1.A point moves in a straight line so that its displacement $x$ at time $t$ is given by $x^2 = t^2 + 1$. Show that its acceleration is inversely proportional to $x^3$. Determine the constant of proportionality.
Sol: Differentiating $x^2 = t^2 + 1$ w.r.t $t$: $2x \frac{dx}{dt} = 2t \implies x v = t \implies v = \frac{t}{x}$.
Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{t}{x}\right) = \frac{x(1) - t(v)}{x^2}$. Substitute $v = t/x$:
$a = \frac{x - t(t/x)}{x^2} = \frac{x^2 - t^2}{x^3}$.
From the given equation, $x^2 - t^2 = 1$. Therefore, $a = \frac{1}{x^3}$.
Thus, $a \propto \frac{1}{x^3}$ and the constant of proportionality is $\mathbf{k = 1}$.
2.A particle moves along the $x$-axis. Its coordinate is given by $x = 10 + 8t - 3t^2$. Another particle moves along the $y$-axis with coordinate $y = 5 - 8t^3$. Find the magnitude of the relative velocity of the second particle with respect to the first at $t = 1 \text{ s}$.
Sol: Velocity of first particle: $v_x = \frac{dx}{dt} = 8 - 6t$. At $t=1$, $\vec{v}_1 = (8 - 6(1))\hat{i} = 2\hat{i} \text{ m/s}$.
Velocity of second particle: $v_y = \frac{dy}{dt} = -24t^2$. At $t=1$, $\vec{v}_2 = -24(1)^2\hat{j} = -24\hat{j} \text{ m/s}$.
Relative velocity $\vec{v}_{21} = \vec{v}_2 - \vec{v}_1 = -24\hat{j} - 2\hat{i}$.
Magnitude $|\vec{v}_{21}| = \sqrt{(-2)^2 + (-24)^2} = \sqrt{4 + 576} = \sqrt{580} \approx \mathbf{24.08 \text{ m/s}}$.
3.An ant crawls on a vertical cylinder of radius $R$ and height $H$. It starts from the bottom edge and reaches the diametrically opposite top edge. If it takes exactly two complete helical turns to reach the top, find the magnitude of its net displacement and the total distance covered.
Sol: Distance: Unrolling the cylinder's surface creates a rectangle of width $2\pi R$ and height $H$. For 2 complete turns, the horizontal distance traversed on the unrolled sheet is $2 \times 2\pi R = 4\pi R$. By Pythagorean theorem, total path length (distance) $= \sqrt{H^2 + (4\pi R)^2} = \mathbf{\sqrt{H^2 + 16\pi^2 R^2}}$.
Displacement: The start and end points are diametrically opposite. The straight-line distance through 3D space connects the bottom edge to the diametrically opposite top edge. Vertical separation is $H$, horizontal separation across the circular base is $2R$. Net displacement $= \mathbf{\sqrt{H^2 + 4R^2}}$.
4.A particle moves such that its position vector is $\vec{r} = (A\sin(\omega t))\hat{i} + (A\cos(\omega t))\hat{j} + (Bt)\hat{k}$. Find the ratio of the magnitude of its velocity to the magnitude of its acceleration.
Sol: Velocity $\vec{v} = \frac{d\vec{r}}{dt} = (A\omega\cos\omega t)\hat{i} - (A\omega\sin\omega t)\hat{j} + B\hat{k}$.
Magnitude $|\vec{v}| = \sqrt{(A\omega\cos\omega t)^2 + (-A\omega\sin\omega t)^2 + B^2} = \sqrt{A^2\omega^2(\cos^2\omega t + \sin^2\omega t) + B^2} = \sqrt{A^2\omega^2 + B^2}$.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = (-A\omega^2\sin\omega t)\hat{i} - (A\omega^2\cos\omega t)\hat{j}$.
Magnitude $|\vec{a}| = \sqrt{(-A\omega^2\sin\omega t)^2 + (-A\omega^2\cos\omega t)^2} = A\omega^2$.
Ratio $= \frac{|\vec{v}|}{|\vec{a}|} = \mathbf{\frac{\sqrt{A^2\omega^2 + B^2}}{A\omega^2}}$.
5.A person walks on a horizontal plane. First he walks $a$ meters due East, then $2a$ meters due North, then $3a$ meters due West, then $4a$ meters due South, and so on in a spiral pattern. What is the magnitude of his displacement after 10 such distinct segments?
Sol: Taking East as $+x$ and North as $+y$ axis.
After 10 segments, the components are:
$x = a \text{ (E)} - 3a \text{ (W)} + 5a \text{ (E)} - 7a \text{ (W)} + 9a \text{ (E)} = a - 3a + 5a - 7a + 9a = 5a$.
$y = 2a \text{ (N)} - 4a \text{ (S)} + 6a \text{ (N)} - 8a \text{ (S)} + 10a \text{ (N)} = 2a - 4a + 6a - 8a + 10a = 6a$.
Net displacement $= \sqrt{x^2 + y^2} = \sqrt{(5a)^2 + (6a)^2} = \sqrt{25a^2 + 36a^2} = \mathbf{\sqrt{61}a}$. (Note: Check phrasing carefully. If the pattern resets, calculate sequentially.)
6.If the displacement of a particle is given by $s = a e^{\alpha t} + b e^{-\beta t}$, where $a, b, \alpha, \beta$ are positive constants, prove that the acceleration is always positive and proportional to the original displacement only if $\alpha = \beta$.
Sol: Velocity $v = \frac{ds}{dt} = a\alpha e^{\alpha t} - b\beta e^{-\beta t}$.
Acceleration $A_{cc} = \frac{dv}{dt} = a\alpha^2 e^{\alpha t} + b\beta^2 e^{-\beta t}$.
Since $a, b, \alpha, \beta$ are positive, exponentials are positive, thus $A_{cc} > 0$ (always positive).
For $A_{cc} \propto s$, we need $a\alpha^2 e^{\alpha t} + b\beta^2 e^{-\beta t} = k(a e^{\alpha t} + b e^{-\beta t})$.
Comparing coefficients, $k = \alpha^2$ and $k = \beta^2$. This is possible only if $\alpha^2 = \beta^2 \implies \mathbf{\alpha = \beta}$ (since both are positive).
7.The position of a particle is given by $x(t) = t^3 - 6t^2 + 9t + 5$. Find the total distance travelled by the particle in the interval $t=0$ to $t=4 \text{ s}$. Be careful to locate any turning points.
Sol: Velocity $v = \frac{dx}{dt} = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)$.
Particle turns (stops) at $t=1\text{ s}$ and $t=3\text{ s}$. We must calculate distance in intervals: $0\to1$, $1\to3$, $3\to4$.
Positions: $x(0) = 5$, $x(1) = 1 - 6 + 9 + 5 = 9$, $x(3) = 27 - 54 + 27 + 5 = 5$, $x(4) = 64 - 96 + 36 + 5 = 9$.
Distance $0\to1 = |9 - 5| = 4\text{ m}$. Distance $1\to3 = |5 - 9| = 4\text{ m}$. Distance $3\to4 = |9 - 5| = 4\text{ m}$.
Total distance $= 4 + 4 + 4 = \mathbf{12 \text{ m}}$. (Note: Original key may contain typo, 12m is mathematically robust.)
8.A point moves such that its distance from the origin at time $t$ is $r = t^2 e^{-t}$. Find the time at which the velocity of the point away from the origin is maximum.
Sol: Velocity $v = \frac{dr}{dt} = 2t e^{-t} - t^2 e^{-t} = e^{-t}(2t - t^2)$.
