Uniformly Accelerated Motion Questions in English

(A) Let $u$ be the initial velocity of the body.
Distance covered in $2 \, s$ is given by $S = ut + \frac{1}{2}gt^2$.
$S = u(2) + \frac{1}{2}(10)(2)^2 = 2u + 20$ ... $(i)$
Distance covered in the $3^{rd}$ second is given by $S_{3^{rd}} = u + \frac{g}{2}(2n - 1)$,where $n = 3$.
$S_{3^{rd}} = u + \frac{10}{2}(2 \times 3 - 1) = u + 5(5) = u + 25$ ... (ii)
According to the problem,the distance covered in $2 \, s$ is equal to the distance covered in the $3^{rd}$ second,so $S = S_{3^{rd}}$.
$2u + 20 = u + 25 \Rightarrow u = 5 \, m/s$.
Substituting $u = 5$ into equation $(i)$:
$S = 2(5) + 20 = 10 + 20 = 30 \, m$.

(C) Given acceleration $a = \frac{dv}{dt} = bt$.
Integrating with respect to time $t$:
$\int dv = \int bt \, dt \Rightarrow v = \frac{1}{2}bt^2 + C_1$.
At $t = 0$,$v = v_0$,so $C_1 = v_0$.
Thus,the velocity is $v = \frac{1}{2}bt^2 + v_0$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \frac{1}{2}bt^2 + v_0$.
Integrating with respect to time $t$:
$x = \int (\frac{1}{2}bt^2 + v_0) dt = \frac{1}{2}b(\frac{t^3}{3}) + v_0t + C_2$.
At $t = 0$,$x = 0$,so $C_2 = 0$.
Therefore,the distance travelled is $x = v_0t + \frac{1}{6}bt^3$.

(D) Given the differential equation: $\frac{dv}{dt} = 6 - 3v$.
Rearranging the terms: $\frac{dv}{6 - 3v} = dt$.
Integrating both sides: $\int \frac{dv}{6 - 3v} = \int dt$.
This yields: $-\frac{1}{3} \ln(6 - 3v) = t + C$.
$\ln(6 - 3v) = -3t + C'$.
At $t = 0$,$v = 0$,so $\ln(6) = C'$.
Substituting $C'$: $\ln(6 - 3v) = -3t + \ln(6) \Rightarrow \ln(\frac{6 - 3v}{6}) = -3t$.
Taking the exponential: $\frac{6 - 3v}{6} = e^{-3t} \Rightarrow 1 - 0.5v = e^{-3t} \Rightarrow v(t) = 2(1 - e^{-3t}) \, m/s$.
Terminal speed $(t \to \infty)$: $v = 2(1 - 0) = 2.0 \, m/s$.
Initial acceleration $(t = 0)$: $a = \frac{dv}{dt} = 6 - 3(0) = 6.0 \, m/s^2$.
Since all statements are correct,the answer is $(d)$.

(D) Let the car accelerate at a constant rate $\alpha$ for time $t_1$. The maximum velocity $v$ attained is given by $v = u + \alpha t_1$. Since it starts from rest,$u = 0$,so $v = \alpha t_1$.
Next,the car decelerates at a constant rate $\beta$ for the remaining time $(t - t_1)$ until it comes to rest. Using the equation $v = u + at$ for the deceleration phase,we have $0 = v - \beta(t - t_1)$.
Substituting $v = \alpha t_1$ into the equation,we get $0 = \alpha t_1 - \beta t + \beta t_1$.
Rearranging the terms,we get $(\alpha + \beta) t_1 = \beta t$,which gives $t_1 = \frac{\beta}{\alpha + \beta} t$.
Substituting $t_1$ back into the expression for $v$,we get $v = \alpha \left( \frac{\beta}{\alpha + \beta} t \right) = \frac{\alpha \beta}{\alpha + \beta} t$.

(D) The stopping distance $S$ of a vehicle is given by the relation $v^2 = u^2 + 2as$. Since the final velocity $v = 0$,we have $0 = u^2 - 2a|S|$,which implies $S = \frac{u^2}{2|a|}$.
Since the deceleration $a$ is constant,the stopping distance is directly proportional to the square of the initial velocity: $S \propto u^2$.
Given $S_1 = 20\,m$ at $u_1 = 60\,km/h$ and $u_2 = 120\,km/h$.
Using the ratio: $\frac{S_2}{S_1} = \left(\frac{u_2}{u_1}\right)^2 = \left(\frac{120}{60}\right)^2 = (2)^2 = 4$.
Therefore,$S_2 = 4 \times S_1 = 4 \times 20\,m = 80\,m$.

