Uniformly Accelerated Motion Questions in English

(B) The distance travelled by an object in the $n^{th}$ second starting from rest is given by the formula $S_n = u + \frac{a}{2}(2n-1)$.
Since the block starts from rest,the initial velocity $u = 0$.
Therefore,the distance travelled in the $n^{th}$ interval is $S_n = \frac{a}{2}(2n-1)$.
Similarly,the distance travelled in the $(n+1)^{th}$ interval is $S_{n+1} = \frac{a}{2}(2(n+1)-1) = \frac{a}{2}(2n+2-1) = \frac{a}{2}(2n+1)$.
Now,calculating the ratio $\frac{S_n}{S_{n+1}}$:
$\frac{S_n}{S_{n+1}} = \frac{\frac{a}{2}(2n-1)}{\frac{a}{2}(2n+1)} = \frac{2n-1}{2n+1}$.

(B) Given the relation: $t = m x^{2} + n x$.
Differentiating with respect to $x$:
$\frac{dt}{dx} = 2mx + n$.
Since velocity $v = \frac{dx}{dt}$,we have $\frac{1}{v} = \frac{dt}{dx} = 2mx + n$.
Thus,$v = (2mx + n)^{-1}$.
Now,differentiate $v$ with respect to $t$ to find acceleration $a$:
$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx}$.
$\frac{dv}{dx} = -1(2mx + n)^{-2} \cdot (2m) = -2m(2mx + n)^{-2}$.
Since $(2mx + n) = \frac{1}{v}$,then $(2mx + n)^{-2} = v^{2}$.
Therefore,$a = v \cdot (-2m \cdot v^{2}) = -2mv^{3}$.
Retardation is the negative of acceleration,so retardation $= 2mv^{3}$.

(C) Given: Initial velocity $u = 0$. Let the constant acceleration be $a$.
For the first time interval $t$,the distance travelled is $s_1 = 10 \, m$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$10 = 0(t) + \frac{1}{2}at^2 \implies 10 = \frac{1}{2}at^2$ --- (Equation $1$)
For the total time interval $2t$,let the total distance travelled be $s_2 = 10 + x$,where $x$ is the distance travelled in the next $t \, s$.
$10 + x = 0(2t) + \frac{1}{2}a(2t)^2$
$10 + x = \frac{1}{2}a(4t^2) = 4 \left( \frac{1}{2}at^2 \right)$ --- (Equation $2$)
Substituting Equation $1$ into Equation $2$:
$10 + x = 4(10)$
$10 + x = 40$
$x = 30 \, m$.

(D) The stopping distance $d$ is given by the formula $d = \frac{v^2}{2a}$,where $v$ is the initial velocity and $a$ is the magnitude of deceleration.
Since the braking acceleration $a$ remains constant,the stopping distance is directly proportional to the square of the initial velocity: $d \propto v^2$.
Let the initial speed be $v_1 = 150 \ km/h$ and the initial stopping distance be $d_1 = 27 \ m$.
The new speed is $v_2 = \frac{1}{3} v_1$.
Therefore,the new stopping distance $d_2$ is given by $d_2 = (\frac{1}{3})^2 \times d_1 = \frac{1}{9} \times 27 \ m = 3 \ m$.

(C) Let the particle start from point $A$ with initial velocity $u=0$. It accelerates with $f$ for a distance $s_1 = \frac{2}{3}d$ to reach point $B$ with velocity $v_1$.
Using the equation $v^2 - u^2 = 2as$,we have:
$v_1^2 - 0^2 = 2f(\frac{2}{3}d) \Rightarrow v_1^2 = \frac{4}{3}fd \Rightarrow v_1 = 2\sqrt{\frac{fd}{3}}$.
Time taken for the first part $t_1$ is given by $v = u + at$:
$v_1 = 0 + ft_1 \Rightarrow t_1 = \frac{v_1}{f} = \frac{2}{f}\sqrt{\frac{fd}{3}} = 2\sqrt{\frac{d}{3f}}$.
For the second part of the journey from $B$ to $C$,the distance is $s_2 = \frac{1}{3}d$,initial velocity is $v_1$,and final velocity is $v_2 = 0$. Let the retardation be $a'$.
Using $v_2^2 - v_1^2 = 2a's_2$:
$0 - \frac{4}{3}fd = 2a'(\frac{1}{3}d) \Rightarrow a' = -2f$.
The time taken for the second part $t_2$ is given by $v_2 = v_1 + a't_2$:
$0 = v_1 - 2ft_2 \Rightarrow t_2 = \frac{v_1}{2f} = \frac{2\sqrt{fd/3}}{2f} = \sqrt{\frac{d}{3f}}$.
The total time $t = t_1 + t_2 = 2\sqrt{\frac{d}{3f}} + \sqrt{\frac{d}{3f}} = 3\sqrt{\frac{d}{3f}} = \sqrt{\frac{9d}{3f}} = \sqrt{\frac{3d}{f}}$.

