} = \frac{11}{6}$
   • $F(4) = \frac{64}{3} - 40 + 16 = \frac{64 - 72}{3} = -\frac{8}{3}$
   • $F(5) = \frac{125}{3} - \frac{125}{2} + 20 = -\frac{5}{6}$

   $$\text{Distance} = [F(1) - F(0)] - [F(4) - F(1)] + [F(5) - F(4)]$$ $$= \frac{11}{6} - \left(-\frac{8}{3} - \frac{11}{6}\right) + \left(-\frac{5}{6} + \frac{8}{3}\right)$$ $$= \frac{11}{6} + \frac{16 + 11}{6} + \frac{-5 + 16}{6}$$ $$= \frac{11 + 27 + 11}{6} = \frac{49}{6} \approx 8.17 \text{ m}$$

1. $$v(t) = \int a(t)\,dt = \int (6t - 12)\,dt = 3t^2 - 12t + C$$

   Using $v(0) = 9$: $C = 9$

   $$v(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)$$

2. $v(t) = 0$ when $t = 1$ or $t = 3$.

   Sign analysis:

   • $[0, 1]$: $v(0.5) = 3(0.5)(-2.5) < 0$? Let me check: $v(0.5) = 3(0.25) - 6 + 9 = 0.75 + 3 = 3.75 > 0$
   • $[1, 3]$: $v(2) = 3(1)(-1) = -3 < 0$
   • $[3, 4]$: $v(3.5) = 3(2.5)(0.5) = 3.75 > 0$

   Let $F(t) = t^3 - 6t^2 + 9t$ (antiderivative of $v(t)$).

   • $F(0) = 0$
   • $F(1) = 1 - 6 + 9 = 4$
   • $F(3) = 27 - 54 + 27 = 0$
   • $F(4) = 64 - 96 + 36 = 4$

   $$\text{Distance} = \vert F(1) - F(0)\vert + \vert F(3) - F(1)\vert + \vert F(4) - F(3)\vert $$ $$= \vert 4 - 0\vert + \vert 0 - 4\vert + \vert 4 - 0\vert $$ $$= 4 + 4 + 4 = 12 \text{ m}$$

1. Proof: For any $v(t)$, we have $-\vert v(t)\vert \leq v(t) \leq \vert v(t)\vert $.

   Integrating all three parts from $a$ to $b$: $$-\int_a^b \vert v(t)\vert \,dt \leq \int_a^b v(t)\,dt \leq \int_a^b \vert v(t)\vert \,dt$$

   This is exactly the definition of: $$\left\vert \int_a^b v(t)\,dt\right\vert \leq \int_a^b \vert v(t)\vert \,dt$$

   Physical interpretation: The magnitude of net displacement can never exceed total distance traveled. You cannot end up farther from your starting point than the total ground you covered.

2. Equality holds if and only if $v(t)$ does not change sign on $[a, b]$.

   Proof of “if”: Suppose $v(t) \geq 0$ on $[a, b]$. Then $\vert v(t)\vert = v(t)$, so both integrals are identical.

   Similarly if $v(t) \leq 0$: then $\vert v(t)\vert = -v(t)$ and $\int_a^b v(t)\,dt \leq 0$, so: $$\left\vert \int_a^b v(t)\,dt\right\vert = -\int_a^b v(t)\,dt = \int_a^b (-v(t))\,dt = \int_a^b \vert v(t)\vert \,dt$$

   Proof of “only if” (contrapositive): Suppose $v(t)$ changes sign. Then there exist subintervals where $v(t) > 0$ and others where $v(t) < 0$. When computing $\int v(t)\,dt$, these partially cancel. When computing $\int \vert v(t)\vert \,dt$, they add. Since both positive and negative contributions exist, the total distance strictly exceeds $\vert $displacement$\vert $.

3. • Displacement: $\int_0^{2\pi} \sin(t)\,dt = [-\cos(t)]_0^{2\pi} = (-\cos(2\pi)) - (-\cos(0)) = -1 + 1 = 0$

     The displacement is zero. Over one full period, $\sin(t)$ spends equal “time” positive and negative, and by symmetry, the positive and negative areas cancel exactly.

   • Total distance vs |displacement|: Since displacement = 0 and total distance > 0 (the particle is moving), we have: $$\text{Total distance} > \vert \text{displacement}\vert = 0$$

     The inequality is strict. This is because $v(t) = \sin(t)$ changes sign at $t = \pi$, so by part (b), equality cannot hold.

     Verification: $\int_0^{2\pi} \vert \sin(t)\vert \,dt = \int_0^{\pi} \sin(t)\,dt + \int_{\pi}^{2\pi} (-\sin(t))\,dt = 2 + 2 = 4$