(C) Let the thickness of one plank be $s$. If the bullet enters with velocity $u$,then it leaves with velocity $v = u - \frac{u}{20} = \frac{19}{20}u$.
Using the equation of motion $v^2 = u^2 - 2as$,where $a$ is the retardation:
$(\frac{19}{20}u)^2 = u^2 - 2as$
$\frac{361}{400}u^2 = u^2 - 2as$
$2as = u^2 - \frac{361}{400}u^2 = \frac{39}{400}u^2$
$\frac{u^2}{2as} = \frac{400}{39} \approx 10.25$
Now,if $n$ planks are required to stop the bullet,the final velocity $v_f = 0$ after traveling a total distance $ns$:
$0^2 = u^2 - 2a(ns)$
$u^2 = 2ans$
$n = \frac{u^2}{2as} = \frac{400}{39} \approx 10.25$
Since the number of planks must be an integer and $10$ planks are not enough to stop the bullet,we need $11$ planks.

(C) The stopping distance $s$ of a vehicle is given by the formula $s = \frac{u^2}{2a}$,where $u$ is the initial speed and $a$ is the magnitude of deceleration.
From this relation,we can see that the stopping distance is directly proportional to the square of the initial speed,i.e.,$s \propto u^2$.
If the speed is tripled,the new speed becomes $u' = 3u$.
The new stopping distance $s'$ will be $s' \propto (3u)^2 = 9u^2$.
Therefore,$s' = 9s$.
Thus,tripling the speed increases the stopping distance by a factor of $9$.

(A) Let the initial velocity be $u$. After penetrating $s_1 = 3 \, cm$,the velocity becomes $v_1 = u/2$.
Using the equation of motion $v^2 = u^2 + 2as$:
$(u/2)^2 = u^2 + 2a(3)$
$u^2/4 = u^2 + 6a$
$6a = -3u^2/4$
$a = -u^2/8$
Now,for the second part,the initial velocity is $u/2$ and the final velocity is $0$. Let the additional distance covered be $x$.
$0^2 = (u/2)^2 + 2ax$
$0 = u^2/4 + 2(-u^2/8)x$
$u^2/4 = (u^2/4)x$
$x = 1 \, cm$.

(D) Given the relation: $t = \sqrt{x} + 3$.
Rearranging for $x$: $\sqrt{x} = t - 3$.
Squaring both sides: $x = (t - 3)^2$.
Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(t - 3)^2 = 2(t - 3)$.
When the velocity is zero: $2(t - 3) = 0$,which gives $t = 3 \ s$.
Initial position at $t = 0 \ s$: $x_i = (0 - 3)^2 = 9 \ m$.
Final position at $t = 3 \ s$: $x_f = (3 - 3)^2 = 0 \ m$.
Displacement is defined as the change in position: $\Delta x = x_f - x_i = 0 - 9 = -9 \ m$.

(A) Let the initial velocity be $u$ and the thickness of one plank be $s$. After passing through one plank,the velocity becomes $v = u - (1/20)u = (19/20)u$.
Using the equation of motion $v^2 = u^2 + 2as$:
$(19/20u)^2 = u^2 + 2as$
$2as = (361/400)u^2 - u^2 = -(39/400)u^2$.
Let $n$ be the minimum number of planks required to stop the bullet. For the final velocity to be $0$ after passing through $n$ planks,the total distance covered is $ns$:
$0^2 = u^2 + 2a(ns)$
$0 = u^2 + n(2as)$
Substituting $2as = -(39/400)u^2$:
$0 = u^2 + n(-(39/400)u^2)$
$n(39/400) = 1$
$n = 400/39 \approx 10.26$.
Since the number of planks must be an integer,the minimum number of planks required to stop the bullet is $11$.

(C) The position of the particle is given by $x = at^2 - bt^3$.
The velocity $v$ is the first derivative of position with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt}(at^2 - bt^3) = 2at - 3bt^2$.
The acceleration $a_{acc}$ is the derivative of velocity with respect to time:
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(2at - 3bt^2) = 2a - 6bt$.
To find the time when acceleration is zero,set $a_{acc} = 0$:
$2a - 6bt = 0$
$2a = 6bt$
$t = \frac{2a}{6b} = \frac{a}{3b}$.

(C) The position of the particle is given by $y = a + bt + ct^2 - dt^4$.
The velocity $v$ is the first derivative of position with respect to time $t$:
$v = \frac{dy}{dt} = \frac{d}{dt}(a + bt + ct^2 - dt^4) = 0 + b + 2ct - 4dt^3$.
At the initial time $t = 0$,the initial velocity $v_{initial}$ is:
$v_{initial} = b + 2c(0) - 4d(0)^3 = b$.
The acceleration $a_{acc}$ is the derivative of velocity with respect to time $t$:
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(b + 2ct - 4dt^3) = 0 + 2c - 12dt^2$.
At the initial time $t = 0$,the initial acceleration $a_{initial}$ is:
$a_{initial} = 2c - 12d(0)^2 = 2c$.
Thus,the initial velocity is $b$ and the initial acceleration is $2c$.

