Kinematics-1D

Given displacement function:
$s = 2t^3 + 3t^2 + 2t + 8$
Differentiating displacement with respect to time to obtain velocity $v$:
$v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 + 3t^2 + 2t + 8)$
$v = 6t^2 + 6t + 2$
Differentiating velocity with respect to time to obtain acceleration $a$:
$a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 + 6t + 2)$
$a = 12t + 6$
Setting acceleration equal to zero:
$12t + 6 = 0$
$12t = -6$
$t = -\frac{6}{12}$
$t = -\frac{1}{2}$

Given velocity as a function of time:
$v = 4t$
The relation between displacement and velocity is given by:
$s = \int_{t_1}^{t_2} v \, dt$
Substitute the velocity expression into the integral for limits $t = 2\text{ s}$ to $t = 4\text{ s}$:
$s = \int_{2}^{4} 4t \, dt$
Integrating $4t$ with respect to $t$:
$s = [2t^2]_{2}^{4}$
Applying the upper and lower limits:
$s = 2(4)^2 - 2(2)^2$
$s = 2(16) - 2(4)$
$s = 32 - 8$
$s = 24\text{ m}$

Given velocity as a function of time:
$v = 5t - t^2$
The fourth second corresponds to the time interval from $t = 3\text{ s}$ to $t = 4\text{ s}$.
Checking direction of motion during this interval:
At $t = 3$, $v = 5(3) - 3^2 = 15 - 9 = 6\text{ m/s} > 0$
At $t = 4$, $v = 5(4) - 4^2 = 20 - 16 = 4\text{ m/s} > 0$
Since velocity remains positive in the interval $t \in [3, 4]$, distance equals displacement:
$\text{Distance} = \int_{3}^{4} (5t - t^2) \, dt$
Integrating term by term:
$\text{Distance} = \left[ \frac{5t^2}{2} - \frac{t^3}{3} \right]_{3}^{4}$
Evaluating at the upper limit $t = 4$:
$\left( \frac{5(4)^2}{2} - \frac{4^3}{3} \right) = \left( \frac{5(16)}{2} - \frac{64}{3} \right) = \left( 40 - \frac{64}{3} \right) = \frac{120 - 64}{3} = \frac{56}{3}$
Evaluating at the lower limit $t = 3$:
$\left( \frac{5(3)^2}{2} - \frac{3^3}{3} \right) = \left( \frac{5(9)}{2} - \frac{27}{3} \right) = \left( \frac{45}{2} - 9 \right) = \frac{45 - 18}{2} = \frac{27}{2}$
Subtracting the lower limit value from the upper limit value:
$\text{Distance} = \frac{56}{3} - \frac{27}{2}$
Taking a common denominator of 6:
$\text{Distance} = \frac{112 - 81}{6}$
$\text{Distance} = \frac{31}{6}\text{ m}$

Given acceleration as a function of displacement:
$a = -\alpha x^2$
Substitute $a = v \frac{dv}{dx}$:
$v \frac{dv}{dx} = -\alpha x^2$
Separating variables:
$v \, dv = -\alpha x^2 \, dx$
Integrating both sides from $x = 0$ where $v = v_0$ to $x$ where the particle stops ($v = 0$):
$\int_{v_0}^{0} v \, dv = \int_{0}^{x} -\alpha x^2 \, dx$
Evaluating the definite integrals:
$\left[ \frac{v^2}{2} \right]_{v_0}^{0} = \left[ -\frac{\alpha x^3}{3} \right]_{0}^{x}$
$0 - \frac{v_0^2}{2} = -\frac{\alpha x^3}{3}$
$\frac{v_0^2}{2} = \frac{\alpha x^3}{3}$
Solving for $x^3$:
$x^3 = \frac{3 v_0^2}{2 \alpha}$
Taking cube root on both sides:
$x = \left(\frac{3 v_0^2}{2 \alpha}\right)^{\frac{1}{3}}$

