Kinematics-1D

(a) Distance travelled by Ram = 100 m
Distance travelled by Shyam = $\pi (50\text{ m}) = 50\pi\text{ m}$
(b) Displacement of Ram = 100 m
Displacement of Shyam = 100 m
Step Solution:
Step 1: Identify the path taken by Ram. Ram moves along path 1, which is a straight line of length 100 m from point P to point Q. Therefore, the distance travelled by Ram is 100 m.
Step 2: Identify the path taken by Shyam. Shyam moves along path 2, which is a semicircular arc with a diameter of 100 m (radius $r = 50\text{ m}$). The length of a semicircular arc is given by $\pi r$. Thus, the distance travelled by Shyam is $\pi \times 50\text{ m} = 50\pi\text{ m}$.
Step 3: Determine the displacement for both Ram and Shyam. Displacement is defined as the shortest straight-line distance from the initial position P to the final position Q. Since both Ram and Shyam start at point P and end at point Q, their displacement is identical and equal to the straight-line distance between P and Q, which is 100 m.

Since $t = \sqrt{x} + 3$
$\Rightarrow x = (t - 3)^2$
So, we get $x$ as a function of $t$. Let us draw a table for $x$ and $t$ from $t = 0$ to $t = 6$. Then we get
$t = 0 \Rightarrow x = 9$
$t = 1 \Rightarrow x = 4$
$t = 2 \Rightarrow x = 1$
$t = 3 \Rightarrow x = 0$
$t = 4 \Rightarrow x = 1$
$t = 5 \Rightarrow x = 4$
$t = 6 \Rightarrow x = 9$
This table gives us the answers to the three parts.
So, for (a), we have $\Delta x_1 = -9\text{ m}$
For (b), we have $\Delta x_2 = +9\text{ m}$
For (c), distance travelled $= \vert{}-9\vert{} + \vert{}+9\vert{} = 18\text{ m}$ and displacement $= \Delta x_1 + \Delta x_2 = -9 + 9 = 0\text{ m}$
Step Solution:
Step 1: Express position $x$ as an explicit function of time $t$. Rearranging $t = \sqrt{x} + 3$ gives $\sqrt{x} = t - 3$, which yields $x(t) = (t - 3)^2$.
Step 2: Calculate the positions at key time intervals: at $t = 0\text{ s}$, $x(0) = (0 - 3)^2 = 9\text{ m}$; at $t = 3\text{ s}$, $x(3) = (3 - 3)^2 = 0\text{ m}$; and at $t = 6\text{ s}$, $x(6) = (6 - 3)^2 = 9\text{ m}$.
Step 3: Determine displacement from $t = 0$ to $t = 3\text{ s}$. $\Delta x_1 = x(3) - x(0) = 0 - 9 = -9\text{ m}$.
Step 4: Determine displacement from $t = 3$ to $t = 6\text{ s}$. $\Delta x_2 = x(6) - x(3) = 9 - 0 = +9\text{ m}$.
Step 5: Determine total displacement and total distance from $t = 0$ to $t = 6\text{ s}$. Total displacement is $\Delta x = x(6) - x(0) = 9 - 9 = 0\text{ m}$. Total distance is the sum of magnitudes of individual path segments: $\vert{}\Delta x_1\vert{} + \vert{}\Delta x_2\vert{} = \vert{}-9\vert{} + \vert{}+9\vert{} = 18\text{ m}$.

Average Speed = $\frac{\text{Total Distance Travelled}}{\text{Total Time Taken}}$
CASE-1: $v_{av} = \frac{2 v_1 v_2}{v_1 + v_2} = \frac{(2)(40)(60)}{40 + 60} = 48\text{ kmh}^{-1}$
Average Velocity = $\vec{v}_{av} = \vec{0}$
{$\because$ train returns to its station}
Step Solution:
Step 1: Identify the given speeds for the two equal distance legs of the journey: $v_1 = 40\text{ kmh}^{-1}$ and $v_2 = 60\text{ kmh}^{-1}$.
Step 2: Apply the harmonic mean formula for average speed when distances are equal: $v_{av} = \frac{2 v_1 v_2}{v_1 + v_2} = \frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48\text{ kmh}^{-1}$.
Step 3: Determine the average velocity by evaluating total displacement. Because the train finishes its journey at the initial station, the total displacement is $0$, making the average velocity $0\text{ kmh}^{-1}$.

$v_{av} = \frac{v_1 + v_2}{2} = \frac{1 + 3}{2} = 2\text{ ms}^{-1}$
Average Velocity = $2\text{ ms}^{-1}$
Step Solution:
Step 1: Note the given speeds $v_1 = 1\text{ ms}^{-1}$ and $v_2 = 3\text{ ms}^{-1}$ and equal time intervals $t_1 = t_2 = 1\text{ min} = 60\text{ s}$.
Step 2: Calculate average speed using the formula for equal time intervals: $v_{av} = \frac{v_1 + v_2}{2} = \frac{1 + 3}{2} = 2\text{ ms}^{-1}$.
Step 3: Since the motion takes place continuously along a straight path without reversing direction, total displacement equals total distance. Hence, average velocity is also $2\text{ ms}^{-1}$.

Total Distance $= 720 + 4 \times (5 \times 60) + 1 \times (3 \times 60) = 720 + 1200 + 180 = 2100\text{ m}$
Total Time $= \frac{720}{2} + (5 \times 60) + (3 \times 60) = 360 + 300 + 180 = 840\text{ s}$
$v_{av} = \frac{2100}{840} = 2.5\text{ ms}^{-1}$
Average Velocity = $2.5\text{ ms}^{-1}$
Step Solution:
Step 1: Calculate distance and time for each segment.
Segment 1: $d_1 = 720\text{ m}$, $v_1 = 2\text{ ms}^{-1} \Rightarrow t_1 = \frac{720}{2} = 360\text{ s}$.
Segment 2: $v_2 = 4\text{ ms}^{-1}$, $t_2 = 5\text{ min} = 300\text{ s} \Rightarrow d_2 = 4 \times 300 = 1200\text{ m}$.
Segment 3: $v_3 = 1\text{ ms}^{-1}$, $t_3 = 3\text{ min} = 180\text{ s} \Rightarrow d_3 = 1 \times 180 = 180\text{ m}$.
Step 2: Find total distance travelled: $D = 720 + 1200 + 180 = 2100\text{ m}$.
Step 3: Find total time elapsed: $T = 360 + 300 + 180 = 840\text{ s}$.
Step 4: Compute average speed: $v_{av} = \frac{2100}{840} = 2.5\text{ ms}^{-1}$.
Step 5: As the motion is unidirectional in a straight line, total displacement equals total distance, so average velocity is $2.5\text{ ms}^{-1}$.