(A) At $t = 5 \, s$,the velocity of the particle is $v_B = u + at = 0 + (1)(5) = 5 \, m/s$ and its position is $x_B = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(1)(5)^2 = 12.5 \, m$.
Let the additional acceleration be $a'$. The total acceleration for $t > 5 \, s$ is $(1 + a')$.
For the case with additional acceleration,at $t = 10 \, s$ (which is $5 \, s$ after the change):
$v = v_B + (1 + a')(5) = 5 + 5 + 5a' = 10 + 5a'$
$x = x_B + v_B(5) + \frac{1}{2}(1 + a')(5)^2 = 12.5 + 5(5) + 12.5(1 + a') = 12.5 + 25 + 12.5 + 12.5a' = 50 + 12.5a'$
If the additional acceleration had not been provided,$a' = 0$:
$v_0 = 5 + (1)(5) = 10 \, m/s$
$x_0 = 12.5 + 5(5) + \frac{1}{2}(1)(5)^2 = 12.5 + 25 + 12.5 = 50 \, m$
Given $x - x_0 = 12.5 \, m$:
$(50 + 12.5a') - 50 = 12.5 \implies 12.5a' = 12.5 \implies a' = 1 \, m/s^2$.
Then,$v - v_0 = (10 + 5a') - 10 = 5a' = 5(1) = 5 \, m/s$.

(B) The distance travelled by a particle in the $n^{\text{th}}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.
Given that the sum of the distance travelled in the $t^{\text{th}}$ and $(t+1)^{\text{th}}$ seconds is $100 \text{ cm}$.
$S_t + S_{t+1} = 100$
$[u + \frac{a}{2}(2t - 1)] + [u + \frac{a}{2}(2(t+1) - 1)] = 100$
$2u + \frac{a}{2}(2t - 1 + 2t + 2 - 1) = 100$
$2u + \frac{a}{2}(4t) = 100$
$2u + 2at = 100$
$u + at = 50$
Since the velocity after $t$ seconds is given by $v = u + at$,we have $v = 50 \text{ cm/s}$.

(B) Using the first equation of motion,$v = u + at$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the acceleration,and $t$ is the time.
For $t = 2 \ s$,$v = 30 \ m/s$: $30 = u + 2a$ --- (Equation $1$)
For $t = 4 \ s$,$v = 60 \ m/s$: $60 = u + 4a$ --- (Equation $2$)
Subtracting Equation $1$ from Equation $2$:
$(60 - 30) = (u + 4a) - (u + 2a)$
$30 = 2a$
$a = 15 \ m/s^2$
Substituting $a = 15 \ m/s^2$ into Equation $1$:
$30 = u + 2(15)$
$30 = u + 30$
$u = 0 \ m/s$
Therefore,the initial velocity is $0 \ m/s$.

(D) Given the displacement equation: $\sqrt{x} = t + 7$.
Squaring both sides,we get: $x = (t + 7)^2 = t^2 + 14t + 49$.
The velocity $v$ is the rate of change of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(t^2 + 14t + 49) = 2t + 14$.
The acceleration $a$ is the rate of change of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2t + 14) = 2 \, m/s^2$.
Since the acceleration $a = 2 \, m/s^2$ is a constant value,the particle moves with constant acceleration.

(B) The particle starts from rest,so initial velocity $u = 0$. The acceleration is $a = 2 \, m/s^2$.
To find the distance travelled in the $5^{\text{th}}$ half-second,we calculate the difference between the total distance covered in $2.5 \, s$ and the distance covered in $2 \, s$.
The distance formula is $S = ut + \frac{1}{2}at^2$.
For $t = 2.5 \, s$: $S_{2.5} = 0 + \frac{1}{2} \times 2 \times (2.5)^2 = 6.25 \, m$.
For $t = 2 \, s$: $S_2 = 0 + \frac{1}{2} \times 2 \times (2)^2 = 4 \, m$.
Distance in the $5^{\text{th}}$ half-second = $S_{2.5} - S_2 = 6.25 - 4 = 2.25 \, m$.

(C) Let the length of the train be $L$ and its constant acceleration be $a$.
Let the velocity at the first end be $v_1 = u$ and at the second end be $v_2 = 3u$.
Using the equation of motion $v^2 = u^2 + 2as$,for the entire length of the train:
$(3u)^2 = u^2 + 2aL$
$9u^2 = u^2 + 2aL$
$8u^2 = 2aL \implies aL = 4u^2$.
Now,let the velocity at the midpoint be $v_m$. The distance covered by the midpoint from the first end is $L/2$.
Using the equation of motion $v_m^2 = u^2 + 2a(L/2)$:
$v_m^2 = u^2 + aL$
Substitute $aL = 4u^2$ into the equation:
$v_m^2 = u^2 + 4u^2 = 5u^2$
$v_m = \sqrt{5}u$.
Thus,the velocity of the middle point is $\sqrt{5}u$.