Given acceleration as a function of distance:
$a = \sqrt{x}$
Substitute $a = v \frac{dv}{dx}$:
$v \frac{dv}{dx} = x^{\frac{1}{2}}$
Separating variables:
$v \, dv = x^{\frac{1}{2}} \, dx$
Integrating both sides from initial speed $v = v_0$ at $x = 0$ to final speed $v = 2v_0$ at position $x$:
$\int_{v_0}^{2v_0} v \, dv = \int_{0}^{x} x^{\frac{1}{2}} \, dx$
Evaluating the left-hand side integral:
$\left[ \frac{v^2}{2} \right]_{v_0}^{2v_0} = \frac{(2v_0)^2 - v_0^2}{2} = \frac{4v_0^2 - v_0^2}{2} = \frac{3}{2} v_0^2$
Evaluating the right-hand side integral:
$\left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{x} = \frac{2}{3} x^{\frac{3}{2}}$
Equating both sides:
$\frac{2}{3} x^{\frac{3}{2}} = \frac{3}{2} v_0^2$
Solving for $x^{\frac{3}{2}}$:
$x^{\frac{3}{2}} = \frac{9}{4} v_0^2$
Raising both sides to the power $\frac{2}{3}$:
$x = \left(\frac{9}{4} v_0^2\right)^{\frac{2}{3}}$
Expressing $\frac{9}{4}$ as $\left(\frac{3}{2}\right)^2$:
$x = \left[\left(\frac{3}{2}\right)^2 v_0^2\right]^{\frac{2}{3}}$
$x = \left[\left(\frac{3}{2} v_0\right)^2\right]^{\frac{2}{3}}$
$x = \left(\frac{3}{2} v_0\right)^{\frac{4}{3}}$

Given position as a function of time:
$x = \frac{t^3}{6} - t^2 - 9t + 18$
Differentiating position with respect to time to get velocity $v$:
$v = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^3}{6} - t^2 - 9t + 18\right)$
$v = \frac{3t^2}{6} - 2t - 9$
$v = \frac{t^2}{2} - 2t - 9$
Differentiating velocity with respect to time to get acceleration $a$:
$a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{t^2}{2} - 2t - 9\right)$
$a = t - 2$
Setting acceleration to zero to find the time:
$t - 2 = 0$
$t = 2\text{ s}$
Substituting $t = 2\text{ s}$ back into the velocity equation:
$v = \frac{2^2}{2} - 2(2) - 9$
$v = \frac{4}{2} - 4 - 9$
$v = 2 - 4 - 9$
$v = -11\text{ m/s}$

Given displacement function:
$s = 3t^2 - 2$
Differentiating displacement with respect to time to get velocity $v$:
$v = \frac{ds}{dt} = \frac{d}{dt}(3t^2 - 2)$
$v = 6t$
Evaluating the velocity at time $t = 2$:
$v = 6(2)$
$v = 12\text{ ms}^{-1}$

Given temperature function:
$T = T_0 + \alpha t^2 + \beta \sin t$
Differentiating temperature with respect to time $t$ to find the rate of change of temperature:
$\frac{dT}{dt} = \frac{d}{dt}(T_0 + \alpha t^2 + \beta \sin t)$
Since $T_0$ is a constant, its derivative is zero:
$\frac{dT}{dt} = 2\alpha t + \beta \cos t$
Substituting the given values $\alpha = \frac{2}{\pi}$ and $\beta = -4$:
$\frac{dT}{dt} = 2\left(\frac{2}{\pi}\right)t + (-4)\cos t$
$\frac{dT}{dt} = \frac{4t}{\pi} - 4\cos t$
Evaluating the rate of change at time $t = \pi\text{ s}$:
$\left.\frac{dT}{dt}\right\vert{}_{t=\pi} = \frac{4(\pi)}{\pi} - 4\cos(\pi)$
Since $\cos(\pi) = -1$:
$\left.\frac{dT}{dt}\right\vert{}_{t=\pi} = 4 - 4(-1)$
$\left.\frac{dT}{dt}\right\vert{}_{t=\pi} = 4 + 4$
$\left.\frac{dT}{dt}\right\vert{}_{t=\pi} = 8\text{ K/sec}$