(A) Given the relation: $t = \alpha x^2 + \beta x$.
Differentiating with respect to $x$: $\frac{dt}{dx} = 2\alpha x + \beta$.
Since velocity $v = \frac{dx}{dt}$,we have $v = \frac{1}{2\alpha x + \beta}$.
Acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.
Differentiating $v = (2\alpha x + \beta)^{-1}$ with respect to $x$: $\frac{dv}{dx} = -1(2\alpha x + \beta)^{-2} \cdot (2\alpha) = -\frac{2\alpha}{(2\alpha x + \beta)^2}$.
Substituting this into the acceleration formula: $a = v \cdot \left( -\frac{2\alpha}{(2\alpha x + \beta)^2} \right)$.
Since $v = \frac{1}{2\alpha x + \beta}$,then $v^2 = \frac{1}{(2\alpha x + \beta)^2}$.
Therefore,$a = v \cdot (-2\alpha \cdot v^2) = -2\alpha v^3$.
The retardation is the magnitude of negative acceleration,which is $2\alpha v^3$.

(A) Let $s_A$ be the displacement of object $A$ and $s_B$ be the displacement of object $B$ at time $t$.
For object $A$,initial velocity $u_A = 0$ and acceleration is $a$. Using the equation of motion $s = ut + (1/2)at^2$,we get $s_A = (1/2)at^2$.
For object $B$,velocity is constant $v$. Thus,$s_B = vt$.
Since both objects meet at the same position at time $t$,their displacements must be equal: $s_A = s_B$.
Therefore,$(1/2)at^2 = vt$.
Dividing both sides by $t$ (assuming $t \neq 0$),we get $(1/2)at = v$.
Solving for $t$,we find $t = 2v/a$.

(D) Using the equation of motion $v^2 = u^2 + 2as$,where $v = 0$ (final velocity),$u$ is initial velocity,$a$ is retardation,and $s$ is the stopping distance.
Since $0 = u^2 - 2as$,we get $s = \frac{u^2}{2a}$.
Assuming the retardation $a$ is constant,the stopping distance $s$ is directly proportional to the square of the initial velocity: $s \propto u^2$.
Therefore,$\frac{s_2}{s_1} = \left( \frac{u_2}{u_1} \right)^2$.
Given $u_1 = 50 \, km/hr$,$s_1 = 6 \, m$,and $u_2 = 100 \, km/hr$.
$\frac{s_2}{6} = \left( \frac{100}{50} \right)^2 = (2)^2 = 4$.
$s_2 = 4 \times 6 = 24 \, m$.

(D) Given:
Initial velocity $u = 200\; m/s$
Final velocity $v = 100\; m/s$
Distance $s = 10\; cm = 0.1\; m$
Using the third equation of motion: $v^2 = u^2 + 2as$
$a = \frac{v^2 - u^2}{2s}$
$a = \frac{(100)^2 - (200)^2}{2 \times 0.1}$
$a = \frac{10000 - 40000}{0.2} = \frac{-30000}{0.2} = -150000\; m/s^2$
Retardation is the magnitude of negative acceleration,so retardation $= 15 \times 10^4\; m/s^2$.

(A) The distance covered by an object in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
Given that the body starts from rest,the initial velocity $u = 0\,m/s$.
The acceleration $a = 8\,m/s^2$ and the time $n = 5$.
Substituting these values into the formula:
$S_5 = 0 + \frac{8}{2}(2 \times 5 - 1)$
$S_5 = 4 \times (10 - 1)$
$S_5 = 4 \times 9 = 36\,m$.
Therefore,the distance covered in the $5^{th}$ second is $36\,m$.

(C) Using the third equation of motion: $v^2 = u^2 - 2as$.
Since the car stops,the final velocity $v = 0$.
Therefore,$0 = u^2 - 2as$,which gives $s = \frac{u^2}{2a}$.
This implies that the stopping distance $s$ is directly proportional to the square of the initial velocity,i.e.,$s \propto u^2$.
Let the initial velocity be $u_1 = V$ and the stopping distance be $s_1 = 20 \ m$.
When the velocity is doubled,$u_2 = 2V$.
Then,$\frac{s_2}{s_1} = \left( \frac{u_2}{u_1} \right)^2 = \left( \frac{2V}{V} \right)^2 = 4$.
Thus,$s_2 = 4 \times s_1 = 4 \times 20 \ m = 80 \ m$.

(A) The distance $x$ covered by an object is given by the integral of velocity with respect to time: $x = \int_{t_1}^{t_2} v \, dt$.
Given $v = kt$ and $k = 2 \, m/s^2$,we integrate from $t = 0$ to $t = 3 \, s$.
$x = \int_{0}^{3} 2t \, dt$.
$x = 2 \left[ \frac{t^2}{2} \right]_0^3$.
$x = [t^2]_0^3 = 3^2 - 0^2 = 9 \, m$.
Thus,the distance covered is $9 \, m$.

(C) Given acceleration $a = \frac{dv}{dt} = bt$.
Integrating with respect to time to find velocity $v$:
$\int_{v_0}^{v} dv = \int_{0}^{t} bt \, dt$
$v - v_0 = \frac{1}{2}bt^2$
$v = v_0 + \frac{1}{2}bt^2$
Now,since $v = \frac{ds}{dt}$,we integrate to find displacement $s$:
$\int_{0}^{s} ds = \int_{0}^{t} (v_0 + \frac{1}{2}bt^2) dt$
$s = [v_0t + \frac{1}{2}b \cdot \frac{t^3}{3}]_0^t$
$s = v_0t + \frac{1}{6}bt^3$.