(C) Given,acceleration $a = \frac{dv}{dt} = \alpha t^{3/2}$.
Integrating both sides with respect to time $t$ from $0$ to $t$:
$\int_{u}^{v} dv = \int_{0}^{t} a dt$
$\int_{u}^{v} dv = \int_{0}^{t} \alpha t^{3/2} dt$
$[v]_{u}^{v} = \alpha \left[ \frac{t^{3/2 + 1}}{3/2 + 1} \right]_{0}^{t}$
$v - u = \alpha \left[ \frac{t^{5/2}}{5/2} \right]_{0}^{t}$
$v - u = \frac{2}{5} \alpha t^{5/2}$
$v = u + \frac{2}{5} \alpha t^{5/2}$

(B) Given the position equation: $x^2 = t + 2$.
Differentiating both sides with respect to time $t$:
$2x \frac{dx}{dt} = 1 \Rightarrow \frac{dx}{dt} = \frac{1}{2x}$.
Now,differentiate the velocity $v = \frac{dx}{dt} = \frac{1}{2} x^{-1}$ with respect to time $t$ to find acceleration $a$:
$a = \frac{dv}{dt} = \frac{d}{dt} (\frac{1}{2} x^{-1}) = \frac{1}{2} (-1) x^{-2} \frac{dx}{dt}$.
Substitute $\frac{dx}{dt} = \frac{1}{2x}$ into the equation:
$a = -\frac{1}{2x^2} \times \frac{1}{2x} = -\frac{1}{4x^3}$.

(C) Given the position function: $x = -2t^3 + 3t^2 + 5$.
Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(-2t^3 + 3t^2 + 5) = -6t^2 + 6t$.
Set the velocity to zero to find the time $t$: $-6t^2 + 6t = 0 \Rightarrow 6t(1 - t) = 0$. This gives $t = 0 \ s$ or $t = 1 \ s$. Since the particle is moving,we consider $t = 1 \ s$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(-6t^2 + 6t) = -12t + 6$.
Substitute $t = 1 \ s$ into the acceleration equation: $a = -12(1) + 6 = -6 \ m/s^2$.

(D) For a body moving with uniform acceleration $(a = \text{constant})$, the relationship between velocity $(v)$ and position $(x)$ is given by the third equation of motion:
$v^2 = u^2 + 2ax$
Assuming the body starts from rest $(u = 0)$, the equation simplifies to:
$v^2 = 2ax$
Since $2a$ is a constant, we have $v^2 \propto x$, which implies $v \propto \sqrt{x}$.
The graph of $v = k\sqrt{x}$ (where $k$ is a constant) is a parabola opening along the positive $x$-axis. Among the given options, the curve that represents this relationship is shown in option $(D)$.

(C) Given that acceleration $a = kx$.
We know that acceleration can be expressed as $a = v \frac{dv}{dx}$.
Substituting the given expression for $a$:
$v \frac{dv}{dx} = kx$
Rearranging the terms to integrate:
$v \, dv = kx \, dx$
Integrating both sides with initial conditions $v=u$ at $x=0$ and final velocity $v$ at position $x$:
$\int_{u}^{v} v \, dv = \int_{0}^{x} kx \, dx$
Evaluating the integrals:
$\left[ \frac{v^2}{2} \right]_{u}^{v} = k \left[ \frac{x^2}{2} \right]_{0}^{x}$
$\frac{v^2 - u^2}{2} = \frac{kx^2}{2}$
Multiplying by $2$ on both sides:
$v^2 - u^2 = kx^2$
Therefore,$v^2 = u^2 + kx^2$.

(D) For uniformly accelerated motion,the position $(x)$ as a function of time $(t)$ must be a quadratic equation of the form $x = ut + \frac{1}{2}at^2$,where $u$ is the initial velocity and $a$ is the constant acceleration.
Let us analyze option $(d)$:
$\alpha t = \sqrt{\beta + x}$
Squaring both sides:
$(\alpha t)^2 = \beta + x$
$\alpha^2 t^2 = \beta + x$
$x = \alpha^2 t^2 - \beta$
Comparing this with the standard kinematic equation $x = ut + \frac{1}{2}at^2$,we see that the position $x$ is a quadratic function of time $t$ (with $u = 0$ and $a = 2\alpha^2$).
Since the acceleration $a = 2\alpha^2$ is a constant,this equation represents uniformly accelerated motion.
Therefore,the correct option is $(d)$.

(C) Given the velocity $v = \alpha \sqrt{x}$.
We know that acceleration $a = v \frac{dv}{dx}$.
First,find $\frac{dv}{dx}$:
$\frac{dv}{dx} = \alpha \cdot \frac{1}{2\sqrt{x}} = \frac{\alpha}{2\sqrt{x}}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (\alpha \sqrt{x}) \cdot \left( \frac{\alpha}{2\sqrt{x}} \right) = \frac{\alpha^2}{2}$.
Since $\alpha$ is a constant,the acceleration $a = \frac{\alpha^2}{2}$ is also a constant.
Therefore,the graph of acceleration $a$ versus time $t$ is a horizontal straight line parallel to the time axis,which corresponds to option $C$.