Given velocity as a function of position:
$v = x^2 + x$
Differentiating velocity with respect to position $x$:
$\frac{dv}{dx} = \frac{d}{dx}(x^2 + x)$
$\frac{dv}{dx} = 2x + 1$
Using the formula for acceleration $a = v \frac{dv}{dx}$:
$a = (x^2 + x)(2x + 1)$
Evaluating $v$ at $x = 2\text{ m}$:
$v = 2^2 + 2 = 4 + 2 = 6\text{ m/s}$
Evaluating $\frac{dv}{dx}$ at $x = 2\text{ m}$:
$\frac{dv}{dx} = 2(2) + 1 = 4 + 1 = 5\text{ s}^{-1}$
Calculating acceleration at $x = 2\text{ m}$:
$a = (6)(5)$
$a = 30\text{ m/s}^2$

Given equation relating displacement and velocity:
$x = \sqrt{v+1}$
Squaring both sides gives:
$x^2 = v + 1$
Rearranging to express velocity in terms of displacement:
$v = x^2 - 1$
Differentiating velocity with respect to time to find acceleration $a$:
$a = \frac{dv}{dt} = \frac{d}{dt}(x^2 - 1)$
Using the chain rule $\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$:
$a = 2x \cdot \frac{dx}{dt}$
Since $\frac{dx}{dt} = v$:
$a = 2x \cdot v$
Substituting $v = x^2 - 1$ into the acceleration expression:
$a = 2x(x^2 - 1)$
Now, evaluating acceleration at $x = 5\text{ m}$:
$a = 2(5)(5^2 - 1)$
$a = 10(25 - 1)$
$a = 10(24)$
$a = 240\text{ m/s}^2$

Given velocity relationship:
$v = \alpha \sqrt{x}$
Substitute $v = \frac{dx}{dt}$:
$\frac{dx}{dt} = \alpha \sqrt{x}$
Separating variables to integrate:
$\frac{dx}{\sqrt{x}} = \alpha dt$
Integrating both sides from $x = 0$ at $t = 0$ to displacement $x$ at time $t$:
$\int_0^x x^{-1/2} dx = \int_0^t \alpha dt$
$2 \sqrt{x} = \alpha t$
Squaring both sides:
$4x = \alpha^2 t^2$
$x = \frac{1}{4} \alpha^2 t^2$
Differentiating displacement with respect to time to get velocity:
$v = \frac{dx}{dt} = \frac{1}{2} \alpha^2 t$
Differentiating velocity with respect to time to get acceleration:
$a = \frac{dv}{dt} = \frac{d^2 x}{dt^2} = \frac{1}{2} \alpha^2$

Given equation of trajectory:
$y = px - qx^2$
Differentiating with respect to time $t$:
$\frac{dy}{dt} = p \frac{dx}{dt} - 2qx \frac{dx}{dt}$
Differentiating again with respect to time $t$:
$\frac{d^2y}{dt^2} = p \frac{d^2x}{dt^2} - 2qx \frac{d^2x}{dt^2} - 2q \left(\frac{dx}{dt}\right)^2$
Since acceleration is entirely along the negative y-axis, $a_x = \frac{d^2x}{dt^2} = 0$ and $a_y = \frac{d^2y}{dt^2} = -a$:
$-a = p(0) - 2qx(0) - 2q v_x^2$
$-a = -2q v_x^2$
Solving for $v_x^2$:
$v_x^2 = \frac{a}{2q}$
$v_x = \sqrt{\frac{a}{2q}}$
At the origin ($x = 0$), calculating y-component of velocity $v_y$:
$v_y = p v_x - 2q(0) v_x$
$v_y = p \sqrt{\frac{a}{2q}}$
Calculating total velocity $v$ at the origin:
$v = \sqrt{v_x^2 + v_y^2}$
$v = \sqrt{\frac{a}{2q} + p^2 \left(\frac{a}{2q}\right)}$
$v = \sqrt{\frac{a(p^2+1)}{2q}}$