To find maximum velocity, set acceleration $a = \frac{dv}{dt} = 0$.
$a = -e^{-t}(2t - t^2) + e^{-t}(2 - 2t) = e^{-t}(t^2 - 4t + 2) = 0$.
Since $e^{-t} \ne 0$, we have $t^2 - 4t + 2 = 0$. Using quadratic formula:
$t = \frac{4 \pm \sqrt{16 - 8}}{2} = 2 \pm \sqrt{2}$.
Since $t=0 \implies v=0$, the first peak corresponds to max velocity. Thus, $\mathbf{t = 2 - \sqrt{2} \text{ s}}$.
9.A particle covers a distance $L$. It travels the first half of the distance with speed $v_1$. The remaining half is covered with speed $v_2$ for half the time and with speed $v_3$ for the other half of the time. Find the average speed of the particle for the entire journey.
Sol: Time for first half ($L/2$): $t_1 = \frac{L}{2v_1}$.
For the second half ($L/2$), let total time be $t_2$. It travels at $v_2$ for $t_2/2$ and $v_3$ for $t_2/2$.
Distance $= v_2(t_2/2) + v_3(t_2/2) = \frac{L}{2} \implies t_2(v_2 + v_3) = L \implies t_2 = \frac{L}{v_2 + v_3}$.
Total time $T = t_1 + t_2 = \frac{L}{2v_1} + \frac{L}{v_2 + v_3} = L \left( \frac{v_2 + v_3 + 2v_1}{2v_1(v_2 + v_3)} \right)$.
Average speed $= \frac{L}{T} = \mathbf{\frac{2v_1(v_2 + v_3)}{2v_1 + v_2 + v_3}}$.
10.A car moves along a straight line whose speed varies with distance $s$ as $v = \alpha \sqrt{s}$, where $\alpha$ is a constant. Find the average velocity of the car over a total distance $S$.
Sol: By definition, $v = \frac{ds}{dt} \implies dt = \frac{ds}{\alpha \sqrt{s}}$.
Total time $T = \int_0^S \frac{1}{\alpha} s^{-1/2} ds = \frac{1}{\alpha} \left[ 2\sqrt{s} \right]_0^S = \frac{2\sqrt{S}}{\alpha}$.
Average velocity over time (standard definition) $= \frac{\text{Total Distance}}{\text{Total Time}} = \frac{S}{2\sqrt{S}/\alpha} = \mathbf{\frac{\alpha \sqrt{S}}{2}}$.
(Note: If spatial average is asked, $\frac{1}{S}\int_0^S \alpha\sqrt{s} ds = \frac{2}{3}\alpha\sqrt{S}$, but time-average is standard).
11.A point traversing a straight line moves with a constant velocity $v_1$ for $1/n^{\text{th}}$ of the total distance, then with velocity $v_2$ for the next $1/n^{\text{th}}$ of the distance, and so on up to $v_n$. Prove that the average velocity is the harmonic mean of the individual velocities.
Sol: Let total distance be $D$. Each segment is $D/n$.
Time for $i^{\text{th}}$ segment $t_i = \frac{D/n}{v_i} = \frac{D}{n v_i}$.
Total time $T = \sum_{i=1}^n t_i = \sum_{i=1}^n \frac{D}{n v_i} = \frac{D}{n} \sum_{i=1}^n \frac{1}{v_i}$.
Average velocity $V_{avg} = \frac{D}{T} = \frac{D}{\frac{D}{n} \sum \frac{1}{v_i}} = \mathbf{\frac{n}{\sum_{i=1}^n \frac{1}{v_i}}}$. This is the definition of Harmonic Mean.
12.A particle's velocity is given by $v(t) = v_0 e^{-bt}$ where $v_0$ and $b$ are positive constants. Find the average velocity of the particle between $t=0$ and the time when its velocity reduces to $v_0/2$.
Sol: Find time $t_f$ when $v = v_0/2$: $v_0 e^{-bt_f} = v_0/2 \implies e^{-bt_f} = 1/2 \implies -bt_f = -\ln 2 \implies t_f = \frac{\ln 2}{b}$.
Distance covered $S = \int_0^{t_f} v_0 e^{-bt} dt = \left[ \frac{v_0}{-b} e^{-bt} \right]_0^{t_f} = \frac{v_0}{-b} (e^{-\ln 2} - 1) = \frac{v_0}{-b} (\frac{1}{2} - 1) = \frac{v_0}{2b}$.
Average velocity $V_{avg} = \frac{S}{t_f} = \frac{v_0 / 2b}{\ln 2 / b} = \mathbf{\frac{v_0}{2 \ln 2}}$.
13.An object travels continuously in one direction. During the first $N$ seconds, its velocity at time $t$ is given by $v = k t^2$. Calculate the ratio of its average velocity during the first $N$ seconds to its instantaneous velocity at $t = N$.
Sol: Instantaneous velocity at $t=N$: $v(N) = k N^2$.
Distance covered in $N$ seconds: $S = \int_0^N k t^2 dt = k \left[ \frac{t^3}{3} \right]_0^N = \frac{k N^3}{3}$.
Average velocity $V_{avg} = \frac{S}{N} = \frac{k N^3 / 3}{N} = \frac{k N^2}{3}$.
Ratio $= \frac{V_{avg}}{v(N)} = \frac{k N^2 / 3}{k N^2} = \frac{1}{3}$. Ratio is $\mathbf{1 : 3}$.
14.A particle starts from rest and moves with acceleration $a$ which varies with time $t$ as $a = kt$, where $k$ is a constant. Find the average velocity of the particle during the time interval it takes to cover a distance $S$.
Sol: Given $a = \frac{dv}{dt} = kt \implies v = \int kt dt = \frac{kt^2}{2}$ (since $v(0)=0$).
Distance $s = \int v dt = \int \frac{kt^2}{2} dt = \frac{kt^3}{6}$ (since $s(0)=0$).
Time to cover distance $S$: $S = \frac{kt_f^3}{6} \implies t_f = \left(\frac{6S}{k}\right)^{1/3}$.
Average velocity $V_{avg} = \frac{S}{t_f} = \frac{S}{(6S/k)^{1/3}} = S^{1 - 1/3} \left(\frac{k}{6}\right)^{1/3} = \mathbf{\left(\frac{k S^2}{6}\right)^{1/3}}$.
15.The speed of a train varies as $v = v_0 \sin(\pi t / T)$ for $0 \le t \le T$. Find the average speed of the train for this duration and compare it to the maximum speed $v_0$.
Sol: Distance $S = \int_0^T v_0 \sin(\frac{\pi t}{T}) dt = v_0 \left[ -\frac{T}{\pi} \cos(\frac{\pi t}{T}) \right]_0^T = -\frac{v_0 T}{\pi} (\cos\pi - \cos 0) = -\frac{v_0 T}{\pi} (-1 - 1) = \frac{2v_0 T}{\pi}$.
Average speed $V_{avg} = \frac{S}{T} = \frac{2v_0 T / \pi}{T} = \mathbf{\frac{2}{\pi} v_0}$.
It is $\frac{2}{\pi}$ times ($\approx 63.7\%$) of the maximum speed $v_0$.
16.A body moves in a straight line with a velocity varying as the square of the time. If it covers a distance $x$ in time $t$, and $y$ in the next time $t$, prove that $y = 7x$.
Sol: Given $v \propto t^2 \implies v = c t^2$. Position $s(t) = \int c t^2 dt = \frac{c t^3}{3}$ (assuming $s(0)=0$).
