Uniformly Accelerated Motion Questions in English

(A) Let $v_s = 0$ be the speed of the source (siren) and $v_o$ be the speed of the observer (vehicle). The frequency heard by the observer is given by the Doppler effect formula: $f' = f_0 \left( \frac{v - v_o}{v} \right)$, where $v = 330 \,m/s$ is the speed of sound.
Given $f' = 0.94 f_0$, we have $0.94 = \frac{330 - v_o}{330}$.
Solving for $v_o$: $330 \times 0.94 = 330 - v_o \implies 310.2 = 330 - v_o \implies v_o = 19.8 \,m/s$.
The vehicle starts from rest $(u = 0)$ with acceleration $a = 2 \,m/s^2$. Using the equation of motion $v_o^2 = u^2 + 2as$:
$(19.8)^2 = 0^2 + 2(2)s \implies 392.04 = 4s$.
$s = \frac{392.04}{4} = 98.01 \,m$.
Thus, the vehicle has gone nearly $98 \,m$.

(C) Let the total time be $t$ and acceleration be $a$. The initial velocity $u = 0$.
For the first half of the time,$t_{1} = t/2$:
$x_{1} = u t_{1} + \frac{1}{2} a t_{1}^{2} = 0 + \frac{1}{2} a (t/2)^{2} = \frac{1}{8} a t^{2}$.
For the total time $t$,the total distance $x_{total} = x_{1} + x_{2} = \frac{1}{2} a t^{2}$.
Substituting $x_{1} = \frac{1}{8} a t^{2}$ into the total distance equation:
$\frac{1}{8} a t^{2} + x_{2} = \frac{1}{2} a t^{2}$.
$x_{2} = \frac{1}{2} a t^{2} - \frac{1}{8} a t^{2} = \frac{4-1}{8} a t^{2} = \frac{3}{8} a t^{2}$.
Comparing $x_{1}$ and $x_{2}$:
$x_{2} = 3 \times (\frac{1}{8} a t^{2}) = 3 x_{1}$.

(D) Using the third equation of motion, $v^2 - u^2 = 2as$, where $v = 0$ (final velocity), $u$ is initial velocity, $a$ is retardation $(-a)$, and $s$ is distance.
$0^2 - u^2 = 2(-a)s \Rightarrow u^2 = 2as \Rightarrow s = \frac{u^2}{2a}$.
Since the retardation $a$ is constant, $s \propto u^2$.
For the first case: $40 = \frac{(20)^2}{2a} \Rightarrow 2a = \frac{400}{40} = 10 \,m \,s^{-2}$.
For the second case, the new velocity $u' = 2 \times 20 = 40 \,m \,s^{-1}$.
The new distance $s' = \frac{(u')^2}{2a} = \frac{(40)^2}{10} = \frac{1600}{10} = 160 \,m$.

(C) The distance travelled by a body in the $n^{\text{th}}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
For the first second $(n=1)$: $5 = u + \frac{a}{2}(2(1) - 1) \implies 5 = u + \frac{a}{2}$ (Eq. $i$).
For the third second $(n=3)$: $2 = u + \frac{a}{2}(2(3) - 1) \implies 2 = u + \frac{5a}{2}$ (Eq. $ii$).
Subtracting Eq. $i$ from Eq. $ii$: $(2 - 5) = (u + \frac{5a}{2}) - (u + \frac{a}{2}) \implies -3 = 2a \implies a = -1.5 \,m/s^2$.
The negative sign indicates deceleration.
The magnitude of the force is $F = |m \times a| = 4 \,kg \times 1.5 \,m/s^2 = 6 \,N$.

(D) Given: Initial velocity $= v_0$,Final velocity $= v$,Acceleration $= a$,Time interval $= t$.
Using the third equation of motion: $v^2 - v_0^2 = 2as$.
From this,the total distance covered $s$ is given by: $s = \frac{v^2 - v_0^2}{2a}$.
Average velocity is defined as the total displacement divided by the total time interval.
$\bar{v} = \frac{s}{t} = \frac{\frac{v^2 - v_0^2}{2a}}{t}$.
Therefore,$\bar{v} = \frac{v^2 - v_0^2}{2at}$.