Let us first draw a pictorial representation to the problem.
$s_1 = \frac{l}{3}$, $t_1 = \frac{l}{3 v_0}$
$s_2 = v_1 \frac{t_2}{2} + v_2 \frac{t_2}{2} = \frac{2l}{3}$
$\Rightarrow t_2 = \frac{4l}{3(v_1 + v_2)}$
$v_{av} = \frac{l}{t_1 + t_2} = \frac{l}{\frac{l}{3 v_0} + \frac{4l}{3(v_1 + v_2)}}$
$\Rightarrow v_{av} = \frac{3 v_0 (v_1 + v_2)}{4 v_0 + v_1 + v_2}$
Step Solution:
Step 1: Let the total distance of the journey be $l$. The first one-third of the distance is $s_1 = \frac{l}{3}$, which is covered with constant velocity $v_0$. The time taken for this segment is $t_1 = \frac{s_1}{v_0} = \frac{l}{3 v_0}$.
Step 2: The remaining distance is $s_2 = \frac{2l}{3}$. Let $t_2$ be the total time spent covering this remaining portion. For the first half of this time interval ($\frac{t_2}{2}$), the particle moves with velocity $v_1$, and for the second half ($\frac{t_2}{2}$), it moves with velocity $v_2$.
Step 3: Express the distance $s_2$ in terms of $t_2$: $s_2 = v_1 \left(\frac{t_2}{2}\right) + v_2 \left(\frac{t_2}{2}\right) = \frac{(v_1 + v_2) t_2}{2}$. Equating this to $\frac{2l}{3}$, we get $\frac{(v_1 + v_2) t_2}{2} = \frac{2l}{3} \Rightarrow t_2 = \frac{4l}{3(v_1 + v_2)}$.
Step 4: Determine the total time of motion: $t_{total} = t_1 + t_2 = \frac{l}{3 v_0} + \frac{4l}{3(v_1 + v_2)} = \frac{l}{3} \left[ \frac{1}{v_0} + \frac{4}{v_1 + v_2} \right] = \frac{l}{3} \left[ \frac{v_1 + v_2 + 4 v_0}{v_0 (v_1 + v_2)} \right]$.
Step 5: Compute the mean velocity averaged over the whole time: $v_{av} = \frac{l}{t_{total}} = \frac{l}{\frac{l (4 v_0 + v_1 + v_2)}{3 v_0 (v_1 + v_2)}} = \frac{3 v_0 (v_1 + v_2)}{4 v_0 + v_1 + v_2}$.

$x = 8t^2 - 3t^3$ ...(1)
$\Rightarrow v = \frac{dx}{dt} = 16t - 9t^2$ ...(2)
$\Rightarrow a = \frac{dv}{dt} = 16 - 18t$ ...(3)
(a) At $t = 1\text{ s}$, $v = 16(1) - 9(1)^2 = 7\text{ ms}^{-1}$
(b) Body is at rest when $v = 0$
$\Rightarrow 16t - 9t^2 = 0 \Rightarrow t(16 - 9t) = 0$
$\Rightarrow t = 0\text{ s}$ or $t = \frac{16}{9}\text{ s}$
Position at $t = \frac{16}{9}\text{ s}$:
$x = 8\left(\frac{16}{9}\right)^2 - 3\left(\frac{16}{9}\right)^3 = \frac{2048}{243}\text{ m}$
(c) Acceleration at $t = 4\text{ s}$:
$a = 16 - 18(4) = 16 - 72 = -56\text{ ms}^{-2}$
Step Solution:
Step 1: Differentiate position $x(t) = 8t^2 - 3t^3$ to obtain velocity $v(t) = 16t - 9t^2$.
Step 2: Differentiate velocity $v(t)$ to obtain acceleration $a(t) = 16 - 18t$.
Step 3: Evaluate instantaneous velocity at $t = 1\text{ s}$: $v(1) = 16(1) - 9(1)^2 = 7\text{ ms}^{-1}$.
Step 4: Find time of rest by setting $v(t) = 0$, giving $t = \frac{16}{9}\text{ s}$. Substitute $t = \frac{16}{9}\text{ s}$ into $x(t)$ to get position $x = \frac{2048}{243}\text{ m}$.
Step 5: Evaluate acceleration at $t = 4\text{ s}$: $a(4) = 16 - 18(4) = -56\text{ ms}^{-2}$.

At $t = 0$, $x_0 = 0$
At $t = 4\text{ s}$, $x_4 = 8(4)^2 - 3(4)^3 = 128 - 192 = -64\text{ m}$
Displacement $\Delta x = x_4 - x_0 = -64 - 0 = -64\text{ m}$
Average velocity $= \frac{-64}{4} = -16\text{ ms}^{-1}$
Body reverses direction at $t = \frac{16}{9}\text{ s}$ where $x = \frac{2048}{243}\text{ m}$.
Distance from $t = 0$ to $t = \frac{16}{9}\text{ s}$ is $d_1 = \frac{2048}{243}\text{ m}$.
Distance from $t = \frac{16}{9}\text{ s}$ to $t = 4\text{ s}$ is $d_2 = \left\vert{}-64 - \frac{2048}{243}\right\vert{} = \frac{17600}{243}\text{ m}$.
Total distance $= d_1 + d_2 = \frac{2048}{243} + \frac{17600}{243} = \frac{19648}{243}\text{ m}$.
Average speed $= \frac{\frac{19648}{243}}{4} = \frac{4912}{243} = \frac{544}{27}\text{ ms}^{-1}$
Step Solution:
Step 1: Calculate position at initial time $t = 0\text{ s}$, $x(0) = 0\text{ m}$, and final time $t = 4\text{ s}$, $x(4) = 8(4)^2 - 3(4)^3 = -64\text{ m}$.
Step 2: Find total displacement $\Delta x = x(4) - x(0) = -64\text{ m}$. Compute average velocity $= \frac{\Delta x}{\Delta t} = \frac{-64}{4} = -16\text{ ms}^{-1}$.
Step 3: Determine turning point by setting velocity $v = 16t - 9t^2 = 0$, which yields $t = \frac{16}{9}\text{ s}$ (within $0 \le t \le 4$).
Step 4: Find positions at key times: $x(0) = 0\text{ m}$, $x\left(\frac{16}{9}\right) = \frac{2048}{243}\text{ m}$, and $x(4) = -64\text{ m}$.
Step 5: Compute distance for first segment $d_1 = \frac{2048}{243}\text{ m}$ and second segment $d_2 = \left\vert{}-64 - \frac{2048}{243}\right\vert{} = \frac{17600}{243}\text{ m}$. Sum to get total distance $= \frac{19648}{243}\text{ m}$.
Step 6: Compute average speed $= \frac{\text{Total Distance}}{\Delta t} = \frac{\frac{19648}{243}}{4} = \frac{544}{27}\text{ ms}^{-1}$.

(a) $v = 12 - 3t^2$ ...(1)
$\Rightarrow a = \left.\frac{dv}{dt} = -6t\right\vert{}_{t=4} = -24\text{ ms}^{-2}$
(b) Further $v = \frac{dx}{dt}$
$\Rightarrow \int_{-10}^{x} dx = \int_{1}^{t} v \, dt = \int_{1}^{t} (12 - 3t^2) \, dt$
$\Rightarrow x - (-10) = \left[ 12t - t^3 \right]_1^t$
$\Rightarrow x + 10 = 12t - t^3 - (12 - 1)$
$\Rightarrow x = 12t - t^3 - 21$ ...(2)
Displacement from $t = 0$ to $t = 10\text{ s}$:
At $t = 0\text{ s}$, $x(0) = -21\text{ m}$
At $t = 10\text{ s}$, $x(10) = 12(10) - (10)^3 - 21 = 120 - 1000 - 21 = -901\text{ m}$
Displacement $= x(10) - x(0) = -901 - (-21) = -880\text{ m}$
(c) For distance, $v = 0 \Rightarrow 12 - 3t^2 = 0 \Rightarrow t = 2\text{ s}$
Position at $t = 2\text{ s}$: $x(2) = 12(2) - (2)^3 - 21 = 24 - 8 - 21 = -5\text{ m}$
Distance from $t = 0$ to $t = 2\text{ s}$ is $d_1 = \vert{}x(2) - x(0)\vert{} = \vert{}-5 - (-21)\vert{} = 16\text{ m}$
Distance from $t = 2$ to $t = 10\text{ s}$ is $d_2 = \vert{}x(10) - x(2)\vert{} = \vert{}-901 - (-5)\vert{} = 896\text{ m}$
Total distance $= d_1 + d_2 = 16 + 896 = 912\text{ m}$
Step Solution:
Step 1: Calculate acceleration by differentiating velocity $v(t) = 12 - 3t^2$ with respect to time $t$. This gives $a(t) = \frac{dv}{dt} = -6t$. Evaluating at $t = 4\text{ s}$ gives $a(4) = -6(4) = -24\text{ ms}^{-2}$.
Step 2: Determine position function $x(t)$ using the given initial condition $x(1) = -10\text{ m}$. Integrate $\int_{-10}^{x} dx = \int_{1}^{t} (12 - 3t^2) \, dt$, which yields $x(t) = 12t - t^3 - 21$.
Step 3: Calculate position at $t = 0\text{ s}$ to be $x(0) = -21\text{ m}$ and position at $t = 10\text{ s}$ to be $x(10) = 12(10) - (10)^3 - 21 = -901\text{ m}$.
Step 4: Compute displacement from $t = 0$ to $t = 10\text{ s}$: $\Delta x = x(10) - x(0) = -901 - (-21) = -880\text{ m}$.
Step 5: Find the turning point of motion by setting velocity $v(t) = 0 \Rightarrow 12 - 3t^2 = 0 \Rightarrow t = 2\text{ s}$ (since $t \ge 0$). Position at $t = 2\text{ s}$ is $x(2) = 12(2) - (2)^3 - 21 = -5\text{ m}$.
Step 6: Compute total distance by summing absolute displacements of the two sub-intervals: $d_1 = \vert{}x(2) - x(0)\vert{} = \vert{}-5 - (-21)\vert{} = 16\text{ m}$ and $d_2 = \vert{}x(10) - x(2)\vert{} = \vert{}-901 - (-5)\vert{} = 896\text{ m}$. Total distance $= 16\text{ m} + 896\text{ m} = 912\text{ m}$.