Distance in first interval $t$: $x = s(t) - s(0) = \frac{c t^3}{3}$.
Distance in next interval $t$ (from $t$ to $2t$): $y = s(2t) - s(t) = \frac{c(2t)^3}{3} - \frac{c t^3}{3} = \frac{8c t^3}{3} - \frac{c t^3}{3} = 7 \left(\frac{c t^3}{3}\right)$.
Therefore, $y = 7(x) \implies \mathbf{y = 7x}$. (Proved).
17.The deceleration experienced by a moving motor boat after its engine is cut off is given by $dv/dt = -kv^3$, where $k$ is a constant. If $v_0$ is the magnitude of the velocity at cut-off, find the velocity as a function of time $t$ and distance $x$.
Sol: As function of t: $\frac{dv}{dt} = -kv^3 \implies \int_{v_0}^v v^{-3} dv = \int_0^t -k dt \implies \left[\frac{v^{-2}}{-2}\right]_{v_0}^v = -kt$
$-\frac{1}{2v^2} + \frac{1}{2v_0^2} = -kt \implies \frac{1}{v^2} = 2kt + \frac{1}{v_0^2} = \frac{1 + 2kv_0^2 t}{v_0^2} \implies \mathbf{v(t) = \frac{v_0}{\sqrt{1 + 2kv_0^2 t}}}$.
As function of x: $v\frac{dv}{dx} = -kv^3 \implies \int_{v_0}^v v^{-2} dv = \int_0^x -k dx \implies \left[-\frac{1}{v}\right]_{v_0}^v = -kx$
$-\frac{1}{v} + \frac{1}{v_0} = -kx \implies \frac{1}{v} = kx + \frac{1}{v_0} = \frac{kv_0 x + 1}{v_0} \implies \mathbf{v(x) = \frac{v_0}{1 + kv_0 x}}$.
18.A particle of mass $m$ moves in a straight line under a retarding force which gives it an acceleration $a = -k \sqrt{v}$, where $v$ is velocity. If the initial velocity is $u$, find the total distance covered by the particle before coming to rest.
Sol: Acceleration $a = v\frac{dv}{dx} = -k\sqrt{v}$.
Separating variables: $v \cdot v^{-1/2} dv = -k dx \implies \sqrt{v} dv = -k dx$.
Integrate from start ($x=0, v=u$) to rest ($x=S, v=0$): $\int_u^0 v^{1/2} dv = \int_0^S -k dx$.
$\left[ \frac{2}{3} v^{3/2} \right]_u^0 = -kS \implies 0 - \frac{2}{3} u^{3/2} = -kS \implies \mathbf{S = \frac{2u^{3/2}}{3k}}$.
19.A particle moves along a straight line such that its velocity $v$ depends on displacement $x$ as $v = \beta x^{-2n}$. Find the acceleration of the particle as a function of time, given that $x(0) = 0$.
Sol: First find $x(t)$: $v = \frac{dx}{dt} = \beta x^{-2n} \implies x^{2n} dx = \beta dt \implies \int_0^x x^{2n} dx = \int_0^t \beta dt$
$\frac{x^{2n+1}}{2n+1} = \beta t \implies x = ((2n+1)\beta t)^{\frac{1}{2n+1}}$.
Now find acceleration $a(x) = v\frac{dv}{dx} = (\beta x^{-2n})(-2n\beta x^{-2n-1}) = -2n\beta^2 x^{-4n-1}$.
Substitute $x(t)$ into $a(x)$:
$a(t) = -2n\beta^2 \left( ((2n+1)\beta t)^{\frac{1}{2n+1}} \right)^{-4n-1} = \mathbf{-2n\beta^2 ((2n+1)\beta t)^{-\frac{4n+1}{2n+1}}}$.
20.The acceleration of a particle is defined by the relation $a = -k/x^2$. The particle starts with no initial velocity at $x = x_0$. Determine the velocity of the particle when $x = x_0/2$.
Sol: Using $a = v\frac{dv}{dx} = -\frac{k}{x^2}$. Integrate w.r.t $x$: $\int_0^v v dv = \int_{x_0}^{x_0/2} -k x^{-2} dx$.
$\left[ \frac{v^2}{2} \right]_0^v = -k \left[ -\frac{1}{x} \right]_{x_0}^{x_0/2} \implies \frac{v^2}{2} = k \left( \frac{1}{x_0/2} - \frac{1}{x_0} \right)$.
$\frac{v^2}{2} = k \left( \frac{2}{x_0} - \frac{1}{x_0} \right) = \frac{k}{x_0} \implies v^2 = \frac{2k}{x_0} \implies \mathbf{v = \sqrt{\frac{2k}{x_0}}}$.
21.A point moves rectilinearly with deceleration whose modulus depends on the velocity $v$ of the particle as $w = a\sqrt{v}$, where $a$ is a positive constant. At the initial moment, the velocity of the point is equal to $v_0$. What distance will it traverse before it stops? What time will it take to cover that distance?
Sol: Deceleration $\implies \frac{dv}{dt} = -a\sqrt{v}$.
Time: $\int_{v_0}^0 v^{-1/2} dv = \int_0^T -a dt \implies \left[ 2\sqrt{v} \right]_{v_0}^0 = -aT \implies 0 - 2\sqrt{v_0} = -aT \implies \mathbf{T = \frac{2\sqrt{v_0}}{a}}$.
Distance: $v\frac{dv}{dx} = -a\sqrt{v} \implies \int_{v_0}^0 v^{1/2} dv = \int_0^S -a dx \implies \left[ \frac{2}{3} v^{3/2} \right]_{v_0}^0 = -aS$
$0 - \frac{2}{3}v_0^{3/2} = -aS \implies \mathbf{S = \frac{2v_0^{3/2}}{3a}}$.
22.A point mass moves along the $x$-axis. At time $t=0$, $x=0$ and $v=v_0$. The acceleration is given by $a = -cx^3$ where $c$ is a positive constant. Find the maximum distance the point can move along the positive $x$-axis.
Sol: Max distance occurs when $v$ becomes 0.
$a = v\frac{dv}{dx} = -cx^3 \implies \int_{v_0}^0 v dv = \int_0^{x_{max}} -cx^3 dx$.
$\left[ \frac{v^2}{2} \right]_{v_0}^0 = -c \left[ \frac{x^4}{4} \right]_0^{x_{max}} \implies -\frac{v_0^2}{2} = -\frac{c x_{max}^4}{4}$.
$x_{max}^4 = \frac{2v_0^2}{c} \implies \mathbf{x_{max} = \left(\frac{2v_0^2}{c}\right)^{1/4}}$.
23.For a particle moving in a straight line, the relation between time $t$ and position $x$ is $t = \alpha x^2 + \beta x$. Show that the retardation of the particle is proportional to the cube of its instantaneous velocity. Evaluate the constant of proportionality.
Sol: Differentiate $t = \alpha x^2 + \beta x$ w.r.t $t$:
$1 = 2\alpha x \frac{dx}{dt} + \beta \frac{dx}{dt} \implies 1 = (2\alpha x + \beta) v \implies v = (2\alpha x + \beta)^{-1}$.
Differentiate again w.r.t $t$: $a = \frac{dv}{dt} = -1(2\alpha x + \beta)^{-2} \cdot \frac{d}{dt}(2\alpha x + \beta) = -(2\alpha x + \beta)^{-2} \cdot 2\alpha v$.
Since $(2\alpha x + \beta)^{-1} = v$, we substitute $(2\alpha x + \beta)^{-2} = v^2$:
$a = -(v^2)(2\alpha v) = -2\alpha v^3$.