Given velocity as a function of time:
$v = 3t^2 - 6t$
Total time interval is $t = 0$ to $t = 4\text{ s}$, which gives $\Delta t = 4 - 0 = 4\text{ s}$.
Calculating displacement $\Delta s$ by integrating velocity over the interval $t = 0$ to $t = 4\text{ s}$:
$\Delta s = \int_{0}^{4} (3t^2 - 6t) \, dt$
Integrating term by term:
$\Delta s = \left[ t^3 - 3t^2 \right]_{0}^{4}$
Evaluating at upper limit $t = 4\text{ s}$:
$\Delta s = 4^3 - 3(4)^2$
$\Delta s = 64 - 3(16)$
$\Delta s = 64 - 48 = 16\text{ m}$
Calculating average velocity:
$v_{\text{avg}} = \frac{\Delta s}{\Delta t} = \frac{16}{4}$
$v_{\text{avg}} = 4\text{ ms}^{-1}$

Given velocity as a function of time:
$v = 3t^2 - 6t = 3t(t - 2)$
Finding the turning point by setting $v = 0$:
$3t(t - 2) = 0 \implies t = 0\text{ s}$ and $t = 2\text{ s}$
For the interval $t \in [0, 2]$, $v \le 0$, so the magnitude of velocity is $\vert{}v\vert{} = 6t - 3t^2$.
For the interval $t \in [2, 4]$, $v \ge 0$, so the magnitude of velocity is $\vert{}v\vert{} = 3t^2 - 6t$.
Calculating total distance travelled by splitting the integral at $t = 2\text{ s}$:
$\text{Total Distance} = \int_{0}^{2} (6t - 3t^2) \, dt + \int_{2}^{4} (3t^2 - 6t) \, dt$
Evaluating the first integral from $t = 0$ to $t = 2$:
$\left[ 3t^2 - t^3 \right]_{0}^{2} = (3(2)^2 - 2^3) = (12 - 8) = 4\text{ m}$
Evaluating the second integral from $t = 2$ to $t = 4$:
$\left[ t^3 - 3t^2 \right]_{2}^{4} = (4^3 - 3(4)^2) - (2^3 - 3(2)^2) = (64 - 48) - (8 - 12) = 16 - (-4) = 20\text{ m}$
Calculating total distance:
$\text{Total Distance} = 4 + 20 = 24\text{ m}$
Calculating total time taken:
$\Delta t = 4 - 0 = 4\text{ s}$
Calculating average speed:
$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{24}{4}$
$\text{Average Speed} = 6\text{ ms}^{-1}$

Given position coordinates of the particle:
$x = 3 \cos 2t$
$y = 4 \sin 2t$
Position vector as a function of time:
$\vec{r} = (3 \cos 2t)\hat{i} + (4 \sin 2t)\hat{j}$
Differentiating position with respect to time to get velocity vector:
$\vec{v} = \frac{d\vec{r}}{dt} = (-6 \sin 2t)\hat{i} + (8 \cos 2t)\hat{j}$
Evaluating position vector at time $t = \frac{\pi}{4}$:
$\vec{r} = 3 \cos\left(\frac{\pi}{2}\right)\hat{i} + 4 \sin\left(\frac{\pi}{2}\right)\hat{j}$
Since $\cos\left(\frac{\pi}{2}\right) = 0$ and $\sin\left(\frac{\pi}{2}\right) = 1$:
$\vec{r} = 4\hat{j}$
Evaluating velocity vector at time $t = \frac{\pi}{4}$:
$\vec{v} = -6 \sin\left(\frac{\pi}{2}\right)\hat{i} + 8 \cos\left(\frac{\pi}{2}\right)\hat{j}$
$\vec{v} = -6\hat{i}$
Calculating the dot product of position vector and velocity vector:
$\vec{r} \cdot \vec{v} = (4\hat{j}) \cdot (-6\hat{i}) = 0$
Calculating the cosine of the angle $\theta$:
$\cos\theta = \frac{\vec{r} \cdot \vec{v}}{\vert{}\vec{r}\vert{}\vert{}\vec{v}\vert{}} = \frac{0}{(4)(6)} = 0$
$\theta = \frac{\pi}{2}$