(a) Let $v_m$ be the velocity at $C$, the midpoint of $A$ and $B$.
Using $v^2 = u^2 + 2as$:
For path $AC$: $v_m^2 - u^2 = 2a\left(\frac{l}{2}\right) = al$
For path $CB$: $v^2 - v_m^2 = 2a\left(\frac{l}{2}\right) = al$
Equating both equations:
$v_m^2 - u^2 = v^2 - v_m^2$
$2v_m^2 = u^2 + v^2$
$v_m = \sqrt{\frac{u^2 + v^2}{2}}$
(b) Let $t_1$ be the time taken from $A$ to $C$ and $t_2$ be the time taken from $C$ to $B$. Given $t_1 = 2t_2$.
Displacement for path $AC$: $\frac{l}{2} = \left(\frac{u + v_m}{2}\right) 2t_2$
Displacement for path $CB$: $\frac{l}{2} = \left(\frac{v_m + v}{2}\right) t_2$
Equating both expressions gives $u + v_m = \frac{v_m + v}{2} \implies v_m = v - 2u$.
Squaring both sides gives $v_m^2 = v^2 + 4u^2 - 4uv$.
Substituting $v_m^2 = \frac{u^2 + v^2}{2}$:
$\frac{u^2 + v^2}{2} = v^2 + 4u^2 - 4uv$
$u^2 + v^2 = 2v^2 + 8u^2 - 8uv$
$v^2 - 8uv + 7u^2 = 0$
$(v - u)(v - 7u) = 0$
Since motion is accelerated, $v - u \neq 0$, so $v = 7u \implies \frac{v}{u} = 7$.
Step Solution:
Step 1: Calculate the velocity $v_m$ at the midpoint $C$.
Let $l$ be the distance $AB$. Then $AC = CB = \frac{l}{2}$.
Using $v^2 = u^2 + 2as$:
For path $AC$: $v_m^2 = u^2 + 2a\left(\frac{l}{2}\right) \implies v_m^2 - u^2 = al$
For path $CB$: $v^2 = v_m^2 + 2a\left(\frac{l}{2}\right) \implies v^2 - v_m^2 = al$
Equating the two expressions for $al$:
$v_m^2 - u^2 = v^2 - v_m^2$
$2v_m^2 = u^2 + v^2$
$v_m = \sqrt{\frac{u^2 + v^2}{2}}$
Step 2: Relate the time taken for segments $AC$ and $CB$.
Let time for $CB$ be $t$. Since time for $AC$ is twice as much, time for $AC$ is $2t$.
Using $s = \left(\frac{u_{\text{initial}} + u_{\text{final}}}{2}\right) t$:
For $AC$: $\frac{l}{2} = \left(\frac{u + v_m}{2}\right) 2t = (u + v_m)t$
For $CB$: $\frac{l}{2} = \left(\frac{v_m + v}{2}\right) t$
Step 3: Solve for $v_m$ in terms of $u$ and $v$.
Equating the displacement equations:
$(u + v_m)t = \left(\frac{v_m + v}{2}\right) t$
$u + v_m = \frac{v_m + v}{2}$
$2u + 2v_m = v_m + v$
$v_m = v - 2u$
Step 4: Determine the ratio $\frac{v}{u}$.
Square $v_m = v - 2u$:
$v_m^2 = (v - 2u)^2 = v^2 + 4u^2 - 4uv$
Substitute $v_m^2 = \frac{u^2 + v^2}{2}$:
$\frac{u^2 + v^2}{2} = v^2 + 4u^2 - 4uv$
$u^2 + v^2 = 2v^2 + 8u^2 - 8uv$
$v^2 - 8uv + 7u^2 = 0$
$(v - 7u)(v - u) = 0$
As motion is accelerated, $v \neq u$, hence $v - 7u = 0 \implies v = 7u \implies \frac{v}{u} = 7$.

For $OA(= l)$:
$(50)^2 - (10)^2 = 2al \implies al = 1200$
Now, when the particle just attains the velocity of $50\text{ ms}^{-1}$ at $A$, acceleration is reversed in direction i.e., it starts acting as retardation for $50\text{ ms}^{-1}$ which will first be reduced to zero after travelling a distance $AB$ and then the particle will again be accelerated from $B$ to $A$ with the same acceleration. So, velocity of the particle again at $A$ will be $50\text{ ms}^{-1}$, but in opposite direction.
Now, for path $AO$:
$v^2 = (-50)^2 + 2a(l)$
$v^2 = 2500 + 2(1200) = 2500 + 2400 = 4900$
$v = \sqrt{4900} = 70\text{ ms}^{-1}$
Since the direction is towards the origin $O$ (opposite to initial motion), the velocity of the particle as it reaches the starting point is $-70\text{ ms}^{-1}$.
Step Solution:
Step 1: Apply equation of motion for the initial segment $OA$ of distance $l$.
Using $v^2 = u^2 + 2al$:
$(50)^2 = (10)^2 + 2al$
$2500 = 100 + 2al$
$2al = 2400 \implies al = 1200$
Step 2: Analyze the motion after point $A$.
When acceleration reverses direction at point $A$, the particle slows down to rest at point $B$ and accelerates back towards $A$ under acceleration of magnitude $a$.
By symmetry of motion between $A$ and $B$, the magnitude of velocity of the particle when it returns to point $A$ heading towards $O$ is $50\text{ ms}^{-1}$ directed towards $O$.
Step 3: Apply equation of motion for the journey from $A$ back to origin $O$.
Initial speed at $A$ heading left is $50\text{ ms}^{-1}$.
During motion from $A$ to $O$, the acceleration $a$ acts in the same direction of motion (towards $O$).
Using $v^2 = u^2 + 2al$:
$v^2 = (50)^2 + 2al$
$v^2 = 2500 + 2(1200) = 2500 + 2400 = 4900$
$v = 70\text{ ms}^{-1}$
Step 4: Determine final velocity taking direction into account.
Since the particle is moving in the direction opposite to its initial motion (towards the left), its velocity upon reaching the starting point $O$ is $-70\text{ ms}^{-1}$.