(B) Let the total distance covered be $d$.
For the total journey:
Initial velocity $= u$,final velocity $= 0$,acceleration $= -a$,time $= T$.
Using $v = u + at$:
$0 = u - aT \Rightarrow u = aT$
Total distance $d = uT - \frac{1}{2}aT^2 = (aT)T - \frac{1}{2}aT^2 = \frac{1}{2}aT^2$.
For the first half of the journey:
Distance $= \frac{d}{2} = \frac{1}{4}aT^2$.
Let the time taken be $t$.
Using $s = ut - \frac{1}{2}at^2$:
$\frac{1}{4}aT^2 = (aT)t - \frac{1}{2}at^2$
Dividing by $a$:
$\frac{T^2}{4} = Tt - \frac{t^2}{2}$
$T^2 = 4Tt - 2t^2 \Rightarrow 2t^2 - 4Tt + T^2 = 0$.
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$t = \frac{4T \pm \sqrt{16T^2 - 8T^2}}{4} = \frac{4T \pm \sqrt{8T^2}}{4} = \frac{4T \pm 2\sqrt{2}T}{4} = T \pm \frac{T}{\sqrt{2}}$.
Since $t < T$,we take $t = T(1 - \frac{1}{\sqrt{2}})$.
Thus,the correct option is $(b)$.

(C) The initial position of the particle is $(x_0, y_0) = (3, 7)$.
The initial velocity is $u = 0$ as the body starts from rest.
The acceleration is given as $\vec{a} = 4 \hat{i}$,which means $a_x = 4$ and $a_y = 0$.
Using the equation of motion $s = s_0 + ut + \frac{1}{2}at^2$ for both axes:
For the $x$-coordinate:
$x = x_0 + u_x t + \frac{1}{2} a_x t^2$
$x = 3 + (0)(3) + \frac{1}{2} \times 4 \times (3)^2$
$x = 3 + 0 + 2 \times 9 = 3 + 18 = 21$.
For the $y$-coordinate:
$y = y_0 + u_y t + \frac{1}{2} a_y t^2$
$y = 7 + (0)(3) + \frac{1}{2} \times 0 \times (3)^2$
$y = 7 + 0 + 0 = 7$.
Thus,the final coordinates after $3 \,s$ are $(21, 7)$.

(A) Given: Mass $M = 500 \, kg$,initial velocity $u = 2 \, ms^{-1}$,acceleration $a = 2 \, ms^{-2}$,and distance $s = 6 \, m$.
Using the kinematic equation $v^2 = u^2 + 2as$ to find the final velocity $v$:
$v^2 = (2)^2 + 2(2)(6)$
$v^2 = 4 + 24 = 28 \, m^2s^{-2}$.
The kinetic energy $KE$ is given by the formula $KE = \frac{1}{2} Mv^2$.
Substituting the values:
$KE = \frac{1}{2} \times 500 \times 28$
$KE = 250 \times 28 = 7000 \, J$.
Converting to kilojoules: $7000 \, J = 7 \, kJ$.

(D) Initial velocity $u = 20 \; m/s$. Final velocity $v = 0$. Distance $S_1 = 500 \; m$.
Using the third equation of motion,$v^2 = u^2 - 2aS_1$:
$0 = (20)^2 - 2 \cdot a \cdot 500$
$1000a = 400 \Rightarrow a = 0.4 \; m/s^2$.
Now,if the brakes are applied at half the distance,$S_2 = 250 \; m$:
$v^2 = u^2 - 2aS_2$
$v^2 = (20)^2 - 2 \cdot 0.4 \cdot 250$
$v^2 = 400 - 200 = 200$
$v = \sqrt{200} \; m/s$.
Comparing with $\sqrt{x} \; m/s$,we get $x = 200$.

(A) Given: Mass $m = 5\,g = 0.005\,kg$,Linear momentum $p = 0.3\,kg\,m/s$,Time $t = 5\,s$.
We know that linear momentum $p = mv$,where $v$ is the velocity.
Substituting the values: $0.005 \times v = 0.3$.
Solving for velocity: $v = \frac{0.3}{0.005} = \frac{300}{5} = 60\,m/s$.
Assuming the body moves with constant velocity,the distance covered $d = v \times t$.
$d = 60\,m/s \times 5\,s = 300\,m$.

(D) Given:
Initial velocity $u = 10.0 \, m/s$
Final velocity $v = 60.0 \, m/s$
Acceleration $a = 2.0 \, m/s^2$
Using the first equation of motion:
$v = u + at$
Substituting the values:
$60.0 = 10.0 + (2.0)t$
$60.0 - 10.0 = 2.0t$
$50.0 = 2.0t$
$t = \frac{50.0}{2.0} = 25.0 \, s$
Therefore,the time taken is $25 \, s$.