Thus, retardation ($-a$) is proportional to $v^3$. Constant of proportionality is $\mathbf{2\alpha}$.
24.An object moving in a straight line experiences an acceleration $a = A \cos(\omega t)$ where $A$ and $\omega$ are constants. At $t=0$, $x=0$ and $v=v_0$. Find the position $x(t)$. Under what condition will the particle never reverse its direction?
Sol: Velocity $v(t) = \int A\cos(\omega t) dt = \frac{A}{\omega}\sin(\omega t) + C$. At $t=0, v=v_0 \implies C=v_0 \implies v(t) = v_0 + \frac{A}{\omega}\sin(\omega t)$.
Position $x(t) = \int (v_0 + \frac{A}{\omega}\sin(\omega t)) dt = v_0 t - \frac{A}{\omega^2}\cos(\omega t) + C'$.
At $t=0, x=0 \implies 0 = 0 - \frac{A}{\omega^2} + C' \implies C' = \frac{A}{\omega^2}$.
$\mathbf{x(t) = v_0 t + \frac{A}{\omega^2}(1 - \cos(\omega t))}$.
For no reversal, $v(t) \ge 0$ always $\implies v_0 + \frac{A}{\omega}\sin(\omega t) \ge 0$. Min value of sine is -1. Thus, $v_0 - \frac{A}{\omega} \ge 0 \implies \mathbf{v_0 \ge \frac{A}{\omega}}$.
25.Velocity $v$ of a particle moving along a straight line varies with distance $x$ as $v = a - bx$, where $a$ and $b$ are positive constants. Find the time at which the velocity becomes zero.
Sol: $v = \frac{dx}{dt} = a - bx \implies \int_0^x \frac{dx}{a-bx} = \int_0^t dt \implies \left[ -\frac{1}{b}\ln(a-bx) \right]_0^x = t$.
$-\frac{1}{b}(\ln(a-bx) - \ln a) = t \implies \ln\left(\frac{a-bx}{a}\right) = -bt \implies a-bx = a e^{-bt}$.
Substitute into velocity eq: $v = a - (a - ae^{-bt}) = ae^{-bt}$.
For velocity to become strictly zero, $ae^{-bt} = 0$, which occurs mathematically as $\mathbf{t \to \infty}$.
26.A particle is moving such that its velocity is $v = c \ln(x)$. Find the acceleration as a function of $x$. Also find the velocity as a function of time if $x(0) = x_0$.
Sol: Acceleration $a = v\frac{dv}{dx} = (c\ln x) \cdot \frac{d}{dx}(c\ln x) = (c\ln x)\left(\frac{c}{x}\right) = \mathbf{\frac{c^2 \ln x}{x}}$.
For $v(t)$: $\frac{dx}{dt} = c\ln x \implies \int_{x_0}^x \frac{dx}{\ln x} = \int_0^t c dt \implies ct = \int_{x_0}^x \frac{dx}{\ln x}$.
The integral of $1/\ln x$ is the special logarithmic integral function $\text{li}(x)$.
So, $ct = \text{li}(x) - \text{li}(x_0)$. You can implicitly determine $x(t)$ from this, and thus $v(t) = \mathbf{c \ln(x(t))}$.
27.Two cars A and B are simultaneously travelling on a straight road in the same direction. Car A starts from rest with acceleration $a_1$. Car B is travelling with constant velocity $u$. Find the maximum distance by which Car A lags behind Car B, and the time when A overtakes B.
Sol: Positions: $x_A(t) = \frac{1}{2}a_1 t^2$, $x_B(t) = ut$.
Lag $L(t) = x_B - x_A = ut - \frac{1}{2}a_1 t^2$.
To find max lag, set $dL/dt = 0 \implies u - a_1 t = 0 \implies t = \frac{u}{a_1}$.
Max lag distance $L_{max} = u(\frac{u}{a_1}) - \frac{1}{2}a_1(\frac{u}{a_1})^2 = \frac{u^2}{a_1} - \frac{u^2}{2a_1} = \mathbf{\frac{u^2}{2a_1}}$.
A overtakes B when $x_A = x_B \implies \frac{1}{2}a_1 t^2 = ut \implies \mathbf{t = \frac{2u}{a_1}}$ (since $t \ne 0$).
28.A train of length $l$ moves with a constant acceleration $a$. A person standing on a platform notes that the front of the train passes him with velocity $u$ and the rear with velocity $v$. What was the velocity of the train when exactly half of its length had passed the person?
Sol: From third equation of motion over total length $l$: $v^2 - u^2 = 2al \implies al = \frac{v^2 - u^2}{2}$.
Let velocity at midpoint be $v_m$. It covers distance $l/2$: $v_m^2 - u^2 = 2a(l/2) = al$.
Substitute $al$: $v_m^2 - u^2 = \frac{v^2 - u^2}{2} \implies v_m^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{2u^2 + v^2 - u^2}{2} = \frac{u^2 + v^2}{2}$.
Thus, $v_m = \mathbf{\sqrt{\frac{u^2 + v^2}{2}}}$.
29.A particle starts from rest and moves with uniform acceleration. It covers $x$ meters in the $p^{\text{th}}$ second and $y$ meters in the $q^{\text{th}}$ second. Prove that its acceleration is $\frac{2(x-y)}{p-q}$.
Sol: Distance in $n^{\text{th}}$ second: $S_n = u + \frac{a}{2}(2n - 1)$. Since starts from rest, $u=0$.
$x = \frac{a}{2}(2p - 1) \implies \frac{2x}{a} = 2p - 1 \quad \text{---(1)}$
$y = \frac{a}{2}(2q - 1) \implies \frac{2y}{a} = 2q - 1 \quad \text{---(2)}$
Subtracting (2) from (1): $\frac{2x}{a} - \frac{2y}{a} = (2p - 1) - (2q - 1) \implies \frac{2(x-y)}{a} = 2(p-q)$.
Rearranging for $a$: $a(p-q) = (x-y) \implies \mathbf{a = \frac{x-y}{p-q}}$.
(Note: Mathematical proof yields $\frac{x-y}{p-q}$. If key indicates $\frac{2(x-y)}{p-q}$, it contains an error with the factor of 2. Validated rigorously).
30.A car starting from rest accelerates at the rate $f$ through a distance $S$, then continues at constant speed for time $t$, and then decelerates at the rate $f/2$ to come to rest. If the total distance traversed is $15S$, then determine $S$ in terms of $f$ and $t$.
Sol: Phase 1 (Accel): Distance $S_1 = S$. Final velocity $v = \sqrt{2fS}$.
Phase 2 (Constant v): Speed $v$, time $t$. Distance $S_2 = v t = \sqrt{2fS} \cdot t$.
Phase 3 (Decel): Decelerates from $v$ to 0 at $f/2$. $0 - v^2 = 2(-f/2)S_3 \implies S_3 = v^2/f$. Since $v^2 = 2fS$, $S_3 = 2fS/f = 2S$.
Total distance $= S_1 + S_2 + S_3 = S + \sqrt{2fS}t + 2S = 3S + \sqrt{2fS}t$.
Given total is $15S$: $15S = 3S + \sqrt{2fS}t \implies 12S = \sqrt{2fS}t$.
Squaring both sides: $144 S^2 = 2fS t^2 \implies 72 S = f t^2 \implies \mathbf{S = \frac{1}{72} f t^2}$.
31.Two particles start moving simultaneously from the same point in the same direction. The first particle has initial velocity $u$ and uniform acceleration $a_1$. The second particle has initial velocity $0$ and uniform acceleration $a_2 > a_1$. Find the maximum separation between them before the second overtakes the first.
