Question: A particle has velocity $v = t^2 - 5t + 4$ m/s. Find the total distance travelled in the first $5$ seconds.

Step 1. Find when $v = 0$: $t^2 - 5t + 4 = (t-1)(t-4) = 0 \Rightarrow t = 1, 4$.

Step 2. Sign of $v$:

• $0 \leq t < 1$: $v > 0$ (moving forward)
• $1 < t < 4$: $v < 0$ (moving backward)
• $4 < t < 5$: $v > 0$ (moving forward)

Step 3. Total distance = sum of |displacement| over each direction phase. Antiderivative:

Phase 1: $\left[\dfrac{t^3}{3} - \dfrac{5t^2}{2} + 4t\right]_0^1 = \dfrac{1}{3} - \dfrac{5}{2} + 4 = \dfrac{11}{6}$.

Phase 2: $\left[\dfrac{t^3}{3} - \dfrac{5t^2}{2} + 4t\right]_1^4 = \left(\dfrac{64}{3} - 40 + 16\right) - \dfrac{11}{6} = -\dfrac{8}{3} - \dfrac{11}{6} = -\dfrac{9}{2}$. Magnitude: $\dfrac{9}{2}$.

Phase 3: $\left[\cdots\right]_4^5 = \left(\dfrac{125}{3} - \dfrac{125}{2} + 20\right) - \left(\dfrac{64}{3} - 40 + 16\right) = \dfrac{125 - 64}{3} - \dfrac{125}{2} + 20 - (-24) = \dfrac{61}{3} - \dfrac{125}{2} + 44 \approx 0.83$. Use exact: $\dfrac{122 - 375 + 264}{6} = \dfrac{11}{6}$.

Step 4. Total: $\dfrac{11}{6} + \dfrac{9}{2} + \dfrac{11}{6} = \dfrac{11 + 27 + 11}{6} = \dfrac{49}{6}$ m.

Answer: Total distance $= \dfrac{49}{6} \approx 8.17$ m.

Tip: Total distance ≠ |total displacement| if the particle changes direction. Always split at zeros of $v$, integrate each phase, sum the magnitudes.

Question: A particle's velocity is $v = 10 e^{-0.4t} - 4$ m/s. Find (a) when it first comes to rest, (b) its limiting velocity as $t \to \infty$.

Step 1. Rest when $v = 0$:

Step 2. Limiting velocity as $t \to \infty$:

Answer: (a) Rest at $t \approx 2.29$ s. (b) Limiting velocity $= -4$ m/s (particle moves in negative direction asymptotically).

Tip: For exponential models, use $\ln$ to solve for $t$. For "limiting" or "asymptotic" velocity, take $t \to \infty$ — the exponential term decays to $0$.

Question: A particle moves in a straight line so that its displacement from a fixed point $O$, $t$ seconds after passing $O$, is $s = 4\ln(2t + 1)$ metres. Find expressions for the velocity $v$ and acceleration $a$, find $v$ when $t = 2$, and explain why the particle is never at rest for $t \ge 0$.

Step 1. Velocity is $\dfrac{ds}{dt}$ — chain rule on $\ln(2t+1)$ (derivative of the inside, $2$, over the inside):

Step 2. Acceleration is $\dfrac{dv}{dt}$ — write $v = 8(2t+1)^{-1}$ and differentiate:

Step 3. At $t = 2$:

Step 4. For $t \ge 0$, $2t + 1 \ge 1 > 0$, so $v = \dfrac{8}{2t+1} > 0$ for every $t$ — velocity is never zero, so the particle is never at rest. Since $a < 0$ throughout, $v$ is always decreasing: the particle keeps moving away from $O$ but more and more slowly, with $v \to 0$ as $t \to \infty$ without ever reaching it.

⚠ Watch out: $v > 0$ with $a < 0$ is not a contradiction — negative acceleration here means slowing down, not moving backwards. And "$v \to 0$" is not "$v = 0$": do not claim the particle eventually stops. "At rest" needs $v = 0$ at an actual value of $t$, and $\frac{8}{2t+1} = 0$ has no solution.

Question: A particle has velocity $v = 4\cos 3t + 2$ m/s. Find the total distance travelled in the first $\dfrac{\pi}{2}$ seconds.