(B) Let the retardation produced by the wooden block be $a$. Using the equation of motion $v^2 = u^2 + 2as$:
For the first $24\,cm$ $(s_1 = 24\,cm)$:
$(\frac{u}{3})^2 = u^2 - 2a(24)$
$\frac{u^2}{9} = u^2 - 48a$
$48a = u^2 - \frac{u^2}{9} = \frac{8u^2}{9}$
$a = \frac{8u^2}{9 \times 48} = \frac{u^2}{54}.........(1)$
Now,let the total length of the block be $L$. For the entire motion until the bullet comes to rest $(v=0)$:
$0^2 = u^2 - 2aL$
$2aL = u^2$
$L = \frac{u^2}{2a}.........(2)$
Substituting the value of $a$ from $(1)$ into $(2)$:
$L = \frac{u^2}{2 \times (u^2/54)} = \frac{54}{2} = 27\,cm$.

(C) Let the initial velocity be $u$ and the constant retardation be $a$. Using the equation of motion $v^2 - u^2 = 2aS$:
After traveling $S_1 = 4 \ cm = 4 \times 10^{-2} \ m$,the velocity becomes $v_1 = u - \frac{1}{3}u = \frac{2}{3}u$.
Substituting into the equation: $(\frac{2}{3}u)^2 - u^2 = 2(-a)(4 \times 10^{-2})$
$\frac{4}{9}u^2 - u^2 = -8a \times 10^{-2}$
$-\frac{5}{9}u^2 = -8a \times 10^{-2} \implies a = \frac{5u^2}{72 \times 10^{-2}} \dots(1)$
Now,for the remaining distance $x = D \times 10^{-3} \ m$,the initial velocity is $\frac{2}{3}u$ and the final velocity is $0$:
$0^2 - (\frac{2}{3}u)^2 = 2(-a)(x)$
$-\frac{4}{9}u^2 = -2ax \implies x = \frac{4u^2}{18a} = \frac{2u^2}{9a} \dots(2)$
Substituting $a$ from $(1)$ into $(2)$:
$x = \frac{2u^2}{9} \times \frac{72 \times 10^{-2}}{5u^2} = \frac{2 \times 8 \times 10^{-2}}{5} = \frac{16}{5} \times 10^{-2} = 3.2 \times 10^{-2} \ m = 32 \times 10^{-3} \ m$.
Comparing with $D \times 10^{-3} \ m$,we get $D = 32$.

(B) Given that the body starts from rest,the initial velocity $u = 0$. The displacement $S$ covered in time $t$ with constant acceleration $a$ is given by $S = \frac{1}{2}at^2$.
For the first $(p-1)$ seconds,the displacement is $S_1 = \frac{1}{2}a(p-1)^2$.
For the first $p$ seconds,the displacement is $S_2 = \frac{1}{2}ap^2$.
We need to find the time $t$ such that the total displacement is $S_1 + S_2 = \frac{1}{2}at^2$.
Substituting the expressions for $S_1$ and $S_2$:
$\frac{1}{2}a(p-1)^2 + \frac{1}{2}ap^2 = \frac{1}{2}at^2$
Dividing both sides by $\frac{1}{2}a$:
$(p-1)^2 + p^2 = t^2$
$p^2 - 2p + 1 + p^2 = t^2$
$2p^2 - 2p + 1 = t^2$
$t = \sqrt{2p^2 - 2p + 1} \ s$.

(B) Given position: $x = t^3 - 6t^2 + 20t + 15 \ m$.
Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = 3t^2 - 12t + 20 \ m/s$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = 6t - 12 \ m/s^2$.
Set acceleration to zero to find the time: $6t - 12 = 0 \implies t = 2 \ s$.
Substitute $t = 2 \ s$ into the velocity equation: $v = 3(2)^2 - 12(2) + 20 = 12 - 24 + 20 = 8 \ m/s$.

(B) Let the initial velocity at time $t$ be $u$ and acceleration be $a$.
Given that the increase in velocity in $1 \ s$ is $50 \ m/s$,we have $v = u + a(1) = u + 50$.
Thus,$a = 50 \ m/s^2$.
The displacement in the interval $t$ to $t+1$ is given by $s = ut + \frac{1}{2}at^2$. For the interval of $1 \ s$ starting at $t$,the displacement is $s = u(1) + \frac{1}{2}a(1)^2 = 125$.
Substituting $a = 50$,we get $u + 25 = 125$,which implies $u = 100 \ m/s$.
The distance travelled in the $(t+2)^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$,where $n = t+2$.
However,the question asks for the distance in the $(t+2)^{th}$ second relative to the start of the motion. Since $u$ is the velocity at time $t$,the distance in the next second (the $(t+1)^{th}$ second) is $125 \ m$. The distance in the $(t+2)^{th}$ second is $S = (u+a) + \frac{a}{2} = 100 + 50 + 25 = 175 \ m$.