Given position function:
$x(t) = \frac{t^3}{3} - \frac{5t^2}{2} + 6t$
Differentiating position with respect to time to find velocity $v$:
$v = \frac{dx}{dt} = t^2 - 5t + 6$
Finding the turning points by setting velocity to zero:
$t^2 - 5t + 6 = 0$
$(t - 2)(t - 3) = 0$
The particle changes direction at $t = 2\text{ s}$ and $t = 3\text{ s}$.
Calculating position at key time instants $t = 0, 2, 3, 4\text{ s}$:
At $t = 0$:
$x(0) = 0\text{ m}$
At $t = 2$:
$x(2) = \frac{2^3}{3} - \frac{5(2)^2}{2} + 6(2) = \frac{8}{3} - 10 + 12 = \frac{8}{3} + 2 = \frac{14}{3}\text{ m}$
At $t = 3$:
$x(3) = \frac{3^3}{3} - \frac{5(3)^2}{2} + 6(3) = 9 - \frac{45}{2} + 18 = 27 - \frac{45}{2} = \frac{9}{2}\text{ m}$
At $t = 4$:
$x(4) = \frac{4^3}{3} - \frac{5(4)^2}{2} + 6(4) = \frac{64}{3} - 40 + 24 = \frac{64}{3} - 16 = \frac{16}{3}\text{ m}$
Calculating total distance by summing distances over each sub-interval:
$\text{Distance} = \vert{}x(2) - x(0)\vert{} + \vert{}x(3) - x(2)\vert{} + \vert{}x(4) - x(3)\vert{}$
Distance in interval $t = 0$ to $t = 2$:
$\left\vert{}\frac{14}{3} - 0\right\vert{} = \frac{14}{3}\text{ m}$
Distance in interval $t = 2$ to $t = 3$:
$\left\vert{}\frac{9}{2} - \frac{14}{3}\right\vert{} = \left\vert{}\frac{27 - 28}{6}\right\vert{} = \left\vert{}-\frac{1}{6}\right\vert{} = \frac{1}{6}\text{ m}$
Distance in interval $t = 3$ to $t = 4$:
$\left\vert{}\frac{16}{3} - \frac{9}{2}\right\vert{} = \left\vert{}\frac{32 - 27}{6}\right\vert{} = \frac{5}{6}\text{ m}$
Adding all intervals together:
$\text{Total Distance} = \frac{14}{3} + \frac{1}{6} + \frac{5}{6} = \frac{14}{3} + 1 = \frac{17}{3}\text{ m}$
Wait, let us re-verify:
Interval 1: $x(2) - x(0) = \frac{14}{3} - 0 = \frac{14}{3}$
Interval 2: $\vert{}x(3) - x(2)\vert{} = \vert{}\frac{9}{2} - \frac{14}{3}\vert{} = \frac{1}{6}$
Interval 3: $\vert{}x(4) - x(3)\vert{} = \vert{}\frac{16}{3} - \frac{9}{2}\vert{} = \frac{5}{6}$
Total distance $= \frac{14}{3} + \frac{1}{6} + \frac{5}{6} = \frac{28}{6} + \frac{1}{6} + \frac{5}{6} = \frac{34}{6} = \frac{17}{3}\text{ m}$
Let me correct Option B to $\frac{17}{3}\text{ m}$ to match the derivation.