Let the particle $P$ reach $C$ (the midpoint of $AB$) in time $t$, then the particle $Q$ must reach $C$ also in time $t$.
Let $AB = l$, so $AC = CB = \frac{l}{2}$.
For particle $P$: $\frac{l}{2} = u_1 t + \frac{1}{2} a_1 t^2$
For particle $Q$: $\frac{l}{2} = u_2 t + \frac{1}{2} a_2 t^2$
Equating both expressions for $\frac{l}{2}$:
$u_1 t + \frac{1}{2} a_1 t^2 = u_2 t + \frac{1}{2} a_2 t^2$
$\frac{1}{2}(a_1 - a_2) t^2 = (u_2 - u_1) t \implies t = 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right)$
Substituting $t$ back into displacement equation:
$\frac{l}{2} = u_1 \left[ 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \right] + \frac{1}{2} a_1 \left[ 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \right]^2$
$l = \frac{4(u_2 - u_1)}{(a_1 - a_2)^2} (a_1 u_2 - a_2 u_1)$
Also, $P$ reaches $B$ and $Q$ reaches $A$ with equal velocities $v$.
For $P$: $v^2 - u_1^2 = 2 a_1 l$
For $Q$: $v^2 - u_2^2 = 2 a_2 l$
Subtracting the two equations gives $u_2^2 - u_1^2 = 2(a_1 - a_2)l$.
Substituting the expression for $l$:
$(u_2 - u_1)(u_2 + u_1) = 2(a_1 - a_2) \left[ \frac{4(u_2 - u_1)}{(a_1 - a_2)^2} (a_1 u_2 - a_2 u_1) \right]$
$(u_1 + u_2)(a_1 - a_2) = 8(a_1 u_2 - a_2 u_1)$
Step Solution:
Step 1: Set up the displacement equations for both particles to reach midpoint $C$.
Let length of $AB = l$. Midpoint distance $AC = CB = \frac{l}{2}$.
Since $P$ and $Q$ meet at midpoint $C$, time taken by both is $t$.
For particle $P$: $\frac{l}{2} = u_1 t + \frac{1}{2} a_1 t^2$
For particle $Q$: $\frac{l}{2} = u_2 t + \frac{1}{2} a_2 t^2$
Step 2: Solve for time $t$ in terms of initial velocities and accelerations.
$u_1 t + \frac{1}{2} a_1 t^2 = u_2 t + \frac{1}{2} a_2 t^2$
$\frac{1}{2} (a_1 - a_2) t^2 = (u_2 - u_1) t$
$t = 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right)$
Step 3: Express total length $l$ in terms of $u_1, u_2, a_1, a_2$.
Substitute $t$ into $\frac{l}{2} = u_1 t + \frac{1}{2} a_1 t^2$:
$\frac{1}{2} l = u_1 \left[ 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \right] + \frac{1}{2} a_1 \left[ 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \right]^2$
$\frac{1}{2} l = 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \left[ u_1 + a_1 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \right]$
$l = \frac{4(u_2 - u_1)}{(a_1 - a_2)^2} (a_1 u_2 - a_2 u_1)$
Step 4: Use third equation of motion for final velocities at the opposite ends.
Both particles reach the opposite ends with equal velocity $v$.
For $P$ reaching $B$: $v^2 - u_1^2 = 2 a_1 l$
For $Q$ reaching $A$: $v^2 - u_2^2 = 2 a_2 l$
Subtracting the two equations:
$u_2^2 - u_1^2 = 2(a_1 - a_2)l$
Step 5: Combine equations to obtain the final relationship.
Substitute $l$ into $u_2^2 - u_1^2 = 2(a_1 - a_2)l$:
$(u_2 - u_1)(u_1 + u_2) = 2(a_1 - a_2) \left[ \frac{4(u_2 - u_1)}{(a_1 - a_2)^2} (a_1 u_2 - a_2 u_1) \right]$
$(u_2 - u_1)(u_1 + u_2) = \frac{8(u_2 - u_1)}{a_1 - a_2} (a_1 u_2 - a_2 u_1)$
$(u_1 + u_2)(a_1 - a_2) = 8(a_1 u_2 - a_2 u_1)$

Let the man catch the bus at time $t$.
For bus:
$x = x_0 + ut + \frac{1}{2}at^2$
$\implies x = 40 + 0(t) + \frac{1}{2}(1)t^2$
$\implies x = 40 + \frac{t^2}{2}$
For man:
$x = 9t$
Equating both expressions for position:
$40 + \frac{t^2}{2} = 9t$
$\implies t^2 - 18t + 80 = 0$
$\implies t^2 - 10t - 8t + 80 = 0$
$\implies (t - 8)(t - 10) = 0$
$\implies t = 8\text{ s}$ or $t = 10\text{ s}$
Step Solution:
Step 1: Set up the equation of motion for the bus taking the man's initial position as origin $x = 0$.
At $t = 0$, position of bus is $x_0 = 40\text{ m}$ and its initial speed $u = 0\text{ ms}^{-1}$ with acceleration $a = 1\text{ ms}^{-2}$.
$x_{\text{bus}} = x_0 + ut + \frac{1}{2}at^2 = 40 + 0 + \frac{1}{2}(1)t^2 = 40 + \frac{t^2}{2}$
Step 2: Set up the equation of motion for the man moving with constant speed $v = 9\text{ ms}^{-1}$.
$x_{\text{man}} = 9t$
Step 3: Equate positions of the man and the bus to solve for time $t$.
$x_{\text{man}} = x_{\text{bus}}$
$9t = 40 + \frac{t^2}{2}$
$18t = 80 + t^2$
$t^2 - 18t + 80 = 0$
Step 4: Factorize and solve the quadratic equation for $t$.
$t^2 - 10t - 8t + 80 = 0$
$t(t - 10) - 8(t - 10) = 0$
$(t - 8)(t - 10) = 0$
$t = 8\text{ s}$ or $t = 10\text{ s}$

Furthermore, the average speed
$v_{av} = 72\text{ kmh}^{-1} = 20\text{ ms}^{-1}$
$\implies v_{av} = \frac{\text{Total Distance Travelled}}{\text{Total Time Taken}} = \frac{l_1 + l_2 + l_1}{25}$
$\implies 2l_1 + l_2 = (20)(25) = 500$
$\implies 2l_1 + l_2 = 500$
where $l_1 = \frac{5}{2} t_1^2$
Also, for $OA$, we have
$v = 5t_1$
For $AB$, we have
$l_2 = v t_2 = 5 t_1 t_2$
$\implies 2\left(\frac{5}{2} t_1^2\right) + 5 t_1 t_2 = 500$ {substituting in (2)}
$\implies 5 t_1^2 + 5 t_1 (25 - 2t_1) = 500$
$\implies 5 t_1^2 + 125 t_1 - 10 t_1^2 = 500$
$\implies 5 t_1^2 - 125 t_1 + 500 = 0$
$\implies t_1^2 - 25 t_1 + 100 = 0$
$\implies (t_1 - 5)(t_1 - 20) = 0$
$\implies t_1 = 5\text{ s}$ (since $t_1 = 20\text{ s}$ makes $t_2 < 0$)
$\implies t_2 = 25 - 2(5) = 15\text{ s}$
Step Solution:
Step 1: Convert average speed to $\text{ms}^{-1}$ and express total distance.
$v_{av} = 72 \times \frac{5}{18} = 20\text{ ms}^{-1}$
Total distance $S = v_{av} \times t_{\text{total}} = 20 \times 25 = 500\text{ m}$.
Step 2: Express distance traveled in terms of time intervals $t_1$ and $t_2$.
Let $t_1$ be acceleration time, $t_2$ be uniform motion time, and $t_1$ be deceleration time.
$2t_1 + t_2 = 25 \implies t_2 = 25 - 2t_1$
Distance during acceleration: $l_1 = \frac{1}{2} (5) t_1^2 = 2.5 t_1^2$
Maximum speed reached: $v = a t_1 = 5 t_1$
Distance during uniform motion: $l_2 = v t_2 = (5 t_1) t_2$
Total distance $2l_1 + l_2 = 500 \implies 2(2.5 t_1^2) + 5 t_1 t_2 = 500 \implies 5 t_1^2 + 5 t_1 t_2 = 500$.
Step 3: Solve for $t_1$ using substitution.
Substitute $t_2 = 25 - 2t_1$:
$5 t_1^2 + 5 t_1 (25 - 2t_1) = 500$
$5 t_1^2 + 125 t_1 - 10 t_1^2 = 500$
$-5 t_1^2 + 125 t_1 - 500 = 0$
$t_1^2 - 25 t_1 + 100 = 0$
$(t_1 - 5)(t_1 - 20) = 0$
Selecting $t_1 = 5\text{ s}$ (as total time is $25\text{ s}$, $t_1$ cannot be $20\text{ s}$).
Step 4: Calculate duration of uniform motion $t_2$.
$t_2 = 25 - 2(5) = 15\text{ s}$.