(B) Given the velocity of the particle as a function of position: $v = 4\sqrt{x}$.
Acceleration $a$ can be expressed in terms of position $x$ using the formula: $a = v \frac{dv}{dx}$.
First,differentiate $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(4x^{1/2}) = 4 \cdot \frac{1}{2} x^{-1/2} = 2x^{-1/2} = \frac{2}{\sqrt{x}}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (4\sqrt{x}) \cdot (\frac{2}{\sqrt{x}})$.
$a = 4 \cdot 2 = 8 \ m/s^2$.
Thus,the acceleration of the particle is constant at $8 \ m/s^2$.

(B) The distance traveled by a body in the $n^{\text{th}}$ second is given by the formula $S_n = u + \frac{a}{2}(2n - 1)$.
For the $n^{\text{th}}$ second: $102.5 = u + \frac{a}{2}(2n - 1) \quad \dots(1)$
For the $(n+2)^{\text{th}}$ second: $115.0 = u + \frac{a}{2}(2(n+2) - 1) = u + \frac{a}{2}(2n + 3) \quad \dots(2)$
Subtracting equation $(1)$ from equation $(2)$:
$115.0 - 102.5 = [u + \frac{a}{2}(2n + 3)] - [u + \frac{a}{2}(2n - 1)]$
$12.5 = \frac{a}{2} (2n + 3 - 2n + 1)$
$12.5 = \frac{a}{2} (4)$
$12.5 = 2a$
$a = \frac{12.5}{2} = 6.25 \ m/s^2$.
Therefore,the acceleration is $6.25 \ m/s^2$.

(B) Initial velocity $u = 72 \,km/h = 72 \times \frac{5}{18} \,m/s = 20 \,m/s$.
Final velocity $v = 0 \,m/s$ (as the bus comes to a halt).
Time $t = 4 \,s$.
Using the first equation of motion, $v = u + at$, we find the acceleration:
$0 = 20 + a(4) \Rightarrow 4a = -20 \Rightarrow a = -5 \,m/s^2$.
The distance $s$ travelled can be calculated using the second equation of motion, $s = ut + \frac{1}{2}at^2$:
$s = (20)(4) + \frac{1}{2}(-5)(4)^2$
$s = 80 - \frac{1}{2}(5)(16)$
$s = 80 - 40 = 40 \,m$.

(B) The distance traveled by an object in the $n^{th}$ second is given by the formula $S_n = u + \frac{a}{2}(2n-1)$.
Since the body starts from rest,$u = 0$,so $S_n = \frac{a}{2}(2n-1)$.
For $n=10$,$S_{10} = \frac{a}{2}(2(10)-1) = \frac{19a}{2}$.
For $n-1=9$,$S_{9} = \frac{a}{2}(2(9)-1) = \frac{17a}{2}$.
The ratio is $\frac{S_{n-1}}{S_n} = \frac{17a/2}{19a/2} = \frac{17}{19}$.
Given the ratio is $1 - \frac{2}{x}$,we have $1 - \frac{2}{x} = \frac{17}{19}$.
This implies $\frac{2}{x} = 1 - \frac{17}{19} = \frac{2}{19}$.
Therefore,$x = 19$.

(C) Given the displacement equation: $x^2 = 1 + t^2$.
Differentiating both sides with respect to $t$: $2x \frac{dx}{dt} = 2t$,which simplifies to $x v = t$,where $v$ is the velocity.
Differentiating $x v = t$ with respect to $t$ using the product rule: $x \frac{dv}{dt} + v \frac{dx}{dt} = 1$.
Since $\frac{dv}{dt} = a$ (acceleration) and $\frac{dx}{dt} = v$,we get $x a + v^2 = 1$.
Substituting $v = \frac{t}{x}$ into the equation: $x a + (\frac{t}{x})^2 = 1$.
$x a = 1 - \frac{t^2}{x^2} = \frac{x^2 - t^2}{x^2}$.
Since $x^2 - t^2 = 1$ from the original equation,we have $x a = \frac{1}{x^2}$.
Therefore,$a = \frac{1}{x^3} = x^{-3}$.
Comparing this with $x^{-n}$,we find $n = 3$.

(A) Let the acceleration of car $P$ be $a_P = kt$,where $k$ is a constant.
Let the acceleration of car $Q$ be $a_Q = a$,where $a$ is a constant.
The relative acceleration of $Q$ with respect to $P$ is $a_{QP} = a_Q - a_P = a - kt$.
The relative velocity $v_{QP}$ is the integral of relative acceleration: $v_{QP} = u_{QP} + at - \frac{1}{2}kt^2$.
The relative displacement $s_{QP}$ is the integral of relative velocity: $s_{QP} = u_{QP}t + \frac{1}{2}at^2 - \frac{1}{6}kt^3$.
For the cars to cross,the relative displacement $s_{QP}$ must be zero.
Setting $s_{QP} = 0$ gives $t(u_{QP} + \frac{1}{2}at - \frac{1}{6}kt^2) = 0$.
One solution is $t = 0$ (given).
The other crossings occur when the quadratic equation $u_{QP} + \frac{1}{2}at - \frac{1}{6}kt^2 = 0$ has real roots for $t > 0$.
$A$ quadratic equation can have at most $2$ positive roots.
Therefore,the total number of crossings is $1$ (at $t=0$) $+ 2$ (from the quadratic) $= 3$.