Step 1. When is $v = 0$? $4\cos 3t = -2 \Rightarrow \cos 3t = -\dfrac{1}{2}$.

$3t = \dfrac{2\pi}{3}, \dfrac{4\pi}{3} \Rightarrow t = \dfrac{2\pi}{9}, \dfrac{4\pi}{9}$.

Both are within $[0, \pi/2]$ since $\pi/2 = \dfrac{4.5\pi}{9}$. So zeros at $t = \dfrac{2\pi}{9}, \dfrac{4\pi}{9}$.

Step 2. Sign of $v$:

• $[0, 2\pi/9)$: at $t=0$, $v = 4 + 2 = 6 > 0$. Forward.
• $(2\pi/9, 4\pi/9)$: at $t = \pi/3$, $v = 4\cos\pi + 2 = -2 < 0$. Backward.
• $(4\pi/9, \pi/2]$: at $t = \pi/2$, $v = 4\cos(3\pi/2) + 2 = 2 > 0$. Forward.

Step 3. Antiderivative: $s = \int v\,dt = \dfrac{4}{3}\sin 3t + 2t$.

Phase 1 displacement: $\left[\dfrac{4}{3}\sin 3t + 2t\right]_0^{2\pi/9} = \dfrac{4}{3}\sin\dfrac{2\pi}{3} + \dfrac{4\pi}{9} = \dfrac{4}{3} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{4\pi}{9} = \dfrac{2\sqrt{3}}{3} + \dfrac{4\pi}{9}$.

Phase 2: $\left[\cdots\right]_{2\pi/9}^{4\pi/9} = \dfrac{4}{3}\sin\dfrac{4\pi}{3} + \dfrac{8\pi}{9} - \dfrac{4}{3}\sin\dfrac{2\pi}{3} - \dfrac{4\pi}{9} = -\dfrac{2\sqrt{3}}{3} + \dfrac{4\pi}{9} - \dfrac{2\sqrt{3}}{3} = -\dfrac{4\sqrt{3}}{3} + \dfrac{4\pi}{9}$. Magnitude: $\dfrac{4\sqrt{3}}{3} - \dfrac{4\pi}{9}$ (since $\dfrac{4\sqrt{3}}{3} \approx 2.31 > \dfrac{4\pi}{9} \approx 1.40$).

Phase 3: $\left[\cdots\right]_{4\pi/9}^{\pi/2} = (\dfrac{4}{3}\sin\dfrac{3\pi}{2} + \pi) - (\dfrac{4}{3}\sin\dfrac{4\pi}{3} + \dfrac{8\pi}{9}) = -\dfrac{4}{3} + \pi + \dfrac{2\sqrt{3}}{3} - \dfrac{8\pi}{9} = \dfrac{2\sqrt{3}}{3} - \dfrac{4}{3} + \dfrac{\pi}{9}$.

Step 4. Sum magnitudes: total distance.

Tip: Trig motion problems: find $v = 0$ via the relevant inverse trig, identify all roots within the interval, split, integrate.

Question: Acceleration $a = 6t - 20$ m/s². Given $v(1) = 8$ m/s and $s(0) = -6$ m, find $s(t)$.

Step 1. Integrate $a$ for $v$:

Apply $v(1) = 8$: $8 = 3 - 20 + C_1 = -17 + C_1 \Rightarrow C_1 = 25$.

So $v(t) = 3t^2 - 20t + 25$.

Step 2. Integrate $v$ for $s$:

Apply $s(0) = -6$: $-6 = 0 + C_2 \Rightarrow C_2 = -6$.

Answer: $s(t) = t^3 - 10t^2 + 25t - 6$.

Tip: Two integrations from $a$ to $s$ need two initial conditions (one on $v$, one on $s$). Pin them in order.

Question: Velocity $v = pt - qt^2$. Maximum speed is $48$ cm/s at $t = 2$. Show that $p = 48$ and $q = 12$.

Step 1. Maximum velocity condition: $\dfrac{dv}{dt} = 0$ at $t = 2$.

At $t = 2$: $p - 4q = 0 \Rightarrow p = 4q$.