(A) The displacement $x$ is given as a function of time $t$ by the equation: $t = \sqrt{x} + 3$.
Rearranging for $\sqrt{x}$,we get: $\sqrt{x} = t - 3$.
Squaring both sides,we obtain the expression for displacement: $x = (t - 3)^2 = t^2 - 6t + 9$.
The velocity $v$ of the particle is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(t^2 - 6t + 9) = 2t - 6$.
To find the displacement when velocity is zero,we set $v = 0$:
$2t - 6 = 0 \Rightarrow 2t = 6 \Rightarrow t = 3 \ s$.
Now,substitute $t = 3 \ s$ into the displacement equation:
$x = (3 - 3)^2 = 0^2 = 0 \ m$.
Thus,the displacement of the particle when its velocity is zero is $0 \ m$.

(A) Given: Initial position $x_i = 3 \ cm$,final position $x_f = -5 \ cm$,initial velocity $\vec{u} = 12 \hat{i} \ cm \ s^{-1}$,and time $t = 2 \ s$.
Displacement $\vec{s} = x_f - x_i = (-5 - 3) \hat{i} = -8 \hat{i} \ cm$.
Using the equation of motion $\vec{s} = \vec{u}t + \frac{1}{2} \vec{a}t^2$:
$-8 \hat{i} = (12 \hat{i})(2) + \frac{1}{2} \vec{a} (2)^2$
$-8 \hat{i} = 24 \hat{i} + 2 \vec{a}$
$2 \vec{a} = -8 \hat{i} - 24 \hat{i} = -32 \hat{i}$
$\vec{a} = -16 \ cm \ s^{-2} \hat{i}$.

(C) From the third equation of kinematics,we have $v^2 - u^2 = 2as$.
Assuming the object starts from rest,the initial velocity $u = 0$.
Substituting this into the equation,we get $v^2 = 2as$,which can be rearranged as $s = \frac{v^2}{2a}$.
Since $a$ is a uniform (constant) acceleration,the relationship between displacement $s$ and velocity $v$ is $s \propto v^2$.
This represents a parabola opening along the $s$-axis.
Graph $C$ shows a parabolic curve where $s$ increases with the square of $v$,starting from the origin $(0, 0)$,which matches the derived relationship $s = \frac{1}{2a} v^2$.

(D) The velocity of the particle is given by $v = e^{-\beta x}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = e^{-\beta x}$.
Rearranging the terms to separate the variables,we get $e^{\beta x} dx = dt$.
Integrating both sides with the initial condition that at $t = 0$,$x = 0$:
$\int_0^x e^{\beta x} dx = \int_0^t dt$
$\left[ \frac{e^{\beta x}}{\beta} \right]_0^x = [t]_0^t$
$\frac{1}{\beta} (e^{\beta x} - e^0) = t - 0$
$\frac{1}{\beta} (e^{\beta x} - 1) = t$
$e^{\beta x} - 1 = \beta t$
$e^{\beta x} = 1 + \beta t$
Taking the natural logarithm on both sides:
$\beta x = \log(1 + \beta t)$
$x = \frac{1}{\beta} \log(1 + \beta t)$

(C) Given,the displacement equation is $3 s = 9 t + 5 t^2$.
Dividing the entire equation by $3$,we get $s = 3 t + (5/3) t^2$.
The standard equation of motion for a body with constant acceleration is $s = u t + (1/2) a t^2$,where $u$ is the initial velocity and $a$ is the acceleration.
Comparing the given equation $s = 3 t + (5/3) t^2$ with the standard equation $s = u t + (1/2) a t^2$,we find that $(1/2) a = 5/3$.
Solving for $a$,we get $a = 2 \times (5/3) = 10/3 \ m s^{-2}$.

(B) Given the displacement equation: $S = 30t + 5t^2$.
Velocity $v$ is the rate of change of displacement with respect to time: $v = \frac{dS}{dt}$.
Differentiating $S$ with respect to $t$: $v = \frac{d}{dt}(30t + 5t^2) = 30 + 10t$.
Initial velocity is the velocity at $t = 0$.
Substituting $t = 0$ into the velocity equation: $v = 30 + 10(0) = 30 \ m \ s^{-1}$.
Therefore,the initial velocity is $30 \ m \ s^{-1}$.