(D) The velocity of the particle is given by $v = \frac{ds}{dt} = at$.
To find the distance $s$ covered in the first $4 \, s$,we integrate the velocity with respect to time from $t = 0$ to $t = 4 \, s$:
$s = \int_{0}^{4} v \, dt = \int_{0}^{4} at \, dt$
$s = a \left[ \frac{t^2}{2} \right]_{0}^{4}$
$s = a \left( \frac{4^2}{2} - \frac{0^2}{2} \right) = a \left( \frac{16}{2} \right) = 8a$.
Thus,the distance covered is $8a$.

(D) Using the third equation of motion: $v^2 = u^2 + 2as$.
Since the final velocity $v = 0$,we have $0 = u^2 - 2as$,which implies $s = \frac{u^2}{2a}$.
Assuming the deceleration $a$ is constant,the stopping distance $s$ is directly proportional to the square of the initial velocity: $s \propto u^2$.
Therefore,$\frac{s_2}{s_1} = \left( \frac{u_2}{u_1} \right)^2$.
Given $u_1 = 30 \, km/hr$,$s_1 = 8 \, m$,and $u_2 = 60 \, km/hr$:
$\frac{s_2}{8} = \left( \frac{60}{30} \right)^2 = (2)^2 = 4$.
$s_2 = 4 \times 8 = 32 \, m$.

(A) The position of the particle is given by $x = 40 + 12t - t^3$.
Velocity $v$ is the rate of change of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(40 + 12t - t^3) = 12 - 3t^2$.
For the particle to come to rest,its velocity must be zero: $v = 0$.
$12 - 3t^2 = 0 \implies 3t^2 = 12 \implies t^2 = 4 \implies t = 2 \text{ s}$ (since time cannot be negative).
The distance traveled by the particle from $t = 0$ to $t = 2$ is given by the change in position:
$x(0) = 40 + 12(0) - (0)^3 = 40 \text{ m}$.
$x(2) = 40 + 12(2) - (2)^3 = 40 + 24 - 8 = 56 \text{ m}$.
Distance traveled = $x(2) - x(0) = 56 - 40 = 16 \text{ m}$.

(C) Given: At time $t = 0$,velocity $v = 0$.
Acceleration $f = f_0(1 - t/T)$.
At the instant when $f = 0$,we have $0 = f_0(1 - t/T)$. Since $f_0$ is a constant,$1 - t/T = 0$,which implies $t = T$.
We know that acceleration $f = \frac{dv}{dt}$,so $dv = f dt$.
Integrating both sides from $t = 0$ to $t = T$:
$\int_{0}^{v_x} dv = \int_{0}^{T} f_0(1 - t/T) dt$
$v_x = f_0 \left[ t - \frac{t^2}{2T} \right]_{0}^{T}$
$v_x = f_0 \left( T - \frac{T^2}{2T} \right) = f_0 \left( T - \frac{T}{2} \right) = \frac{1}{2} f_0 T$.

(C) Given that the particle starts from rest,the initial velocity $u = 0$.
Since the force is constant,the acceleration $a$ is also constant.
The distance $S$ covered in time $t$ is given by the equation of motion: $S = ut + \frac{1}{2}at^2$.
For $t = 10 \ s$,the distance $S_1 = 0(10) + \frac{1}{2}a(10)^2 = 50a$.
For $t = 20 \ s$,the distance $S_2 = 0(20) + \frac{1}{2}a(20)^2 = 200a$.
Comparing the two,$S_2 = 200a = 4(50a) = 4S_1$.
Therefore,$S_2 = 4S_1$.

(B) Given: Initial velocity $u = 10 \, m/s$,final velocity $v = 20 \, m/s$,and distance $s = 135 \, m$.
Using the equation of motion $v^2 - u^2 = 2as$:
$(20)^2 - (10)^2 = 2 \times a \times 135$
$400 - 100 = 270a$
$300 = 270a$
$a = \frac{300}{270} = \frac{10}{9} \, m/s^2$.
Now,using the equation $v = u + at$ to find $t$:
$20 = 10 + (\frac{10}{9})t$
$10 = (\frac{10}{9})t$
$t = 9 \, s$.

(A) The distance travelled by a particle in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
Given,initial velocity $u = 0 \ m/s$,acceleration $a = \frac{4}{3} \ m/s^2$,and time $n = 3 \ s$.
Substituting these values into the formula:
$S_3 = 0 + \frac{4/3}{2}(2(3) - 1)$
$S_3 = \frac{4}{6}(6 - 1)$
$S_3 = \frac{2}{3}(5)$
$S_3 = \frac{10}{3} \ m$.
Therefore,the distance travelled in the third second is $\frac{10}{3} \ m$.