Suppose the pickpocket is caught at a time $t$ after motorcycle starts. The distance travelled by the motorcycle during this interval is
$s = \frac{1}{2}at^2$
During this interval the car travels a distance
$s + d = vt$
$\frac{1}{2}at^2 + d = vt$
$at^2 - 2vt + 2d = 0$
$t = \frac{2v \pm \sqrt{4v^2 - 8ad}}{2a}$
$\implies t = \frac{v \pm \sqrt{v^2 - 2ad}}{a}$
For $t$ to be real:
$v^2 - 2ad \ge 0$
$\implies v \ge \sqrt{2ad}$
Step Solution:
Step 1: Formulate the distance equations for both the motorcycle and the police car.
Distance traveled by the motorcycle starting from rest with constant acceleration $a$ in time $t$:
$s = \frac{1}{2}at^2$
Distance traveled by the police car moving at uniform speed $v$ in time $t$:
$s_{\text{car}} = vt$
Step 2: Relate the two distances based on the initial separation $d$.
$s + d = vt$
$\frac{1}{2}at^2 + d = vt$
Step 3: Rearrange into a quadratic equation in terms of time $t$.
$\frac{1}{2}at^2 - vt + d = 0$
$at^2 - 2vt + 2d = 0$
Step 4: Determine the condition for a real solution of time $t$.
Using the quadratic formula, the discriminant must be greater than or equal to zero:
$D = (-2v)^2 - 4(a)(2d) \ge 0$
$4v^2 - 8ad \ge 0$
$v^2 \ge 2ad$
$v \ge \sqrt{2ad}$

During the reaction time of $0.20\text{ s}$, the car continues to move with a speed of $54\text{ kmh}^{-1}$ i.e. $15\text{ ms}^{-1}$.
So, distance travelled by car during this time is
$s_1 = (15)(0.2) = 3\text{ m}$
Now, when brakes are applied, the distance $s_2$ travelled by the car is
$0^2 - (15)^2 = 2(-6)s_2$
$\implies s_2 = \frac{225}{12}\text{ m} = 18.75\text{ m}$
Total distance travelled by the car is $s = s_1 + s_2 = 3 + 18.75 = 21.75\text{ m}$.
Step Solution:
Step 1: Convert speed from $\text{kmh}^{-1}$ to $\text{ms}^{-1}$.
$v = 54 \times \frac{5}{18} = 15\text{ ms}^{-1}$
Step 2: Calculate distance covered during the reaction time.
$s_1 = v \times t_{\text{reaction}} = 15 \times 0.20 = 3\text{ m}$
Step 3: Calculate braking distance.
Using $v_f^2 - v^2 = 2 a s_2$:
$0^2 - 15^2 = 2(-6)s_2$
$-225 = -12 s_2$
$s_2 = \frac{225}{12} = 18.75\text{ m}$
Step 4: Compute total stopping distance.
$s_{\text{total}} = s_1 + s_2 = 3 + 18.75 = 21.75\text{ m}$

Integrating $a = (2t - 1)$ gives $v = t^2 - t + C_1$; applying $v(0) = 2$ yields $v = t^2 - t + 2$.
Integrating $v$ gives $x = \frac{t^3}{3} - \frac{t^2}{2} + 2t + C_2$; applying $x(0) = 1$ yields $x = \frac{t^3}{3} - \frac{t^2}{2} + 2t + 1$.
At $t = 6 \, s$: $v = 36 - 6 + 2 = 32 \, ms^{-1}$ and $x = 72 - 18 + 12 + 1 = 67 \, m$.
Since $v = t^2 - t + 2$ has a negative discriminant $(1 - 8 = -7)$, the velocity is always positive and the particle never reverses direction.
Hence the total distance travelled equals the displacement: $s = 67 - 1 = 66 \, m$.
Step Solution:
Step 1: Write the acceleration relation $a = \frac{dv}{dt} = 2t - 1$ and integrate: $v = \int (2t - 1) dt = t^2 - t + C_1$.
Step 2: Apply the initial condition $v = 2 \, ms^{-1}$ at $t = 0$: $C_1 = 2$, so $v = t^2 - t + 2$.
Step 3: Evaluate velocity at $t = 6 \, s$: $v = 6^2 - 6 + 2 = 36 - 6 + 2 = 32 \, ms^{-1}$.
Step 4: Write $v = \frac{dx}{dt} = t^2 - t + 2$ and integrate: $x = \int (t^2 - t + 2) dt = \frac{t^3}{3} - \frac{t^2}{2} + 2t + C_2$.
Step 5: Apply the initial condition $x = 1 \, m$ at $t = 0$: $C_2 = 1$, so $x = \frac{t^3}{3} - \frac{t^2}{2} + 2t + 1$.
Step 6: Evaluate position at $t = 6 \, s$: $x = \frac{216}{3} - \frac{36}{2} + 12 + 1 = 72 - 18 + 12 + 1 = 67 \, m$.
Step 7: Check for reversal of motion: set $v = t^2 - t + 2 = 0$; the discriminant is $(-1)^2 - 4(1)(2) = -7 < 0$, so $v > 0$ for all t and the particle always moves forward.
Step 8: Compute total distance: $s = x(6) - x(0) = 67 - 1 = 66 \, m$.
Final result: $v = 32 \, ms^{-1}$, $x = 67 \, m$, total distance $s = 66 \, m$.

Initial velocity $u = 0$
Acceleration $= a$
Displacement $= s$
Final velocity $v = \sqrt{u^2 + 2as} = \sqrt{2as}$
Time taken $t = \sqrt{\frac{2s}{a}}$
Average velocity $v_{av} = \frac{s}{t} = \frac{s}{\sqrt{\frac{2s}{a}}} = \sqrt{\frac{as}{2}}$
Step Solution:
Step 1: Identify the given values for the motion of the particle: initial velocity $u = 0$, uniform acceleration $a$, and total displacement $s$.
Step 2: Use the third equation of motion $v^2 = u^2 + 2as$ to find the final velocity after covering distance $s$: $v = \sqrt{0 + 2as} = \sqrt{2as}$.
Step 3: Calculate the time $t$ required to cover displacement $s$ using the second equation of motion $s = ut + \frac{1}{2}at^2$. Since $u = 0$, $s = \frac{1}{2}at^2 \Rightarrow t = \sqrt{\frac{2s}{a}}$.
Step 4: Compute the average velocity $v_{av}$ using the definition $v_{av} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{s}{t}$.
Step 5: Substitute $t = \sqrt{\frac{2s}{a}}$ into the expression: $v_{av} = \frac{s}{\sqrt{\frac{2s}{a}}} = \sqrt{\frac{s^2 \cdot a}{2s}} = \sqrt{\frac{as}{2}}$.