(B) Given the relation: $t = x^2 + x$.
Differentiating both sides with respect to $x$: $\frac{dt}{dx} = 2x + 1$.
Since velocity $v = \frac{dx}{dt}$,we have $\frac{1}{v} = 2x + 1$,which implies $v = (2x + 1)^{-1}$.
Acceleration $a$ is given by $a = v \frac{dv}{dx}$.
Differentiating $v$ with respect to $x$: $\frac{dv}{dx} = -1(2x + 1)^{-2} \times 2 = -2(2x + 1)^{-2}$.
Substituting $v$ and $\frac{dv}{dx}$ into the acceleration formula: $a = (2x + 1)^{-1} \times [-2(2x + 1)^{-2}]$.
Therefore,$a = -2(2x + 1)^{-3} = -\frac{2}{(2x + 1)^3}$.

(C) The position of the particle is given by $x(t) = 10 + 6t - 3t^2$.
To find the velocity,we differentiate the position with respect to time: $v(t) = \frac{dx}{dt} = 6 - 6t$.
Setting $v(t) = 0$,we find the particle stops at $t = 1 \ s$.
Since the interval given is $t = 1 \ s$ to $t = 4 \ s$,the particle moves in only one direction (negative direction) after $t = 1 \ s$.
At $t = 1 \ s$,the position is $x(1) = 10 + 6(1) - 3(1)^2 = 10 + 6 - 3 = 13 \ m$.
At $t = 4 \ s$,the position is $x(4) = 10 + 6(4) - 3(4)^2 = 10 + 24 - 48 = -14 \ m$.
The distance travelled is the magnitude of the displacement: $d = |x(4) - x(1)| = |-14 - 13| = |-27| = 27 \ m$.

(B) Let the acceleration be $\alpha = 2 \ m/s^2$ and retardation be $\beta = 4 \ m/s^2$. The total time is $t = 3 \ s$.
Let $t_1$ be the time of acceleration and $t_2$ be the time of retardation. Then $t_1 + t_2 = 3$.
Since the car starts from rest and comes to rest,the maximum velocity $v$ reached is $v = \alpha t_1 = \beta t_2$.
Substituting the values,$2t_1 = 4t_2$,which implies $t_1 = 2t_2$.
Substituting into the time equation: $2t_2 + t_2 = 3$,so $3t_2 = 3$,which gives $t_2 = 1 \ s$ and $t_1 = 2 \ s$.
The maximum velocity is $v = 2 \times 2 = 4 \ m/s$.
The total displacement $d$ is the area under the velocity-time graph,which is a triangle: $d = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \ m$.

(B) Given:
Initial velocity $u = 20 \ m/s$
Final velocity $v = 80 \ m/s$
Distance $s = 200 \ m$
For uniform acceleration,the average velocity is given by $\frac{u+v}{2}$.
The distance covered is $s = \text{average velocity} \times t$.
$s = \left(\frac{u+v}{2}\right) t$
$200 = \left(\frac{20+80}{2}\right) t$
$200 = \left(\frac{100}{2}\right) t$
$200 = 50t$
$t = \frac{200}{50} = 4 \ s$.

(B) Using the third equation of motion, $v^2 = u^2 + 2as$. Since the vehicle comes to rest, $v = 0$, so $0 = u^2 - 2as$, which gives $s = \frac{u^2}{2a}$.
Since the retardation $a$ is constant for the same vehicle, $s \propto u^2$.
Given $u_1 = 15 \,km/hr$ and $s_1 = 5 \,m$.
Given $u_2 = 45 \,km/hr = 3 \times u_1$.
Therefore, the new distance $s_2$ is given by $s_2 = s_1 \times (\frac{u_2}{u_1})^2$.
$s_2 = 5 \,m \times (3)^2 = 5 \,m \times 9 = 45 \,m$.

(D) Let the velocity of body $A$ be $v_A = u$ (constant).
Let the velocity of body $B$ be $v_B = at$ (starting from rest with acceleration $a$).
Their velocities become equal when $v_A = v_B$,which implies $u = at$. Thus,the time taken is $t = \frac{u}{a}$.
The distance covered by body $A$ in time $t$ is $s_A = u \cdot t = u \left( \frac{u}{a} \right) = \frac{u^2}{a}$.
The distance covered by body $B$ in time $t$ is $s_B = \frac{1}{2} a t^2 = \frac{1}{2} a \left( \frac{u}{a} \right)^2 = \frac{u^2}{2a}$.
The distance between them is $d = s_A - s_B = \frac{u^2}{a} - \frac{u^2}{2a} = \frac{u^2}{2a}$.