(A) Given the position equation: $x = (t + 5)^{-1}$ ... $(i)$
Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(t + 5)^{-1} = -(t + 5)^{-2}$ ... (ii)
Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}[-(t + 5)^{-2}] = 2(t + 5)^{-3}$ ... (iii)
From equation (ii),we have $v = -(t + 5)^{-2}$,which implies $v^{3/2} = [-(t + 5)^{-2}]^{3/2} = -(t + 5)^{-3}$ (taking magnitude or considering the relation).
Substituting this into equation (iii),we get $a = -2v^{3/2}$.
Thus,the acceleration is proportional to $(velocity)^{3/2}$.

(D) Given the position equation: $x = 8 + 12t - t^3$.
Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(8 + 12t - t^3) = 12 - 3t^2$.
When the velocity becomes zero: $12 - 3t^2 = 0 \implies 3t^2 = 12 \implies t^2 = 4 \implies t = 2 \, s$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(12 - 3t^2) = -6t$.
At $t = 2 \, s$,the acceleration is $a = -6(2) = -12 \, m/s^2$.
Retardation is the magnitude of negative acceleration,so retardation = $12 \, m/s^2$.

(B) The velocity of the particle is given by $v = At + Bt^2$.
Since $v = \frac{ds}{dt}$,we have $ds = (At + Bt^2) dt$.
To find the distance travelled between $t = 1 \ s$ and $t = 2 \ s$,we integrate the velocity with respect to time:
$s = \int_{1}^{2} (At + Bt^2) dt$
$s = \left[ \frac{At^2}{2} + \frac{Bt^3}{3} \right]_{1}^{2}$
$s = \left( \frac{A(2)^2}{2} + \frac{B(2)^3}{3} \right) - \left( \frac{A(1)^2}{2} + \frac{B(1)^3}{3} \right)$
$s = \left( 2A + \frac{8B}{3} \right) - \left( \frac{A}{2} + \frac{B}{3} \right)$
$s = (2A - \frac{A}{2}) + (\frac{8B}{3} - \frac{B}{3})$
$s = \frac{3}{2}A + \frac{7}{3}B$
Since the velocity does not change sign in the interval $[1, 2]$,the distance is equal to the magnitude of displacement.

(C) Using the equation of motion $v^2 = u^2 - 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the retardation,and $s$ is the distance.
Since the bullet comes to rest,$v = 0$.
Therefore,$0 = u^2 - 2as$,which implies $u^2 = 2as$.
Assuming the retardation $a$ is constant for the same block,we get $u^2 \propto s$.
This gives the ratio: $\frac{u_2^2}{u_1^2} = \frac{s_2}{s_1}$.
Given $u_1 = 200 \, cm/s$,$s_1 = 4 \, cm$,and $s_2 = 9 \, cm$.
Substituting the values: $\frac{u_2^2}{200^2} = \frac{9}{4}$.
Taking the square root on both sides: $\frac{u_2}{200} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
$u_2 = 200 \times \frac{3}{2} = 300 \, cm/s$.

(D) Using the equation of motion $v^2 = u^2 + 2as$:
For the first distance $S$,the final velocity is $2u$:
$(2u)^2 = u^2 + 2aS \Rightarrow 4u^2 = u^2 + 2aS \Rightarrow 2aS = 3u^2$.
Now,for the total distance $2S$,let the final velocity be $v'$:
$(v')^2 = u^2 + 2a(2S) = u^2 + 2(2aS)$.
Substituting $2aS = 3u^2$ into the equation:
$(v')^2 = u^2 + 2(3u^2) = u^2 + 6u^2 = 7u^2$.
Therefore,$v' = \sqrt{7} u$.

(D) Given that the body starts from rest,initial velocity $u = 0$.
Let the uniform acceleration be $a$.
After $n$ seconds,the velocity is $\upsilon = u + an = 0 + an$,so $a = \frac{\upsilon}{n}$.
The displacement in $n$ seconds is $S_n = \frac{1}{2}an^2$.
The displacement in $(n - 2)$ seconds is $S_{n-2} = \frac{1}{2}a(n - 2)^2$.
The displacement in the last two seconds is $\Delta S = S_n - S_{n-2}$.
$\Delta S = \frac{1}{2}an^2 - \frac{1}{2}a(n - 2)^2 = \frac{a}{2}[n^2 - (n - 2)^2]$.
Using the identity $x^2 - y^2 = (x - y)(x + y)$:
$\Delta S = \frac{a}{2}[n - (n - 2)][n + (n - 2)] = \frac{a}{2}[2][2n - 2] = a(2n - 2)$.
Substituting $a = \frac{\upsilon}{n}$:
$\Delta S = \frac{\upsilon}{n}(2n - 2) = \frac{2\upsilon(n - 1)}{n}$.