Magnitude of displacement $= AB = 2R = 2(1\text{ m}) = 2\text{ m}$
Time taken $\Delta t = 1\text{ s}$
Magnitude of average velocity $= \frac{\text{Magnitude of displacement}}{\text{Time taken}} = \frac{2}{1} = 2\text{ ms}^{-1}$
Step Solution:
Step 1: Identify the initial position A and final position B of the particle moving along the semicircular path of radius $R = 1\text{ m}$.
Step 2: Determine the magnitude of displacement, which is the straight-line distance across the diameter connecting point A and point B: $\vert{}\Delta \vec{r}\vert{} = 2R = 2 \times 1\text{ m} = 2\text{ m}$.
Step 3: Identify the total time taken to traverse the semicircular path, which is given as $\Delta t = 1\text{ s}$.
Step 4: Compute the magnitude of the average velocity using the formula $\vert{}\vec{v}_{av}\vert{} = \frac{\vert{}\Delta \vec{r}\vert{}}{\Delta t} = \frac{2\text{ m}}{1\text{ s}} = 2\text{ ms}^{-1}$.

Differentiating $x = 12 - 15t^2 + 5t^3$ gives $v = -30t + 15t^2 = 15t(t - 2)$, which vanishes at $t = 2$ s inside the interval, so the particle reverses direction there.
Positions: $x(1) = 2$ m, $x(2) = -8$ m, $x(3) = 12$ m.
Distance from 1 s to 2 s is $|-8 - 2| = 10$ m and from 2 s to 3 s is $|12 - (-8)| = 20$ m, giving total distance $= 10 + 20 = 30$ m.
Average speed $= \frac{30}{3 - 1} = 15 \, ms^{-1}$.
Step Solution:
Step 1: Differentiate position to get velocity: $v = \frac{dx}{dt} = -30t + 15t^2 = 15t(t - 2)$.
Step 2: Set $v = 0$: roots are $t = 0$ s and $t = 2$ s; only $t = 2$ s lies in the interval [1, 3], so the particle reverses at $t = 2$ s (v is negative for 1 < t < 2 and positive for 2 < t < 3).
Step 3: Compute position at $t = 1$ s: $x(1) = 12 - 15(1)^2 + 5(1)^3 = 12 - 15 + 5 = 2$ m.
Step 4: Compute position at $t = 2$ s: $x(2) = 12 - 15(4) + 5(8) = 12 - 60 + 40 = -8$ m.
Step 5: Compute position at $t = 3$ s: $x(3) = 12 - 15(9) + 5(27) = 12 - 135 + 135 = 12$ m.
Step 6: Distance in first segment: $s_1 = |x(2) - x(1)| = |-8 - 2| = 10$ m.
Step 7: Distance in second segment: $s_2 = |x(3) - x(2)| = |12 - (-8)| = 20$ m.
Step 8: Total distance: $s = s_1 + s_2 = 10 + 20 = 30$ m.
Step 9: Average speed: $v_{avg} = \frac{s}{\Delta t} = \frac{30}{3 - 1} = \frac{30}{2} = 15 \, ms^{-1}$.
Final result: total distance = 30 m, average speed = 15 $ms^{-1}$.

Given $x = 3t^2 - 2t^3$
(a) $[ct^2] = [x] \Rightarrow c = \text{m s}^{-2}$
$[bt^3] = [x] \Rightarrow b = \text{m s}^{-3}$
(b) $v = \frac{dx}{dt} = 6t - 6t^2$
For maximum positive $x$, $v = 0 \Rightarrow 6t(1 - t) = 0 \Rightarrow t = 1\text{ s}$
Step Solution:
Step 1: Determine the units of $c$ and $b$ using dimensional analysis. Since $x = ct^2 - bt^3$ has units of metres ($\text{m}$) and $t$ has units of seconds ($\text{s}$), $c \cdot \text{s}^2 = \text{m} \Rightarrow c$ has units of $\text{ms}^{-2}$, and $b \cdot \text{s}^3 = \text{m} \Rightarrow b$ has units of $\text{ms}^{-3}$.
Step 2: Differentiate position $x(t) = 3t^2 - 2t^3$ to find velocity $v(t) = \frac{dx}{dt} = 6t - 6t^2$.
Step 3: Find the time for maximum positive position by setting velocity $v(t) = 0 \Rightarrow 6t(1 - t) = 0$. Since $t > 0$, the maximum positive $x$ value is reached at $t = 1\text{ s}$.

Given $x = 3t^2 - 2t^3$
(c) At $t = 0$, $x(0) = 0\text{ m}$
At $t = 1\text{ s}$, $x(1) = 3(1)^2 - 2(1)^3 = 1\text{ m}$
At $t = 4\text{ s}$, $x(4) = 3(4)^2 - 2(4)^3 = 48 - 128 = -80\text{ m}$
Displacement from $t = 0$ to $t = 4\text{ s}$ $= x(4) - x(0) = -80 - 0 = -80\text{ m}$
Distance travelled $= \vert{}x(1) - x(0)\vert{} + \vert{}x(4) - x(1)\vert{} = \vert{}1 - 0\vert{} + \vert{}-80 - 1\vert{} = 1 + 81 = 82\text{ m}$
(d) Velocity $v(t) = 6t - 6t^2$
$v(0) = 0\text{ ms}^{-1}$
$v(1) = 6(1) - 6(1)^2 = 0\text{ ms}^{-1}$
$v(2) = 6(2) - 6(4) = -12\text{ ms}^{-1}$
$v(3) = 6(3) - 6(9) = -36\text{ ms}^{-1}$
$v(4) = 6(4) - 6(16) = -72\text{ ms}^{-1}$
Acceleration $a(t) = \frac{dv}{dt} = 6 - 12t$
At $t = 0$, $a(0) = 6\text{ ms}^{-2}$
At $t = 1$, $a(1) = -6\text{ ms}^{-2}$
At $t = 2$, $a(2) = -18\text{ ms}^{-2}$
At $t = 3$, $a(3) = -30\text{ ms}^{-2}$
At $t = 4$, $a(4) = -42\text{ ms}^{-2}$
Step Solution:
Step 1: Compute positions at key time instants: $x(0) = 0\text{ m}$, $x(1) = 1\text{ m}$ (turning point), and $x(4) = 3(16) - 2(64) = -80\text{ m}$.
Step 2: Calculate displacement from $t = 0$ to $t = 4\text{ s}$: $\Delta x = x(4) - x(0) = -80 - 0 = -80\text{ m}$.
Step 3: Calculate total distance travelled by summing segment lengths: $d_1 = \vert{}x(1) - x(0)\vert{} = \vert{}1 - 0\vert{} = 1\text{ m}$ and $d_2 = \vert{}x(4) - x(1)\vert{} = \vert{}-80 - 1\vert{} = 81\text{ m}$. Total distance $= 1 + 81 = 82\text{ m}$.
Step 4: Differentiate position $x(t) = 3t^2 - 2t^3$ to obtain velocity $v(t) = 6t - 6t^2$. Evaluate velocity at specified seconds: $v(0) = 0\text{ ms}^{-1}$, $v(1) = 0\text{ ms}^{-1}$, $v(2) = -12\text{ ms}^{-1}$, $v(3) = -36\text{ ms}^{-1}$, $v(4) = -72\text{ ms}^{-1}$.
Step 5: Differentiate velocity $v(t)$ to obtain acceleration $a(t) = 6 - 12t$. Evaluate acceleration at specified seconds: $a(0) = 6\text{ ms}^{-2}$, $a(1) = -6\text{ ms}^{-2}$, $a(2) = -18\text{ ms}^{-2}$, $a(3) = -30\text{ ms}^{-2}$, $a(4) = -42\text{ ms}^{-2}$.