Sol: Position of P1: $x_1 = ut + \frac{1}{2}a_1 t^2$. Position of P2: $x_2 = \frac{1}{2}a_2 t^2$.
Separation $\Delta x = x_1 - x_2 = ut - \frac{1}{2}(a_2 - a_1)t^2$.
Max separation occurs when $\frac{d(\Delta x)}{dt} = 0 \implies u - (a_2 - a_1)t = 0 \implies t = \frac{u}{a_2 - a_1}$.
Substitute $t$ into $\Delta x$: $\Delta x_{max} = u\left(\frac{u}{a_2 - a_1}\right) - \frac{1}{2}(a_2 - a_1)\left(\frac{u}{a_2 - a_1}\right)^2 = \frac{u^2}{a_2 - a_1} - \frac{u^2}{2(a_2 - a_1)} = \mathbf{\frac{u^2}{2(a_2 - a_1)}}$.
32.A driver travelling at speed $V$ suddenly sees a wall at a distance $d$ directly in front of him. His reaction time is $t_r$. Find the minimum retardation required to avoid a collision.
Sol: Distance covered during reaction time: $d_{reaction} = V \cdot t_r$.
Distance remaining for braking $d_{brake} = d - V \cdot t_r$.
Using $v^2 - u^2 = 2as$: $0 - V^2 = 2(-a_{min})(d - V t_r)$.
$V^2 = 2a_{min}(d - V t_r) \implies \mathbf{a_{min} = \frac{V^2}{2(d - V t_r)}}$.
33.If a body loses half of its velocity on penetrating $3 \text{ cm}$ in a wooden block, how much will it penetrate more before coming to rest, assuming uniform resistance?
Sol: For first $3\text{ cm}$: $u_1 = u, v_1 = u/2, S = 3\text{ cm}$.
$v_1^2 - u_1^2 = 2aS \implies (u/2)^2 - u^2 = 2a(3) \implies \frac{u^2}{4} - u^2 = 6a \implies -\frac{3u^2}{4} = 6a \implies a = -\frac{u^2}{8}$.
For remaining penetration: $u_2 = u/2, v_2 = 0, S' = ?$
$0 - (u/2)^2 = 2(-\frac{u^2}{8})S' \implies -\frac{u^2}{4} = -\frac{u^2}{4} S' \implies \mathbf{S' = 1 \text{ cm}}$.
34.Two trains are moving on the same track in opposite directions at speeds $u_1$ and $u_2$. They are separated by distance $D$ when the drivers apply their brakes, producing retardations $a_1$ and $a_2$. Prove that they will avert a collision if $D > \frac{u_1^2}{2a_1} + \frac{u_2^2}{2a_2}$.
Sol: Minimum distance required for Train 1 to stop: $0 - u_1^2 = 2(-a_1)d_1 \implies d_1 = \frac{u_1^2}{2a_1}$.
Minimum distance required for Train 2 to stop: $0 - u_2^2 = 2(-a_2)d_2 \implies d_2 = \frac{u_2^2}{2a_2}$.
For no collision to occur, the total stopping distance must be strictly less than the initial separation $D$.
Thus, $D > d_1 + d_2 \implies \mathbf{D > \frac{u_1^2}{2a_1} + \frac{u_2^2}{2a_2}}$. Proved.
35.A point moves with uniform acceleration. In the $11^{\text{th}}$ and $15^{\text{th}}$ seconds it covers distances of $24 \text{ m}$ and $32 \text{ m}$ respectively. Find its initial velocity, acceleration, and the distance covered in the first $10$ seconds.
Sol: Using formula $S_n = u + \frac{a}{2}(2n - 1)$.
$S_{11} = 24 \implies u + \frac{a}{2}(21) = 24 \implies u + 10.5a = 24 \quad \text{---(1)}$
$S_{15} = 32 \implies u + \frac{a}{2}(29) = 32 \implies u + 14.5a = 32 \quad \text{---(2)}$
Subtract (1) from (2): $4a = 8 \implies \mathbf{a = 2 \text{ m/s}^2}$.
Sub $a$ into (1): $u + 10.5(2) = 24 \implies u + 21 = 24 \implies \mathbf{u = 3 \text{ m/s}}$.
Distance in first 10s: $S = ut + \frac{1}{2}at^2 = 3(10) + \frac{1}{2}(2)(100) = 30 + 100 = \mathbf{130 \text{ m}}$.
36.An anti-aircraft gun can fire bullets at $500 \text{ m/s}$. If a fighter plane is flying horizontally at an altitude of $2 \text{ km}$ with a speed of $1000 \text{ km/h}$, what should be the angle of elevation of the gun to hit the plane if the gun is fired exactly when the plane is directly overhead?
Sol: Plane speed $v_p = 1000 \times \frac{5}{18} = \frac{2500}{9} \text{ m/s}$. Bullet speed $v_b = 500 \text{ m/s}$.
Since plane is directly overhead when fired, the bullet must be fired forward at an angle so its horizontal velocity matches the plane's speed. Let $\theta$ be angle of elevation (from horizontal).
Horizontal velocity component $v_{bx} = v_b \cos\theta$.
$v_b \cos\theta = v_p \implies 500 \cos\theta = \frac{2500}{9} \implies \cos\theta = \frac{5}{9}$.
$\theta = \mathbf{\cos^{-1}(5/9)}$. (Which is equivalent to $\sin^{-1}(\frac{\sqrt{56}}{9})$).
37.A parachutist bails out from an aeroplane and after dropping through a distance of $40 \text{ m}$, he opens the parachute and decelerates at $2 \text{ m/s}^2$. If he reaches the ground with a speed of $2 \text{ m/s}$, how long was he in the air? At what height did he bail out? ($g=9.8 \text{ m/s}^2$)
Sol: Phase 1 (Free fall 40m): $v_1^2 = 2g(40) = 2(9.8)(40) = 784 \implies v_1 = 28\text{ m/s}$. Time $t_1 = v_1/g = 28/9.8 = \frac{20}{7} \approx 2.857\text{ s}$.
Phase 2 (Parachute): $u = 28\text{ m/s}$, $v = 2\text{ m/s}$, $a = -2\text{ m/s}^2$.
Time $t_2 = \frac{v - u}{a} = \frac{2 - 28}{-2} = 13\text{ s}$.
Distance $h_2 = \frac{v^2 - u^2}{2a} = \frac{4 - 784}{-4} = \frac{-780}{-4} = 195\text{ m}$.
Total Time = $t_1 + t_2 = 2.857 + 13 = \mathbf{15.86\text{ s}}$. Total Height = $40 + 195 = \mathbf{235\text{ m}}$. (Note: $234\text{m}$ in key may use $g \approx 10$ blending, $235\text{m}$ is exact for $g=9.8$).
38.A ball is dropped from the roof of a tower of height $h$. The total distance covered by it in the last second of its motion is equal to the distance covered by it in the first three seconds. What is the height $h$ of the tower? ($g=10 \text{ m/s}^2$)
Sol: Distance in first 3 seconds: $S_3 = \frac{1}{2}g(3)^2 = \frac{9}{2}g$.
Distance in last second (from $t-1$ to $t$): $S_{last} = \frac{1}{2}g(2t - 1)$.
Equating them: $\frac{1}{2}g(2t - 1) = \frac{9}{2}g \implies 2t - 1 = 9 \implies 2t = 10 \implies t = 5\text{ s}$.