(B) Let the acceleration be $a$. The point starts from rest at $A$ and moves with acceleration $a$ for time $t$ to reach point $B$.
Velocity at $B$ is $v = at$.
Distance $S_{AB} = \frac{1}{2}at^2$.
At $B$,the acceleration becomes $-a$. The point continues to move until its velocity becomes zero at point $C$.
Using $v_C = v_B + (-a)t'$,where $v_C = 0$ and $v_B = at$,we get $0 = at - at'$,so $t' = t$.
Distance $S_{BC} = v_B t' - \frac{1}{2}at'^2 = (at)t - \frac{1}{2}at^2 = \frac{1}{2}at^2$.
Total distance $S_{AC} = S_{AB} + S_{BC} = \frac{1}{2}at^2 + \frac{1}{2}at^2 = at^2$.
Total time to reach $C$ is $t + t = 2t$.
Now,the point returns from $C$ to $A$ under constant acceleration $a$ starting from rest at $C$.
$S_{AC} = \frac{1}{2}at_1^2$,where $t_1$ is the time taken to return.
$at^2 = \frac{1}{2}at_1^2 \implies t_1^2 = 2t^2 \implies t_1 = \sqrt{2}t$.
Total time $T = 2t + t_1 = 2t + \sqrt{2}t = (2 + \sqrt{2})t$.

(B) Let the particle move toward the right with velocity $u = 6\;m/s$. Due to retardation $a = 2\;m/s^2$,its velocity becomes zero after time $t_1$.
From $v = u - at$,we have $0 = 6 - 2t_1$,which gives $t_1 = 3\;s$.
Since the retardation acts for $4\;s$,the particle continues to accelerate in the opposite direction for the remaining $1\;s$ $(4\;s - 3\;s = 1\;s)$.
Distance covered in the first $3\;s$ (from $O$ to $A$): $S_{OA} = ut_1 - \frac{1}{2}at_1^2 = 6(3) - \frac{1}{2}(2)(3)^2 = 18 - 9 = 9\;m$.
Distance covered in the next $1\;s$ (from $A$ to $B$): $S_{AB} = \frac{1}{2}at^2 = \frac{1}{2}(2)(1)^2 = 1\;m$.
Position of $B$ relative to $O$: $S_{OB} = S_{OA} - S_{AB} = 9 - 1 = 8\;m$.
Velocity at $B$ (in return journey): $v_B = at = 2(1) = 2\;m/s$.
From $B$ onwards,the particle moves with a constant velocity of $2\;m/s$ to cover the distance $S_{OB} = 8\;m$.
Time taken to travel from $B$ to $O$: $t_{BO} = \frac{S_{OB}}{v_B} = \frac{8}{2} = 4\;s$.
Total time taken to return to $O$: $T = t_{OA} + t_{AB} + t_{BO} = 3 + 1 + 4 = 8\;s$.

(D) Given that the deceleration $a = -\alpha x^2$. We know that $a = v \frac{dv}{dx}$.
Substituting this into the equation,we get $v \frac{dv}{dx} = -\alpha x^2$.
Separating the variables,we have $v \, dv = -\alpha x^2 \, dx$.
Integrating both sides with limits from initial velocity $v_0$ to final velocity $0$ and position from $0$ to $S$:
$\int_{v_0}^{0} v \, dv = -\alpha \int_{0}^{S} x^2 \, dx$.
Evaluating the integrals:
$[\frac{v^2}{2}]_{v_0}^{0} = -\alpha [\frac{x^3}{3}]_{0}^{S}$.
$0 - \frac{v_0^2}{2} = -\alpha (\frac{S^3}{3})$.
$\frac{v_0^2}{2} = \frac{\alpha S^3}{3}$.
Solving for $S$,we get $S^3 = \frac{3v_0^2}{2\alpha}$,which implies $S = (\frac{3v_0^2}{2\alpha})^{1/3}$.

(A) Let $y$ be the distance of the man from the base of the street lamp and $x$ be the length of his shadow.
Using similar triangles,the ratio of the height of the lamp to the total distance from the base is equal to the ratio of the man's height to the length of his shadow:
$\frac{3h}{x + y} = \frac{h}{x}$
$3x = x + y$
$2x = y$
Differentiating both sides with respect to time $t$:
$2 \frac{dx}{dt} = \frac{dy}{dt}$
Given that the man is walking away at a constant speed $v$,we have $\frac{dy}{dt} = v$.
Therefore,$\frac{dx}{dt} = \frac{v}{2}$.
The rate of increase of the shadow length is independent of the distance from the lamp,so at $10h$,it is still $v/2$.

(B) Given that the rate of change of acceleration is constant: $\frac{da}{dt} = 2 m/s^3$.
Integrating with respect to time $t$ with initial conditions $a(0) = 0$:
$a = \int 2 dt = 2t$.
Since acceleration $a = \frac{dv}{dt}$,we have $\frac{dv}{dt} = 2t$.
Integrating with respect to time $t$ with initial velocity $v(0) = 0$:
$v = \int 2t dt = t^2$.
Since velocity $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = t^2$.
Integrating to find the distance $x$ covered from $t = 0$ to $t = 3 s$:
$x = \int_{0}^{3} t^2 dt = \left[ \frac{t^3}{3} \right]_{0}^{3} = \frac{3^3}{3} - 0 = \frac{27}{3} = 9 m$.