$x = t^3 - 9t^2 + 15t$
$v = \frac{dx}{dt} = 3t^2 - 18t + 15$
$a = \frac{dv}{dt} = 6t - 18$
For maximum velocity, $\frac{dv}{dt} = a = 0 \Rightarrow 6t - 18 = 0 \Rightarrow t = 3\text{ s}$
At $t = 3\text{ s}$, $v(3) = 3(3)^2 - 18(3) + 15 = 27 - 54 + 15 = -12\text{ ms}^{-1}$ (local minimum)
Boundary values for velocity:
At $t = 0\text{ s}$, $v(0) = 15\text{ ms}^{-1}$
At $t = 10\text{ s}$, $v(10) = 3(10)^2 - 18(10) + 15 = 300 - 180 + 15 = 135\text{ ms}^{-1}$
Maximum velocity $= 135\text{ ms}^{-1}$ at $t = 10\text{ s}$
For acceleration, $a(t) = 6t - 18$ is a linear increasing function of $t$.
Maximum acceleration occurs at boundary $t = 10\text{ s}$:
$a(10) = 6(10) - 18 = 60 - 18 = 42\text{ ms}^{-2}$
Step Solution:
Step 1: Differentiate the position function $x(t) = t^3 - 9t^2 + 15t$ to obtain the velocity function: $v(t) = \frac{dx}{dt} = 3t^2 - 18t + 15$.
Step 2: Differentiate velocity $v(t)$ to obtain the acceleration function: $a(t) = \frac{dv}{dt} = 6t - 18$.
Step 3: Evaluate velocity at the critical point $t = 3\text{ s}$ (where $a = 0$) and boundary points $t = 0\text{ s}$ and $t = 10\text{ s}$: $v(3) = -12\text{ ms}^{-1}$, $v(0) = 15\text{ ms}^{-1}$, and $v(10) = 3(100) - 18(10) + 15 = 135\text{ ms}^{-1}$. The maximum velocity in $0 \le t \le 10\text{ s}$ is $135\text{ ms}^{-1}$.
Step 4: Evaluate acceleration across the interval $0 \le t \le 10\text{ s}$. Since $a(t) = 6t - 18$ increases linearly with time, its maximum occurs at the endpoint $t = 10\text{ s}$: $a(10) = 6(10) - 18 = 42\text{ ms}^{-2}$.

Initial position vector $\vec{r}_i = 5\hat{i}\text{ m}$
Final position vector $\vec{r}_f = 5.1\hat{i} + 0.4\hat{j}\text{ m}$
Time interval $\Delta t = 0.02\text{ s}$
Displacement vector $\Delta \vec{r} = \vec{r}_f - \vec{r}_i = (5.1\hat{i} + 0.4\hat{j}) - 5\hat{i} = 0.1\hat{i} + 0.4\hat{j}\text{ m}$
Average velocity vector $\vec{v}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \frac{0.1\hat{i} + 0.4\hat{j}}{0.02} = 5\hat{i} + 20\hat{j}\text{ ms}^{-1}$
Magnitude $\vert{}\vec{v}_{av}\vert{} = \sqrt{5^2 + 20^2} = \sqrt{25 + 400} = \sqrt{425} \approx 20.62\text{ ms}^{-1}$
Angle $\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{20}{5}\right) = \tan^{-1}(4)$
Step Solution:
Step 1: Identify the given initial position vector $\vec{r}_i = 5\hat{i}\text{ m}$ at time $t_i = 0$ and final position vector $\vec{r}_f = 5.1\hat{i} + 0.4\hat{j}\text{ m}$ at time $t_f = 0.02\text{ s}$.
Step 2: Calculate displacement vector $\Delta \vec{r} = \vec{r}_f - \vec{r}_i = (5.1 - 5)\hat{i} + 0.4\hat{j} = 0.1\hat{i} + 0.4\hat{j}\text{ m}$.
Step 3: Determine the average velocity vector $\vec{v}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \frac{0.1\hat{i} + 0.4\hat{j}}{0.02} = 5\hat{i} + 20\hat{j}\text{ ms}^{-1}$.
Step 4: Compute the magnitude of average velocity: $\vert{}\vec{v}_{av}\vert{} = \sqrt{5^2 + 20^2} = \sqrt{425} \approx 20.62\text{ ms}^{-1}$.
Step 5: Compute the direction angle $\theta$ with the positive x-axis: $\tan \theta = \frac{v_y}{v_x} = \frac{20}{5} = 4 \Rightarrow \theta = \tan^{-1}(4)$.

Displacement $\Delta x = x_C - x_A = -6 - (-8) = +2\text{ m}$
Total time $\Delta t = 4 + 5 = 9\text{ s}$
Average velocity $= \frac{x_C - x_A}{\Delta t} = \frac{-6 - (-8)}{9} = \frac{+2}{9} \approx +0.22\text{ ms}^{-1}$
Distance segment 1 $= \vert{}x_B - x_A\vert{} = \vert{}3 - (-8)\vert{} = 11\text{ m}$
Distance segment 2 $= \vert{}x_C - x_B\vert{} = \vert{}-6 - 3\vert{} = 9\text{ m}$
Total distance $= 11 + 9 = 20\text{ m}$
Average speed $= \frac{20}{9} \approx 2.22\text{ ms}^{-1}$
Step Solution:
Step 1: Identify the initial position $x_A = -8\text{ m}$, intermediate position $x_B = +3\text{ m}$, and final position $x_C = -6\text{ m}$.
Step 2: Calculate the time taken for each segment: $t_1 = 4\text{ s}$ and $t_2 = 5\text{ s}$. Total time interval $\Delta t = 4 + 5 = 9\text{ s}$.
Step 3: Calculate the net displacement over the entire 9 s interval: $\Delta x = x_C - x_A = -6 - (-8) = +2\text{ m}$.
Step 4: Compute the average velocity: $v_{av} = \frac{\Delta x}{\Delta t} = \frac{+2\text{ m}}{9\text{ s}} \approx +0.22\text{ ms}^{-1}$.
Step 5: Calculate distance for segment 1 from $A$ to $B$: $d_1 = \vert{}x_B - x_A\vert{} = \vert{}3 - (-8)\vert{} = 11\text{ m}$.
Step 6: Calculate distance for segment 2 from $B$ to $C$: $d_2 = \vert{}x_C - x_B\vert{} = \vert{}-6 - 3\vert{} = 9\text{ m}$.
Step 7: Find the total distance travelled: $D = d_1 + d_2 = 11\text{ m} + 9\text{ m} = 20\text{ m}$.
Step 8: Compute the average speed: $\text{Average Speed} = \frac{\text{Total Distance}}{\Delta t} = \frac{20\text{ m}}{9\text{ s}} \approx 2.22\text{ ms}^{-1}$.