Total height $h = \frac{1}{2}g t^2 = \frac{1}{2}(10)(5^2) = 5(25) = \mathbf{125 \text{ m}}$.
39.Two bodies begin to fall from the same height, but the second falls $T$ seconds after the first. Find the time (after the first body starts falling) at which the distance between the bodies equals $L$.
Sol: Let time elapsed for first body be $t$. Its descent $y_1 = \frac{1}{2}g t^2$.
Time for second body is $t-T$. Its descent $y_2 = \frac{1}{2}g(t-T)^2$.
Distance between them $L = y_1 - y_2 = \frac{1}{2}g [t^2 - (t^2 - 2tT + T^2)] = \frac{1}{2}g(2tT - T^2) = gT(t - T/2)$.
Solve for $t$: $L = gTt - \frac{1}{2}gT^2 \implies gTt = L + \frac{1}{2}gT^2 \implies \mathbf{t = \frac{L}{gT} + \frac{T}{2}}$.
40.A stone is dropped from a balloon ascending with a constant velocity of $12 \text{ m/s}$. The stone reaches the ground in $4 \text{ s}$. Find the height of the balloon at the instant the stone was dropped and the velocity of the stone right before it strikes the ground. ($g=9.8 \text{ m/s}^2$)
Sol: Initial velocity of stone $u = +12\text{ m/s}$ (upwards), $a = -g = -9.8\text{ m/s}^2$, $t = 4\text{ s}$.
Height $H$ of drop point is magnitude of net displacement $S$.
$S = ut + \frac{1}{2}at^2 = 12(4) - \frac{1}{2}(9.8)(4^2) = 48 - 4.9(16) = 48 - 78.4 = -30.4\text{ m}$. Height $= \mathbf{30.4 \text{ m}}$.
Velocity before strike: $v = u + at = 12 - 9.8(4) = 12 - 39.2 = -27.2\text{ m/s}$. Speed $= \mathbf{27.2 \text{ m/s}}$.
41.A ball is thrown vertically upward with speed $u$. If it is at a certain height $h$ at two different times $t_1$ and $t_2$, prove that $t_1 + t_2 = \frac{2u}{g}$ and $t_1 t_2 = \frac{2h}{g}$.
Sol: The equation of motion is $h = ut - \frac{1}{2}gt^2$.
Rearranging forms a quadratic in $t$: $\frac{1}{2}gt^2 - ut + h = 0$.
The two times $t_1, t_2$ are the roots of this quadratic equation.
From properties of roots:
Sum of roots $= t_1 + t_2 = \frac{-(-u)}{\frac{1}{2}g} = \mathbf{\frac{2u}{g}}$.
Product of roots $= t_1 t_2 = \frac{h}{\frac{1}{2}g} = \mathbf{\frac{2h}{g}}$. (Proved).
42.An elevator car is moving upward with uniform acceleration $a$. A passenger drops a coin from a height $h$ above the floor of the elevator. How long will the coin take to strike the floor?
Sol: Analyze from the non-inertial frame of the elevator.
Acceleration of elevator is $a$ upwards. Pseudo acceleration on coin is $a$ downwards.
Effective acceleration due to gravity $g_{eff} = g + a$ (downwards).
Initial relative velocity $u_{rel} = 0$. Distance to fall $S_{rel} = h$.
Using $S = ut + \frac{1}{2}a t^2 \implies h = 0 + \frac{1}{2}(g+a)t^2 \implies \mathbf{t = \sqrt{\frac{2h}{g+a}}}$.
43.A particle is thrown vertically upwards from the surface of the earth. Let $T_P$ be the time taken by the particle to travel from a point $P$ above the earth to its highest point and back to the point $P$. Similarly, let $T_Q$ be the time taken by the particle to travel from another point $Q$ above the earth to its highest point and back to $Q$. Prove that the distance $PQ = \frac{g(T_P^2 - T_Q^2)}{8}$.
Sol: For a particle at height $P$, time to go to max height and back is $T_P$. Time from max height down to $P$ is $T_P/2$.
Distance from max height down to $P$: $h_P = \frac{1}{2}g(T_P/2)^2 = \frac{g T_P^2}{8}$.
Similarly, for point $Q$, distance from max height down to $Q$: $h_Q = \frac{1}{2}g(T_Q/2)^2 = \frac{g T_Q^2}{8}$.
The distance between $P$ and $Q$ is the difference in their distances from the peak (assuming $P$ is lower than $Q$).
$PQ = h_P - h_Q = \frac{g T_P^2}{8} - \frac{g T_Q^2}{8} = \mathbf{\frac{g(T_P^2 - T_Q^2)}{8}}$. (Proved).
44.A stone is dropped from the top of a cliff. It is seen to pass a window $1.5 \text{ m}$ high in $0.1 \text{ s}$. How far above the top of the window was the stone dropped? ($g=10 \text{ m/s}^2$)
Sol: Let velocity of stone at the top of the window be $u$.
Passes window of $1.5\text{m}$ in $0.1\text{s}$: $S = ut + \frac{1}{2}gt^2 \implies 1.5 = u(0.1) + \frac{1}{2}(10)(0.1)^2 = 0.1u + 0.05$.
$0.1u = 1.45 \implies u = 14.5\text{ m/s}$.
Height $h$ above window where it was dropped (from rest): $v^2 = u_{initial}^2 + 2gh \implies (14.5)^2 = 0 + 2(10)h$.
$210.25 = 20h \implies h = \frac{210.25}{20} = \mathbf{10.5125 \text{ m}}$.
45.Water drops are falling from a nozzle of a shower onto the floor, from a height of $9.8 \text{ m}$. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of the second drop from the floor when the first drop strikes the floor.
Sol: Time for 1st drop to reach floor: $t = \sqrt{2H/g} = \sqrt{2(9.8)/9.8} = \sqrt{2}\text{ s}$.
Let the regular interval be $\Delta t$. When 1st drop hits, 3rd starts. The times elapsed for drops 1, 2, 3 are $2\Delta t, \Delta t, 0$.
Thus, $2\Delta t = \sqrt{2} \implies \Delta t = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \text{ s}$.
2nd drop has fallen for $\Delta t = 1/\sqrt{2}\text{ s}$.
Distance fallen by 2nd drop = $\frac{1}{2}g(\Delta t)^2 = \frac{1}{2}(9.8)\left(\frac{1}{\sqrt{2}}\right)^2 = 4.9(0.5) = 2.45\text{ m}$.
Height from floor = Total Height - Distance fallen = $9.8 - 2.45 = \mathbf{7.35 \text{ m}}$.
46.A body falling freely from a given height $H$ hits an inclined plane in its path at a height $h$. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. For what value of $h / H$ will the body take the maximum time to reach the ground?
Sol: Distance fallen before impact = $H - h$. Time taken $t_1 = \sqrt{\frac{2(H-h)}{g}}$.
Impact is elastic, speed remains $v = \sqrt{2g(H-h)}$, but direction is purely horizontal.
Thus, vertical velocity becomes 0. It now falls remaining height $h$ as a horizontal projectile.
Time taken for remainder $t_2 = \sqrt{\frac{2h}{g}}$. Total time $T = t_1 + t_2 = \sqrt{\frac{2}{g}}(\sqrt{H-h} + \sqrt{h})$.
To maximize $T$, set $dT/dh = 0 \implies \sqrt{\frac{2}{g}} \left( \frac{-1}{2\sqrt{H-h}} + \frac{1}{2\sqrt{h}} \right) = 0 \implies \sqrt{h} = \sqrt{H-h}$.
$h = H - h \implies 2h = H \implies \mathbf{h/H = 1/2}$.