(B) The acceleration of the particle is given by $a = \frac{dv}{dt} = \frac{d}{dt}(t^2 - t) = 2t - 1$.
$A$ particle retards when its acceleration and velocity are in opposite directions,i.e.,$a \cdot v < 0$.
Substituting the expressions for $v$ and $a$:
$(2t - 1)(t^2 - t) < 0$
$(2t - 1)t(t - 1) < 0$
Since $t \ge 0$,we analyze the sign of the expression $t(2t - 1)(t - 1) < 0$ for $t > 0$:
$1$. For $0 < t < 1/2$: $t > 0$,$(2t - 1) < 0$,and $(t - 1) < 0$. The product is $(+)(-)(-) = (+)$,which is $> 0$.
$2$. For $1/2 < t < 1$: $t > 0$,$(2t - 1) > 0$,and $(t - 1) < 0$. The product is $(+)(+)(-) = (-)$,which is $< 0$.
$3$. For $t > 1$: $t > 0$,$(2t - 1) > 0$,and $(t - 1) > 0$. The product is $(+)(+)(+) = (+)$,which is $> 0$.
Therefore,the particle retards in the interval $1/2 < t < 1$.

(A) The particle moves under a constant force $F$ directed towards the equilibrium position $E$. This is a case of constant acceleration motion,not simple harmonic motion.
When the particle is at a distance $A$ from $E$,its acceleration is $a = \frac{F}{m}$.
The time taken to reach $E$ from the starting position is calculated using the kinematic equation $s = ut + \frac{1}{2}at^2$. Here,$s = A$,$u = 0$,and $a = \frac{F}{m}$.
$A = 0 + \frac{1}{2} \left(\frac{F}{m}\right) t_1^2 \implies t_1^2 = \frac{2Am}{F} \implies t_1 = \sqrt{\frac{2Am}{F}}$.
This is the time taken to travel from the extreme position to the equilibrium position $E$. Due to symmetry,the particle will travel the same distance $A$ on the other side of $E$ and return to $E$ in the same time $t_1$,and then return to the starting position in another $2t_1$.
The total time period $T$ is $4 \times t_1 = 4\sqrt{\frac{2Am}{F}}$.

(D) The particle moves under a constant force $F$ directed toward $E$. Let $E$ be at $x = 0$. The force is $F = -F$ for $x > 0$ and $F = +F$ for $x < 0$.
Starting from $x = -A$ with initial velocity $v = 0$,the acceleration is $a = F/m$.
The equation of motion is $x(t) = -A + \frac{1}{2}at^2 = -A + \frac{F}{2m}t^2$.
At $x = -A/2$,we have $-A/2 = -A + \frac{F}{2m}t_1^2$,which gives $\frac{A}{2} = \frac{F}{2m}t_1^2$,so $t_1 = \sqrt{\frac{mA}{F}}$.
The velocity at $x = -A/2$ is $v_1 = at_1 = \frac{F}{m} \sqrt{\frac{mA}{F}} = \sqrt{\frac{FA}{m}}$.
From $x = -A/2$ to $x = 0$,the particle moves with initial velocity $v_1$ and acceleration $a = F/m$.
Using $x = v_1 t_2 + \frac{1}{2}at_2^2$,we have $A/2 = \sqrt{\frac{FA}{m}} t_2 + \frac{F}{2m}t_2^2$.
Let $k = \sqrt{\frac{mA}{F}}$. Then $A/2 = \frac{A}{k} t_2 + \frac{A}{2k^2}t_2^2$.
Dividing by $A/2$,we get $1 = \frac{2}{k} t_2 + \frac{1}{k^2}t_2^2$,or $t_2^2 + 2kt_2 - k^2 = 0$.
Solving for $t_2$,$t_2 = \frac{-2k \pm \sqrt{4k^2 + 4k^2}}{2} = k(\sqrt{2}-1)$.
The total time is $t = t_1 + t_2 = k + k(\sqrt{2}-1) = k\sqrt{2} = \sqrt{\frac{2mA}{F}}$.
Since this result is not in the options,the correct answer is $D$.

(B) Given the velocity $v = \frac{dx}{dt} = \alpha \sqrt{x}$.
Rearranging the terms to separate the variables,we get:
$\frac{dx}{\sqrt{x}} = \alpha dt$.
Integrating both sides with the initial condition $x=0$ at $t=0$:
$\int_{0}^{x} x^{-1/2} dx = \int_{0}^{t} \alpha dt$.
Solving the integral:
$[2x^{1/2}]_{0}^{x} = \alpha [t]_{0}^{t}$.
This simplifies to:
$2\sqrt{x} = \alpha t$.
Squaring both sides:
$4x = \alpha^2 t^2$.
Therefore,the displacement $x$ is proportional to $t^2$:
$x \propto t^2$.

(C) We know that velocity $v = \frac{dx}{dt}$,which implies $dx = v \, dt$.
Integrating both sides with initial condition $x = 0$ at $t = 0$:
$x = \int_{0}^{t} v \, dt = \int_{0}^{t} (v_0 + gt + ft^2) \, dt$.
Performing the integration:
$x = \left[ v_0 t + \frac{gt^2}{2} + \frac{ft^3}{3} \right]_{0}^{t} = v_0 t + \frac{gt^2}{2} + \frac{ft^3}{3}$.
To find the displacement at unit time $(t = 1)$,substitute $t = 1$ into the expression:
$x = v_0(1) + \frac{g(1)^2}{2} + \frac{f(1)^3}{3} = v_0 + \frac{g}{2} + \frac{f}{3}$.