(A) Given: Initial velocity $u = 6.25 \,ms^{-1}$, acceleration $a = -2.5 \sqrt{v} \,ms^{-2}$, and final velocity $v = 0$ (at rest).
We know that acceleration $a = \frac{dv}{dt}$.
Substituting the given values:
$\frac{dv}{dt} = -2.5 \sqrt{v}$
Rearranging the terms to integrate:
$\frac{dv}{\sqrt{v}} = -2.5 dt$
Integrating both sides from initial velocity $u$ to final velocity $0$ over time $t$ from $0$ to $T$:
$\int_{6.25}^{0} v^{-1/2} dv = \int_{0}^{T} -2.5 dt$
$[2 \sqrt{v}]_{6.25}^{0} = -2.5 [t]_{0}^{T}$
$2(\sqrt{0} - \sqrt{6.25}) = -2.5(T - 0)$
$2(0 - 2.5) = -2.5T$
$-5 = -2.5T$
$T = \frac{5}{2.5} = 2 \,s$
Therefore, the time taken for the car to come to rest is $2 \,s$.

(B) Let $v$ be the minimum velocity of the student so that they can catch the bus.
If the student catches the bus in time $t$,then the distance travelled by the student in time $t$ is equal to $16 \ m$ plus the distance travelled by the bus in time $t$.
Using the equations of motion:
Distance travelled by student $= v t$
Distance travelled by bus $= u t + \frac{1}{2} a t^2 = 0 \times t + \frac{1}{2} \times 9 \times t^2 = 4.5 t^2$
Equating the distances:
$v t = 16 + 4.5 t^2$
$4.5 t^2 - v t + 16 = 0$
Multiply by $2$ to simplify:
$9 t^2 - 2 v t + 32 = 0$
For the student to catch the bus,the time $t$ must be a real value. Therefore,the discriminant of this quadratic equation must be greater than or equal to zero $(D \geq 0)$:
$(-2 v)^2 - 4 \times 9 \times 32 \geq 0$
$4 v^2 - 1152 \geq 0$
$v^2 \geq 288$
$v \geq \sqrt{288} = \sqrt{144 \times 2} = 12 \sqrt{2} \ m \ s^{-1}$
The minimum velocity is $12 \sqrt{2} \ m \ s^{-1}$.
Comparing this with $\alpha \sqrt{2} \ m \ s^{-1}$,we get $\alpha = 12$.

(B) Given: Initial velocity $u = 0$, acceleration $a = \frac{5}{4} \,ms^{-2}$, and time $n = 3 \,s$.
The formula for the distance travelled in the $n^{th}$ second is given by $S_{n} = u + \frac{a}{2}(2n - 1)$.
Substituting the values into the formula:
$S_{3} = 0 + \frac{5/4}{2}(2 \times 3 - 1)$
$S_{3} = \frac{5}{8}(6 - 1)$
$S_{3} = \frac{5}{8} \times 5$
$S_{3} = \frac{25}{8} \,m$.

(D) Given the relation: $t = \alpha x^2 + \beta x$.
Differentiating both sides with respect to $t$:
$1 = 2 \alpha x \left( \frac{dx}{dt} \right) + \beta \left( \frac{dx}{dt} \right)$.
Since $v = \frac{dx}{dt}$,we have:
$1 = (2 \alpha x + \beta) v \implies v = \frac{1}{2 \alpha x + \beta}$.
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 $x$:
$\frac{dv}{dx} = -1(2 \alpha x + \beta)^{-2} \cdot (2 \alpha) = -2 \alpha v^2$.
Thus,$a = v (-2 \alpha v^2) = -2 \alpha v^3$.
Retardation is defined as negative acceleration:
$\text{Retardation} = -a = -(-2 \alpha v^3) = 2 \alpha v^3$.

(C) The velocity of the particle is given by $v = \sqrt{196 - 16x}$.
Squaring both sides, we get $v^2 = 196 - 16x$.
Differentiating both sides with respect to time $t$, we get:
$\frac{d}{dt}(v^2) = \frac{d}{dt}(196 - 16x)$
$2v \frac{dv}{dt} = -16 \frac{dx}{dt}$
Since acceleration $a = \frac{dv}{dt}$ and velocity $v = \frac{dx}{dt}$, we substitute these into the equation:
$2v \cdot a = -16v$
Dividing both sides by $2v$ (assuming $v \neq 0$):
$a = -8 \text{ m/s}^2$.
Thus, the acceleration of the particle is constant at $-8 \text{ m/s}^2$.