Step 2. Max velocity value at $t = 2$:

Substitute $p = 4q$: $2(4q) - 4q = 4q = 48 \Rightarrow q = 12$.

Then $p = 4 \times 12 = 48$. $\checkmark$

Tip: Maximum/minimum of a velocity model uses $\dfrac{dv}{dt} = 0$ (which is $a = 0$). Two unknown parameters need two conditions: usually "max at $t = t_0$" + "max value is $v_0$".

Question: Particle $A$ moves with velocity $v_A = 6t - t^2$ m/s. Particle $B$ has constant acceleration $a_B = 2$ m/s² and starts from rest. Find when their velocities are equal.

Step 1. $v_B$ from $a_B$:

$B$ starts from rest: $v_B(0) = 0 \Rightarrow C = 0$. So $v_B = 2t$.

Step 2. Set $v_A = v_B$:

Step 3. At $t = 0$: both have velocity $0$ (trivial). At $t = 4$: $v_A = 24 - 16 = 8$, $v_B = 8$. Match. $\checkmark$

Answer: Velocities are equal at $t = 4$ s.

Tip: Multi-particle problems often require integrating one particle's motion to match the other. "Velocity equal" $\Rightarrow v_A = v_B$; "they meet/collide" $\Rightarrow s_A = s_B$ (with same origin).

Question: A particle has velocity $v = t^2 - 4t$ m/s for $t \in [0, 7]$. Explain why total distance from $t = 0$ to $t = 7$ is NOT $|s(7)|$. Compute the average speed.

Step 1. $v = 0$ at $t = 0$ and $t = 4$. So the particle changes direction at $t = 4$:

• $[0, 4]$: $v < 0$ (moving backward)
• $[4, 7]$: $v > 0$ (moving forward)

Step 2. $s(t) = \int v\,dt = \dfrac{t^3}{3} - 2t^2 + C$. With $s(0) = 0$: $s(t) = \dfrac{t^3}{3} - 2t^2$.

$s(7) = \dfrac{343}{3} - 98 = \dfrac{343 - 294}{3} = \dfrac{49}{3} \approx 16.33$ m. This is the NET displacement, not total distance.

Step 3. Compute total distance:

• Phase 1: $|s(4) - s(0)| = \left|\dfrac{64}{3} - 32\right| = \left|-\dfrac{32}{3}\right| = \dfrac{32}{3}$ m.
• Phase 2: $|s(7) - s(4)| = \left|\dfrac{49}{3} - \left(-\dfrac{32}{3}\right)\right| = \dfrac{81}{3} = 27$ m.

Total distance $= \dfrac{32}{3} + 27 = \dfrac{113}{3} \approx 37.67$ m.

Step 4. Average speed $= \dfrac{\text{total distance}}{\text{total time}} = \dfrac{113/3}{7} = \dfrac{113}{21} \approx 5.38$ m/s.

Tip:

• Speed = magnitude of velocity (always positive).
• Average speed = total distance / total time.
• Average velocity = net displacement / total time.

These differ when the particle changes direction.

Question: A particle's displacement is $s = t^3 - 6t^2 + 9t$ m. Find (a) when it's at rest, (b) total distance travelled from $t = 0$ to $t = 4$.

Step 1. Velocity: $v = \dfrac{ds}{dt} = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)$.

Step 2. Rest when $v = 0$: $t = 1$ or $t = 3$.

Step 3. Sign of $v$:

• $[0, 1)$: $v > 0$ (forward)
• $(1, 3)$: $v < 0$ (backward)
• $(3, 4]$: $v > 0$ (forward)

Step 4. Compute displacements:

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

Step 5. Total distance:

• Phase 1: $|s(1) - s(0)| = 4$
• Phase 2: $|s(3) - s(1)| = |0 - 4| = 4$
• Phase 3: $|s(4) - s(3)| = |4 - 0| = 4$

Total distance $= 4 + 4 + 4 = 12$ m.

Answer: (a) Rest at $t = 1$ and $t = 3$. (b) Total distance $= 12$ m.

Tip: When $s(t)$ is given directly: differentiate for $v$, find zeros of $v$ for "rest" times. Distance over an interval where $v$ changes sign = sum of $|s(\text{end}) - s(\text{start})|$ for each direction phase.