47.Two straight railway tracks are at right angles to each other. At time $t=0$, a train A on one track is at distance $a$ from the intersection and moving towards it with speed $u$. Another train B on the other track is at distance $b$ from the intersection and moving away from it with speed $v$. Find the shortest distance between them and the time when they are closest.
Sol: Coordinate positions: $x_A = a - ut$ (towards), $y_B = b + vt$ (away).
Distance squared $S^2 = (a - ut)^2 + (b + vt)^2$.
Min distance when $\frac{dS^2}{dt} = 0 \implies 2(a - ut)(-u) + 2(b + vt)(v) = 0 \implies -au + u^2t + bv + v^2t = 0$.
Time $t = \frac{au - bv}{u^2 + v^2}$. (Assume $au > bv$ for valid positive time).
Substitute $t$ into $S^2$: $S_{min}^2 = (a - u(\frac{au-bv}{u^2+v^2}))^2 + (b + v(\frac{au-bv}{u^2+v^2}))^2$.
This simplifies algebraically to $S_{min}^2 = \frac{(av+bu)^2}{u^2+v^2} \implies \mathbf{S_{min} = \frac{|av + bu|}{\sqrt{u^2+v^2}}}$.
48.A passenger is running at maximum speed of $8 \text{ m/s}$ to catch a train. When he is distance $d$ from the door, the train starts from rest with a constant acceleration of $1 \text{ m/s}^2$. What is the maximum value of $d$ so that he can just board the train?
Sol: Passenger pos $x_p = 8t$. Train door pos $x_t = d + \frac{1}{2}(1)t^2 = d + 0.5t^2$.
To catch train, passenger must reach door: $x_p \ge x_t \implies 8t \ge d + 0.5t^2 \implies 0.5t^2 - 8t + d \le 0$.
For a real solution for $t$, the quadratic must have a non-negative discriminant: $b^2 - 4ac \ge 0$.
$(-8)^2 - 4(0.5)(d) \ge 0 \implies 64 - 2d \ge 0 \implies 2d \le 64 \implies \mathbf{d \le 32 \text{ m}}$.
49.Ship A is sailing towards the north-east with velocity $\vec{v} = 30\hat{i} + 50\hat{j} \text{ km/hr}$. Ship B is at a distance of $80 \text{ km}$ east and $150 \text{ km}$ north of Ship A and is sailing towards the west at $10 \text{ km/hr}$. A will be at minimum distance from B in what time?
Sol: $\vec{V}_A = 30\hat{i} + 50\hat{j}$. $\vec{V}_B = -10\hat{i}$. Relative velocity $\vec{V}_{BA} = \vec{V}_B - \vec{V}_A = -40\hat{i} - 50\hat{j}$.
Initial relative position $\vec{r}_{BA,0} = 80\hat{i} + 150\hat{j}$.
Time to closest approach $t = \frac{-\vec{r}_{BA,0} \cdot \vec{V}_{BA}}{|\vec{V}_{BA}|^2}$.
Dot product $= (80)(-40) + (150)(-50) = -3200 - 7500 = -10700$.
$|\vec{V}_{BA}|^2 = (-40)^2 + (-50)^2 = 1600 + 2500 = 4100$.
$t = \frac{-(-10700)}{4100} = \frac{107}{41} \approx \mathbf{2.6 \text{ hours}}$.
50.On a two-lane road, car A is travelling with a speed of $36 \text{ km/h}$. Two cars B and C approach car A in opposite directions with a speed of $54 \text{ km/h}$ each. At a certain instant, when the distance AB is equal to AC, both being $1 \text{ km}$, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Sol: $V_A = 36\text{ km/h} = 10\text{ m/s}$. $V_B = 54\text{ km/h} = 15\text{ m/s}$ (same direction as A). $V_C = -15\text{ m/s}$ (opposite).
Relative speed of C towards A = $15 - (-10)$? No, C is approaching A, A is moving towards C. Rel speed $= 15 + 10 = 25\text{ m/s}$.
Time for C to reach A $= \frac{1000\text{ m}}{25\text{ m/s}} = 40\text{ s}$.
B must overtake A in $\le 40\text{ s}$. Relative initial velocity of B wrt A $= 15 - 10 = 5\text{ m/s}$. Distance $= 1000\text{ m}$.
$S_{rel} = u_{rel}t + \frac{1}{2}at^2 \implies 1000 = 5(40) + \frac{1}{2}a(40)^2 \implies 1000 = 200 + 800a \implies 800 = 800a \implies \mathbf{a = 1 \text{ m/s}^2}$.
51.A lift starts ascending with constant acceleration $a$. A passenger throws a ball vertically up with a velocity $v_0$ relative to the lift. Find the time of flight of the ball and its maximum height relative to the lift floor.
Sol: Analyze from the non-inertial frame of the lift.
Downward pseudo acceleration $a$ acts on the ball. Effective gravity $g_{eff} = g + a$.
Time of flight is calculated using the relative effective gravity: $T = \mathbf{\frac{2v_0}{g+a}}$.
Maximum height relative to lift floor: $H = \frac{u_{rel}^2}{2 g_{eff}} = \mathbf{\frac{v_0^2}{2(g+a)}}$.
52.A conveyer belt moves with a constant velocity $v_0$. A block is gently placed on it. The coefficient of kinetic friction between the block and belt is $\mu_k$. Find the distance the block slides on the belt before coming to rest relative to the belt, and the total distance moved by the block relative to the ground.
Sol: Relative to belt, block has initial velocity $-v_0$. Retardation force is friction $\mu_k mg \implies a_{rel} = \mu_k g$.
Stopping distance relative to belt $S_{rel} = \frac{v_0^2}{2a_{rel}} = \mathbf{\frac{v_0^2}{2\mu_k g}}$.
Relative to ground, block starts from rest ($u=0$) and accelerates at $\mu_k g$ until it reaches speed $v_0$.
Time taken $t = \frac{v_0}{\mu_k g}$. Distance on ground $S_{grnd} = \frac{1}{2} (\mu_k g) t^2 = \frac{1}{2}(\mu_k g)\left(\frac{v_0}{\mu_k g}\right)^2 = \mathbf{\frac{v_0^2}{2\mu_k g}}$.
53.Two particles $P$ and $Q$ move in a straight line $AB$. $P$ starts from $A$ with velocity $u_1$ and acceleration $f_1$. Simultaneously $Q$ starts from $B$ with velocity $u_2$ and acceleration $f_2$ towards $A$. If they pass one another at the middle point of $AB$ and arrive at the other ends of $AB$ with equal velocities, prove that $(u_1 + u_2)(f_1 - f_2) = 8(f_1 u_2 - f_2 u_1)$.
Sol: Let total distance be $2S$. They meet at $S$. Since they start simultaneously and meet at midpoint, they take the same time $t$.
$S = u_1 t + \frac{1}{2}f_1 t^2$ and $S = u_2 t + \frac{1}{2}f_2 t^2 \implies (u_1 - u_2)t + \frac{1}{2}(f_1 - f_2)t^2 = 0 \implies t = \frac{2(u_2 - u_1)}{f_1 - f_2}$.
Final velocities at ends (dist $2S$) are equal: $v_1^2 = v_2^2 \implies u_1^2 + 2f_1(2S) = u_2^2 + 2f_2(2S) \implies u_1^2 - u_2^2 = 4S(f_2 - f_1)$.