(B) Given: Displacement of the car,$s = 200 \,m$.
Time taken,$t = 10 \,s$.
Final velocity of the car,$v = 0 \,m/s$.
We know that the average velocity is given by $\frac{v + u}{2}$,where $u$ is the initial velocity.
The displacement is given by the product of average velocity and time:
$s = \left(\frac{v + u}{2}\right) \times t$
Substituting the given values:
$200 = \left(\frac{0 + u}{2}\right) \times 10$
$200 = 5u$
$u = \frac{200}{5} = 40 \,m/s$.
Therefore,the initial speed of the car is $40 \,m/s$.

(A) Using the equation of motion $v = u + at$.
Here,$v$ is the final speed,$u$ is the initial speed,and $a$ is the acceleration for time $t$.
Since there is constant retardation,the acceleration is negative,i.e.,$a_{eff} = -a$.
Substituting this into the equation:
$v = u + (-a)t = u - at$.
Since $at > 0$ for $t > 0$,it follows that $v < u$.
Therefore,the speed of the body decreases under constant retardation.

(D) The velocity of the particle is given by the derivative of displacement with respect to time: $v = \frac{ds}{dt}$.
Given $s = 9t^2 - 2t^3$,we differentiate with respect to $t$:
$v = \frac{d}{dt}(9t^2 - 2t^3) = 18t - 6t^2$.
To find the time when the particle attains zero velocity,we set $v = 0$:
$18t - 6t^2 = 0$.
Factoring out $6t$,we get:
$6t(3 - t) = 0$.
This gives two solutions: $t = 0$ and $t = 3$.
Since the particle starts from rest at $t = 0$,the time at which it will attain zero velocity again is $t = 3 \ s$.

(C) Given that the body starts from rest,the initial velocity $u = 0$. Let the uniform acceleration be $a$.
At time $t = n$,the velocity $v = u + at = 0 + an$,so $a = \frac{v}{n}$.
The displacement in the $k^{\text{th}}$ second is given by $S_k = u + \frac{a}{2}(2k - 1)$.
Since $u = 0$,$S_k = \frac{a}{2}(2k - 1)$.
The displacement in the $n^{\text{th}}$ second is $S_n = \frac{a}{2}(2n - 1)$.
The displacement in the $(n-1)^{\text{th}}$ second is $S_{n-1} = \frac{a}{2}(2(n-1) - 1) = \frac{a}{2}(2n - 3)$.
The total displacement in these two seconds is $S_{total} = S_n + S_{n-1} = \frac{a}{2}(2n - 1 + 2n - 3) = \frac{a}{2}(4n - 4) = 2a(n - 1)$.
Substituting $a = \frac{v}{n}$,we get $S_{total} = 2(\frac{v}{n})(n - 1) = \frac{2v(n - 1)}{n}$.

(C) Let the truck move with constant velocity $v_t = 12 \,m/s$ and the car start from rest $(u_c = 0)$ with acceleration $a_c = 2 \,m/s^2$.
Let the time taken for the car to catch up with the truck be $t$.
In time $t$, the distance traveled by the truck is $s_t = v_t \times t = 12t$.
The distance traveled by the car is $s_c = u_c t + \frac{1}{2} a_c t^2 = 0 + \frac{1}{2} \times 2 \times t^2 = t^2$.
For the car to cross the truck, the distances must be equal: $s_c = s_t$.
Therefore, $t^2 = 12t$.
Since $t \neq 0$, we have $t = 12 \,s$.
The distance traveled by the car is $s_c = t^2 = (12)^2 = 144 \,m$.

(A) Let the initial velocity be $u$ and uniform acceleration be $a$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$:
For the first $2 \,s$,$S_1 = 200 \,m$:
$200 = u(2) + \frac{1}{2}a(2)^2 \implies 200 = 2u + 2a \implies u + a = 100$ ...$(1)$
For the first $6 \,s$ (first $2 \,s$ + next $4 \,s$),the total distance $S_2 = 200 + 220 = 420 \,m$:
$420 = u(6) + \frac{1}{2}a(6)^2 \implies 420 = 6u + 18a \implies u + 3a = 70$ ...$(2)$
Subtracting equation $(1)$ from $(2)$:
$(u + 3a) - (u + a) = 70 - 100 \implies 2a = -30 \implies a = -15 \,m/s^2$
Substituting $a$ in equation $(1)$:
$u - 15 = 100 \implies u = 115 \,m/s$
The velocity after $t = 7 \,s$ is given by $v = u + at$:
$v = 115 + (-15)(7) = 115 - 105 = 10 \,m/s$.

