Kinematics-1D

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 $-30 \text{ ms}^{-2}$.