From $S = u_1 t + \frac{1}{2}f_1 t^2$, substitute $t$: $S = \frac{2(u_2-u_1)}{f_1-f_2} [ u_1 + \frac{f_1(u_2-u_1)}{f_1-f_2} ] = 2\frac{u_2-u_1}{(f_1-f_2)^2} [ u_1 f_1 - u_1 f_2 + f_1 u_2 - f_1 u_1 ] = 2\frac{u_2-u_1}{(f_1-f_2)^2} [f_1 u_2 - u_1 f_2]$.
Substitute $S$ into $u_1^2 - u_2^2 = -4S(f_1 - f_2)$: $(u_1-u_2)(u_1+u_2) = -4(f_1-f_2) \left[ \frac{2(u_2-u_1)[f_1u_2 - f_2u_1]}{(f_1-f_2)^2} \right]$.
Cancel $(u_1-u_2)$ and one $(f_1-f_2)$ term, the minus sign absorbs $u_2-u_1$: $(u_1+u_2) = \frac{8(f_1u_2 - f_2u_1)}{f_1-f_2} \implies \mathbf{(u_1+u_2)(f_1-f_2) = 8(f_1u_2 - f_2u_1)}$.
54.An open elevator is ascending with constant acceleration $a=1.2 \text{ m/s}^2$. A boy in the elevator throws a ball vertically upwards with a speed $v_0=5 \text{ m/s}$ relative to the elevator. If the initial height of the ball from the elevator floor is $2 \text{ m}$, find how long the ball takes to hit the floor. ($g=9.8 \text{ m/s}^2$)
Sol: Work in elevator frame. Pseudo gravity $g_{eff} = 9.8 + 1.2 = 11.0\text{ m/s}^2$ (downwards).
Initial velocity relative to floor $u = +5\text{ m/s}$ (up). Displacement $S = -2\text{ m}$ (hits floor below throw point).
$S = ut - \frac{1}{2}g_{eff}t^2 \implies -2 = 5t - \frac{1}{2}(11)t^2 \implies 5.5t^2 - 5t - 2 = 0$.
$t = \frac{5 + \sqrt{25 - 4(5.5)(-2)}}{2(5.5)} = \frac{5 + \sqrt{25 + 44}}{11} = \frac{5 + \sqrt{69}}{11} \approx \frac{5 + 8.306}{11} = \frac{13.306}{11} \approx \mathbf{1.21 \text{ s}}$.
(Note: Original answer key stating 0.81s contains an algebraic error, verified mathematically as 1.21s).
55.The graph of $1/v$ versus displacement $x$ is a straight line passing through the origin with a slope $k$. Determine the relation between acceleration $a$ and velocity $v$.
Sol: Equation of the line: $\frac{1}{v} = kx \implies v = \frac{1}{kx}$.
By definition $a = v\frac{dv}{dx}$.
$\frac{dv}{dx} = \frac{d}{dx}(kx)^{-1} = -(kx)^{-2} \cdot k = -\frac{k}{k^2 x^2} = -\frac{1}{k x^2}$.
Substitute into acceleration: $a = \left(\frac{1}{kx}\right) \left(-\frac{1}{k x^2}\right) = -\frac{1}{k^2 x^3}$.
Since $v = \frac{1}{kx}$, then $v^3 = \frac{1}{k^3 x^3} \implies \frac{1}{x^3} = k^3 v^3$.
Thus, $a = -\frac{1}{k^2} (k^3 v^3) = \mathbf{-k v^3}$.
56.An acceleration-displacement ($a-x$) graph for a particle moving in a straight line is given as a straight line passing through $(0, a_0)$ and $(x_0, 0)$. If the particle starts from rest at $x=0$, find its velocity as a function of $x$.
Sol: Equation of the straight line $a(x) = a_0 - \left(\frac{a_0}{x_0}\right)x$.
Use $a = v\frac{dv}{dx} \implies \int_0^v v dv = \int_0^x a(x) dx \implies \frac{v^2}{2} = \int_0^x \left(a_0 - \frac{a_0}{x_0}x\right) dx$.
$\frac{v^2}{2} = a_0 x - \frac{a_0}{2x_0}x^2 = \frac{a_0}{2} \left(2x - \frac{x^2}{x_0}\right)$.
$v^2 = a_0 \left(2x - \frac{x^2}{x_0}\right) \implies \mathbf{v = \sqrt{a_0 x \left(2 - \frac{x}{x_0}\right)}}$.
57.The velocity-displacement ($v-x$) graph of a particle moving in a straight line is a straight line intercepting the $v$-axis at $v_0$ and the $x$-axis at $x_0$. Prove that the acceleration of the particle varies linearly with displacement $x$, and find its maximum magnitude.
Sol: Equation of the line is $v(x) = v_0 - \left(\frac{v_0}{x_0}\right)x$.
Acceleration $a = v\frac{dv}{dx} = \left[v_0 - \left(\frac{v_0}{x_0}\right)x\right] \left( -\frac{v_0}{x_0} \right) = -\frac{v_0^2}{x_0} + \left(\frac{v_0^2}{x_0^2}\right)x$.
This represents a linear relationship (form $y = mx + c$).
Maximum magnitude of $a$ occurs at $x=0$ (since at $x=x_0$, $a=0$).
$|a_{max}| = \left|-\frac{v_0^2}{x_0} + 0\right| = \mathbf{\frac{v_0^2}{x_0}}$.
58.A particle's velocity $v$ squared ($v^2$) is plotted against displacement $x$ resulting in a straight line with a positive slope $m$ and y-intercept $C$. What are the physical significances of $m$ and $C$? What is the acceleration of the particle?
Sol: Equation from graph: $v^2 = mx + C$.
Comparing with the 3rd equation of kinematics: $v^2 = u^2 + 2ax$.
By direct comparison, the y-intercept $C$ represents the square of initial velocity ($u^2$).
The slope $m$ represents twice the acceleration ($2a$).
Therefore, the acceleration of the particle is $a = \mathbf{m/2}$.
59.A particle starts from rest and undergoes an acceleration $a$ that varies with time $t$ as shown in an $a-t$ graph which is an isosceles triangle with base on the time axis from $t=0$ to $t=2t_0$ and peak $a_0$ at $t=t_0$. Calculate the maximum velocity and total distance covered by the particle in time $2t_0$.
Sol: Max velocity = Total area under $a-t$ curve (since acceleration is always positive, velocity keeps increasing).
Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(2t_0)(a_0) = \mathbf{a_0 t_0}$.
Distance requires integrating $v(t)$. Since $a(t)$ is symmetric and starts/ends at 0, $v(t)$ builds up to $a_0 t_0$.
In first half, $a(t) = a_0 \frac{t}{t_0} \implies v(t) = \frac{a_0 t^2}{2t_0} \implies S_1 = \int_0^{t_0} \frac{a_0 t^2}{2t_0} dt = \frac{a_0 t_0^2}{6}$.
For the second half, integrating symmetrically gives total area under $v-t$ curve. Total Distance $S = \mathbf{a_0 t_0^2}$.
60.If the position of a particle varies with time as $x = A \sin(\omega t)$, sketch the graphs of $x-t$, $v-t$, $a-t$ and $a-x$. State the nature of the $v-x$ graph.
Sol:
$x = A\sin(\omega t)$
$v = \frac{dx}{dt} = A\omega\cos(\omega t)$
$a = \frac{dv}{dt} = -A\omega^2\sin(\omega t) = -\omega^2 x$.
Nature of $v-x$ graph: $\frac{x^2}{A^2} + \frac{v^2}{A^2\omega^2} = \sin^2(\omega t) + \cos^2(\omega t) = 1$. This is an
Ellipse.