(C) Using the third equation of motion,$v_f^2 - v_i^2 = 2as$.
Let the total length of the train be $L$.
When the engine crosses the tree,the initial velocity is $u$. When the last compartment crosses the tree,the train has covered a distance equal to its length $L$,and its final velocity is $v$.
Thus,$v^2 - u^2 = 2aL$,which gives $a = \frac{v^2 - u^2}{2L}$ ....$(1)$
Now,consider the middle compartment. It is at a distance of $\frac{L}{2}$ from the engine.
Let $v_m$ be the velocity of the middle compartment as it crosses the tree.
Using the equation of motion for the distance $\frac{L}{2}$:
$v_m^2 - u^2 = 2a(\frac{L}{2}) = aL$.
Substituting the value of $a$ from equation $(1)$:
$v_m^2 = u^2 + (\frac{v^2 - u^2}{2L}) \times L$
$v_m^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{2u^2 + v^2 - u^2}{2} = \frac{v^2 + u^2}{2}$
Therefore,$v_m = \sqrt{\frac{v^2 + u^2}{2}}$.

(B) The velocity of the particle is given by $v = 4t$. Since the particle starts from rest,the initial velocity $u = 0$.
We know that the distance $S$ covered by a particle is given by the integral of velocity with respect to time:
$S = \int_{0}^{t} v \ dt$
Substituting the given expression for $v$:
$S = \int_{0}^{4} 4t \ dt$
$S = 4 \left[ \frac{t^2}{2} \right]_{0}^{4}$
$S = 2 \times [t^2]_{0}^{4}$
$S = 2 \times (4^2 - 0^2)$
$S = 2 \times 16 = 32 \ m$
Alternatively,using the kinematic equation $v = u + at$,we compare $v = 4t$ with $v = 0 + at$ to find acceleration $a = 4 \ m s^{-2}$.
Using $S = ut + \frac{1}{2}at^2$:
$S = 0(4) + \frac{1}{2}(4)(4)^2 = 2 \times 16 = 32 \ m$.

(A) Given initial velocity $u = 36 \ km/h = 36 \times \frac{5}{18} \ m/s = 10 \ m/s$.
Since the object comes to rest,the final velocity $v = 0 \ m/s$.
The distance covered $s = 200 \ m$.
Using the third equation of motion,$v^2 - u^2 = 2as$.
Substituting the values: $0^2 - (10)^2 = 2 \times a \times 200$.
$-100 = 400a$.
$a = -\frac{100}{400} = -0.25 \ m/s^2$.
The negative sign indicates retardation.
Therefore,the retardation produced by the brakes is $0.25 \ m/s^2$.

(B) Given: Mass of the body, $m = 10 \,kg$.
Initial velocity, $u = 10 \,m \,s^{-1}$.
Time duration, $t = 4 \,s$.
Final velocity, $v = -2 \,m \,s^{-1}$ (since it is in the opposite direction).
Using the first equation of motion, $v = u + at$.
Rearranging for acceleration: $a = \frac{v - u}{t}$.
Substituting the values: $a = \frac{-2 - 10}{4} = \frac{-12}{4} = -3 \,m \,s^{-2}$.
Thus, the acceleration produced is $-3 \,m \,s^{-2}$.

(D) Let $u=0$ be the initial velocity and $a$ be the uniform acceleration of the body.
Distance travelled in the first $t=2 \,s$ is:
$x_1 = ut + \frac{1}{2}at^2 = 0(2) + \frac{1}{2}a(2)^2 = 2a$ --- $(i)$
Velocity at the end of $2 \,s$ is:
$v = u + at = 0 + a(2) = 2a$ --- (ii)
Distance travelled in the next $2 \,s$ (from $t=2 \,s$ to $t=4 \,s$) with initial velocity $v=2a$ is:
$x_2 = vt + \frac{1}{2}at^2 = (2a)(2) + \frac{1}{2}a(2)^2 = 4a + 2a = 6a$
Comparing $x_1$ and $x_2$:
$x_2 = 6a = 3(2a) = 3x_1$
Thus, $x_2 = 3x_1$.