(C) Let the initial velocity be $u = 0$ and uniform acceleration be $a$.
The distance covered in the $n^{\text{th}}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
Since $u = 0$,$S_n = \frac{a}{2}(2n - 1)$.
The total distance travelled in $n$ seconds is given by: $S_{total} = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}an^2 = \frac{1}{2}an^2$.
The ratio of the distance covered in the $n^{\text{th}}$ second to the total distance in $n$ seconds is:
Ratio $= \frac{S_n}{S_{total}} = \frac{\frac{a}{2}(2n - 1)}{\frac{1}{2}an^2} = \frac{2n - 1}{n^2} = \frac{2n}{n^2} - \frac{1}{n^2} = \frac{2}{n} - \frac{1}{n^2}$.
Thus,the correct option is $C$.

(B) Let the initial speed be $V$ and the final speed be $0$. The distance covered is $s$ and the deceleration is $a$. Using the equation of motion $v^2 = u^2 + 2as$,we have $0 = V^2 - 2as$,which gives $s = \frac{V^2}{2a}$.
Also,using $v = u + at$,we have $0 = V - at$,which gives $t = \frac{V}{a}$.
Now,for the new case,the initial speed is $u' = nV$,final speed $v' = 0$,distance $s' = s$,and time $t' = t$. Let the new deceleration be $a'$.
From $v' = u' - a't'$,we have $0 = nV - a't$. Substituting $t = \frac{V}{a}$,we get $0 = nV - a'(\frac{V}{a})$,which implies $a' = na$.
Checking with the distance condition: $s' = u't' - \frac{1}{2}a't'^2$. Substituting $s' = s = \frac{V^2}{2a}$,$u' = nV$,$t' = t = \frac{V}{a}$,and $a' = na$,we get $\frac{V^2}{2a} = (nV)(\frac{V}{a}) - \frac{1}{2}(na)(\frac{V}{a})^2 = \frac{nV^2}{a} - \frac{nV^2}{2a} = \frac{nV^2}{2a}$.
For this to equal $\frac{V^2}{2a}$,we must have $n = 1$. However,the question asks for the deceleration required to satisfy both conditions. Given the constraints,the deceleration must be $n \cdot a$ to satisfy the time condition.

(D) Initial velocity $u = 15 \ m/s$,retardation $a = -0.3 \ m/s^2$,and final velocity $v = 0 \ m/s$ (when it stops).
First,calculate the time taken to stop: $t = \frac{v-u}{a} = \frac{0-15}{-0.3} = 50 \ s$.
Since the vehicle stops in $50 \ s$,which is less than $60 \ s$ (one minute),the displacement after $60 \ s$ is the same as the displacement after $50 \ s$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$s = (15 \times 50) + \frac{1}{2} \times (-0.3) \times (50)^2$
$s = 750 - 0.15 \times 2500 = 750 - 375 = 375 \ m$.
The initial distance from the traffic light was $400 \ m$.
Therefore,the distance from the traffic light after one minute is $400 \ m - 375 \ m = 25 \ m$.

(B) Let the acceleration of the body be $a$ and the distance between points $A$ and $B$ be $d$.
Using the equation of motion $v^2 = u^2 + 2as$:
For the path from $A$ to $B$:
$(30)^2 = (20)^2 + 2ad$
$900 = 400 + 2ad$
$2ad = 500$
$ad = 250$
Let $v_m$ be the velocity at the midpoint of $AB$. The distance from $A$ to the midpoint is $d/2$.
Using the equation of motion for the path from $A$ to the midpoint:
$v_m^2 = (20)^2 + 2a(d/2)$
$v_m^2 = 400 + ad$
Substituting $ad = 250$:
$v_m^2 = 400 + 250 = 650$
$v_m = \sqrt{650} \approx 25.495 \,m/s \approx 25.5 \,m/s$.

(A) For body $A$ starting from rest (initial velocity $u=0$):
$S_A = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}at^2 = \frac{1}{2}at^2$
For body $B$ moving with uniform velocity $V$:
$S_B = Vt$
Since they meet at the same point after time $t$,their displacements must be equal:
$S_A = S_B$
$\frac{1}{2}at^2 = Vt$
Dividing both sides by $t$ (assuming $t \neq 0$):
$\frac{1}{2}at = V$
$t = \frac{2V}{a}$

(B) Given that the distance $s$ is proportional to the square of the time $t$,we have $s \propto t^{2}$.
This can be written as $s = k t^{2}$,where $k$ is a constant.
The velocity $v$ is the first derivative of distance with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(k t^{2}) = 2kt$.
The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2kt) = 2k$.
Since $k$ is a constant,$2k$ is also a constant and is not equal to zero.
Therefore,the acceleration of the body is constant but not zero.

(D) The kinematic equation $x=ut+\frac{1}{2}at^2$ describes the motion of an object under constant acceleration.
This equation is derived from the definitions of velocity and acceleration using calculus or graphical methods.
It is a kinematic relationship and does not directly follow from Newton's laws of motion,which relate force,mass,and acceleration $(F=ma)$.
Therefore,the correct option is $D$.