(C) Given the deceleration equation: $\frac{dv}{dt} = - 2.5\sqrt{v}$.
Separating the variables,we get: $\frac{dv}{\sqrt{v}} = - 2.5 \ dt$.
Integrating both sides from the initial speed $v = 6.25 \ m/s$ to final speed $v = 0 \ m/s$ over time $t$:
$\int_{6.25}^{0} v^{-1/2} \ dv = \int_{0}^{t} - 2.5 \ dt$.
Evaluating the integral:
$[2\sqrt{v}]_{6.25}^{0} = - 2.5t$.
Substituting the limits:
$2(\sqrt{0} - \sqrt{6.25}) = - 2.5t$.
$2(0 - 2.5) = - 2.5t$.
$-5 = - 2.5t$.
$t = \frac{5}{2.5} = 2 \ s$.

(D) Let the speed of each train be $u = 40\ m/s$. The total distance between them is $d = 2.0\ km = 2000\ m$.
To barely avoid a collision,each train must come to rest after covering a distance of $s = d/2 = 1000\ m$.
Using the kinematic equation $v^2 = u^2 + 2as$,where $v = 0$ (final velocity),$u = 40\ m/s$,and $s = 1000\ m$:
$0 = (40)^2 + 2a(1000)$
$0 = 1600 + 2000a$
$2000a = -1600$
$a = -1600 / 2000 = -0.8\ m/s^2$.
The magnitude of the deceleration is $0.8\ m/s^2$.

(B) Given the displacement equation: $x^3 = t^3 + 1$.
Differentiating both sides with respect to time $t$:
$3x^2 \frac{dx}{dt} = 3t^2$
$x^2 v = t^2$,where $v = \frac{dx}{dt}$ is the velocity.
$v = \frac{t^2}{x^2}$.
Differentiating $x^2 v = t^2$ with respect to time $t$ using the product rule:
$2x \frac{dx}{dt} v + x^2 \frac{dv}{dt} = 2t$
Since $a = \frac{dv}{dt}$,we have:
$2x v^2 + x^2 a = 2t$
Substitute $v = \frac{t^2}{x^2}$:
$2x (\frac{t^2}{x^2})^2 + x^2 a = 2t$
$2x \frac{t^4}{x^4} + x^2 a = 2t$
$x^2 a = 2t - \frac{2t^4}{x^3}$
Since $x^3 = t^3 + 1$,we have $t^3 = x^3 - 1$:
$x^2 a = 2t - \frac{2t(t^3)}{x^3} = 2t - \frac{2t(x^3 - 1)}{x^3} = 2t - 2t + \frac{2t}{x^3} = \frac{2t}{x^3}$
$a = \frac{2t}{x^3 \cdot x^2} = \frac{2t}{x^5}$.

(A) To minimize the total time,the train must accelerate at its maximum rate $(a_1 = 10 \ m/s^2)$ to reach the maximum speed $(v_{max} = 10 \ m/s)$ and decelerate at its maximum rate $(a_2 = 5 \ m/s^2)$ to come to rest.
Time taken to accelerate $(t_1)$: $v = u + a_1 t_1 \implies 10 = 0 + 10 t_1 \implies t_1 = 1 \ s$.
Distance covered during acceleration $(d_1)$: $d_1 = \frac{1}{2} a_1 t_1^2 = \frac{1}{2} \times 10 \times (1)^2 = 5 \ m$.
Time taken to decelerate $(t_2)$: $v = u - a_2 t_2 \implies 0 = 10 - 5 t_2 \implies t_2 = 2 \ s$.
Distance covered during deceleration $(d_2)$: $d_2 = v_{max} t_2 - \frac{1}{2} a_2 t_2^2 = 10 \times 2 - \frac{1}{2} \times 5 \times (2)^2 = 20 - 10 = 10 \ m$.
Remaining distance to be covered at constant speed $(d_3)$: $d_3 = 135 - (d_1 + d_2) = 135 - (5 + 10) = 120 \ m$.
Time taken at constant speed $(t_3)$: $t_3 = \frac{d_3}{v_{max}} = \frac{120}{10} = 12 \ s$.
Total time $(T)$: $T = t_1 + t_2 + t_3 = 1 + 2 + 12 = 15 \ s$.

(B) The displacement is given by $x = \frac{k}{b^2}(1 - e^{-bt})$.
To find the velocity $v$,we differentiate $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt} [\frac{k}{b^2}(1 - e^{-bt})] = \frac{k}{b^2} [0 - (-b)e^{-bt}] = \frac{k}{b^2} (b e^{-bt}) = \frac{k}{b} e^{-bt}$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt} [\frac{k}{b} e^{-bt}] = \frac{k}{b} (-b) e^{-bt} = -k e^{-bt}$.
Thus,the acceleration of the particle is $-k e^{-bt}$.