To find the distance travelled, we first determine when the particle reverses its direction by setting $v = 0$.
$3t^2 - 6t = 0 \implies 3t(t - 2) = 0$, so $v = 0$ at $t = 0\text{ s}$ and $t = 2\text{ s}$.
For $0 < t < 2\text{ s}$, $v < 0$, so the particle moves in the negative direction.
For $t > 2\text{ s}$, $v > 0$, so the particle moves in the positive direction.
Displacement from $t = 0$ to $t = 2\text{ s}$ is $x_1 = \int_{0}^{2} (3t^2 - 6t) dt = [t^3 - 3t^2]_{0}^{2} = (8 - 12) = -4\text{ m}$.
Displacement from $t = 2$ to $t = 3.5\text{ s}$ is $x_2 = \int_{2}^{3.5} (3t^2 - 6t) dt = [t^3 - 3t^2]_{2}^{3.5} = (3.5^3 - 3(3.5)^2) - (8 - 12) = (42.875 - 36.75) - (-4) = 6.125 + 4 = 10.125\text{ m}$.
Total Distance = $\vert{}x_1\vert{} + x_2 = \vert{}-4\vert{} + 6.125 = 10.125 + 4 = 14.125\text{ m}$.
Total Displacement = $x_1 + x_2 = -4 + 10.125 = 6.125\text{ m}$.
Average velocity = $\frac{6.125}{3.5} = 1.75\text{ ms}^{-1}$ (or $\frac{9.1875}{3.5} = 2.625\text{ ms}^{-1}$ for total displacement from origin $x(3.5) = 9.1875\text{ m}$).
Average speed = $\frac{14.125}{3.5} \approx 4.036\text{ ms}^{-1}$.
Step Solution:
Step 1: Find the points where the velocity is zero to identify any change in the direction of motion.
Setting $v = 0$:
$3t^2 - 6t = 0$
$3t(t - 2) = 0$
$t = 0\text{ s}$ and $t = 2\text{ s}$.
Step 2: Calculate the displacement for each interval of motion.
For interval $t = 0$ to $t = 2\text{ s}$:
$s_1 = \int_{0}^{2} (3t^2 - 6t) dt = [t^3 - 3t^2]_{0}^{2} = 2^3 - 3(2)^2 = 8 - 12 = -4\text{ m}$.
For interval $t = 2$ to $t = 3.5\text{ s}$:
$s_2 = \int_{2}^{3.5} (3t^2 - 6t) dt = [t^3 - 3t^2]_{2}^{3.5} = (3.5^3 - 3(3.5)^2) - (2^3 - 3(2)^2) = (42.875 - 36.75) - (-4) = 6.125 + 4 = 10.125\text{ m}$.
Step 3: Calculate total distance and total displacement during the time interval $t = 0$ to $t = 3.5\text{ s}$.
Total distance = $\vert{}s_1\vert{} + \vert{}s_2 - s_1 \text{ relative shift}\vert{} = \vert{}-4\vert{} + \vert{}10.125\vert{} = 4 + 10.125 = 14.125\text{ m}$.
Net displacement at $t = 3.5\text{ s}$:
$x(3.5) = [t^3 - 3t^2]_{0}^{3.5} = 3.5^3 - 3(3.5)^2 = 42.875 - 36.75 = 9.1875\text{ m}$.
Step 4: Compute Average Velocity and Average Speed.
Average Velocity = $\frac{\text{Net Displacement}}{\text{Total Time}} = \frac{9.1875\text{ m}}{3.5\text{ s}} = 2.625\text{ ms}^{-1}$.
Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{14.125\text{ m}}{3.5\text{ s}} \approx 4.036\text{ ms}^{-1}$.

Position at $t = 6\text{ s}$:
$x(6) = 1.5(6)^3 - 13.5(6)^2 + 22.5(6) = 324 - 486 + 135 = -27\text{ m}$.
To find distance, differentiate position to get velocity:
$v = \frac{dx}{dt} = 4.5t^2 - 27t + 22.5$.
Set $v = 0$:
$4.5t^2 - 27t + 22.5 = 0 \implies t^2 - 6t + 5 = 0 \implies (t - 1)(t - 5) = 0$.
The particle reverses direction at $t = 1\text{ s}$ and $t = 5\text{ s}$.
Position at $t = 0\text{ s}$: $x(0) = 0\text{ m}$.
Position at $t = 1\text{ s}$: $x(1) = 1.5(1)^3 - 13.5(1)^2 + 22.5(1) = 10.5\text{ m}$.
Position at $t = 5\text{ s}$: $x(5) = 1.5(5)^3 - 13.5(5)^2 + 22.5(5) = 187.5 - 337.5 + 112.5 = -37.5\text{ m}$.
Position at $t = 6\text{ s}$: $x(6) = -27\text{ m}$.
Distance from $t = 0$ to $t = 1$: $\vert{}10.5 - 0\vert{} = 10.5\text{ m}$.
Distance from $t = 1$ to $t = 5$: $\vert{}-37.5 - 10.5\vert{} = 48\text{ m}$.
Distance from $t = 5$ to $t = 6$: $\vert{}-27 - (-37.5)\vert{} = 10.5\text{ m}$.
Total distance = $10.5 + 48 + 10.5 = 69\text{ m}$ (or total path distance summed across key segments).
Step Solution:
Step 1: Calculate position at $t = 6\text{ s}$.
Given $x(t) = 1.5t^3 - 13.5t^2 + 22.5t$.
Substitute $t = 6$:
$x(6) = 1.5(216) - 13.5(36) + 22.5(6) = 324 - 486 + 135 = -27\text{ m}$.
Step 2: Find velocity and turning points.
$v(t) = \frac{dx}{dt} = 4.5t^2 - 27t + 22.5$.
Setting $v(t) = 0$:
$4.5(t^2 - 6t + 5) = 0$
$(t - 1)(t - 5) = 0$
Turning points occur at $t = 1\text{ s}$ and $t = 5\text{ s}$.
Step 3: Evaluate positions at critical time points $t = 0, 1, 5, 6\text{ s}$.
$x(0) = 0\text{ m}$
$x(1) = 1.5(1) - 13.5(1) + 22.5(1) = 10.5\text{ m}$
$x(5) = 1.5(125) - 13.5(25) + 22.5(5) = 187.5 - 337.5 + 112.5 = -37.5\text{ m}$
$x(6) = -27\text{ m}$
Step 4: Compute distance traveled in each interval.
$s_1 = \vert{}x(1) - x(0)\vert{} = \vert{}10.5 - 0\vert{} = 10.5\text{ m}$
$s_2 = \vert{}x(5) - x(1)\vert{} = \vert{}-37.5 - 10.5\vert{} = 48\text{ m}$
$s_3 = \vert{}x(6) - x(5)\vert{} = \vert{}-27 - (-37.5)\vert{} = 10.5\text{ m}$
Total distance $S = s_1 + s_2 + s_3 = 10.5 + 48 + 10.5 = 69\text{ m}$.

Given that velocity $v = 10 \text{ ms}^{-1}$.
The rate of decrease of velocity with respect to displacement is $\frac{dv}{dx} = -3 \text{ ms}^{-1}\text{m}^{-1}$.
Using the formula for acceleration $a = v \frac{dv}{dx}$:
$a = 10 \times (-3) = -30 \text{ ms}^{-2}$.
Thus, the acceleration of the particle at this instant is $-30 \text{ ms}^{-2}$.
Step Solution:
Step 1: Identify the given values from the problem statement.
Given velocity $v = 10 \text{ ms}^{-1}$.
The rate of change of velocity per meter of displacement is given as decreasing at $3 \text{ ms}^{-1}/\text{m}$, which means $\frac{dv}{dx} = -3 \text{ s}^{-1}$.
Step 2: Apply the kinetic relation for acceleration in terms of position.
$a = v \frac{dv}{dx}$
Step 3: Substitute the known values into the equation to calculate acceleration.
$a = (10 \text{ ms}^{-1}) \times (-3 \text{ s}^{-1}) = -30 \text{ ms}^{-2}$
Step 4: State the final calculated acceleration.
The acceleration of the particle at the given instant is $-30 \text{ ms}^{-2}$.