(B) Let the body start from point $A$ at $t=0$ with initial velocity $u=0$ and constant acceleration $a$. At $t=2 \ s$,it reaches point $B$. The distance covered is $s_1 = \frac{1}{2} a (2)^2 = 2a$. The velocity at $B$ is $v_B = a(2) = 2a$.
At $t=2 \ s$,the body reverses its direction but the acceleration $a$ remains in the same direction (acting as deceleration). It stops at point $C$ where its velocity becomes $0$. Let the time taken from $B$ to $C$ be $t'$. Using $v = u + at$,we have $0 = 2a - a(t')$,which gives $t' = 2 \ s$. The distance $BC$ is $s_2 = (2a)(2) - \frac{1}{2} a (2)^2 = 4a - 2a = 2a$.
The total distance from $A$ to $C$ is $AC = s_1 + s_2 = 2a + 2a = 4a$.
Now,the body starts from rest at $C$ and moves towards $A$ with constant acceleration $a$. Let the time taken to cover distance $AC$ be $T$. Using $s = ut + \frac{1}{2}at^2$,we have $4a = 0 + \frac{1}{2} a T^2$,which gives $T^2 = 8$,so $T = 2\sqrt{2} \ s$.
The total time $t_0$ is the sum of time to reach $C$ and time to return to $A$. Time to reach $C$ is $2 \ s + 2 \ s = 4 \ s$. Thus,$t_0 = 4 + 2\sqrt{2} \ s$.

(D) Let $u$ be the initial velocity of the bullet. After passing through one plank,the velocity becomes $v = u - \frac{u}{25} = \frac{24u}{25}$.
Using the kinematic equation $v^2 - u^2 = 2as$,where $s$ is the thickness of one plank and $a$ is the constant retardation:
$\left(\frac{24u}{25}\right)^2 - u^2 = 2as$
$\frac{576u^2}{625} - u^2 = 2as$
$2as = -\frac{49u^2}{625}$.
If $n$ planks are required to stop the bullet,the final velocity becomes $0$ after a total distance $ns$:
$0^2 - u^2 = 2a(ns)$
$-u^2 = n(2as)$
$-u^2 = n \left(-\frac{49u^2}{625}\right)$
$n = \frac{625}{49} \approx 12.75$.
Since the number of planks must be an integer,we need at least $13$ planks to stop the bullet.

(A) Given that kinetic energy $K \propto t$.
Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 \propto t$,which implies $v^2 \propto t$ or $v \propto t^{1/2}$.
Acceleration $a = \frac{dv}{dt}$.
Since $v = kt^{1/2}$ (where $k$ is a constant),differentiating with respect to $t$ gives $a = k \cdot \frac{1}{2} t^{-1/2} = \frac{k}{2\sqrt{t}}$.
Therefore,$a \propto \frac{1}{\sqrt{t}}$.

(C) Let the initial velocity of the bullet be $u$.
After travelling a distance $x$,the velocity becomes $v_1 = u - \frac{1}{3}u = \frac{2u}{3}$.
Using the equation of motion $v^2 = u^2 + 2as$,where $a$ is the retardation (deceleration) inside the target:
$(\frac{2u}{3})^2 = u^2 - 2ax$
$\frac{4u^2}{9} = u^2 - 2ax$
$2ax = u^2 - \frac{4u^2}{9} = \frac{5u^2}{9} \quad \dots (1)$
Now,for the second part of the motion,the bullet starts with velocity $\frac{2u}{3}$ and comes to rest (final velocity $v_2 = 0$) after travelling a further distance $x^{\prime}$:
$0^2 = (\frac{2u}{3})^2 - 2ax^{\prime}$
$2ax^{\prime} = \frac{4u^2}{9} \quad \dots (2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{2ax^{\prime}}{2ax} = \frac{4u^2/9}{5u^2/9}$
$\frac{x^{\prime}}{x} = \frac{4}{5}$.

(A) Given: Initial velocity $u = 0 \ m \ s^{-1}$,Final velocity $v = 10 \ m \ s^{-1}$,Time $t = 2 \ s$.
For constant acceleration,we use the first equation of motion:
$v = u + at$
$10 = 0 + a \times 2$
$a = \frac{10}{2} = 5 \ m \ s^{-2}$.
Now,for the distance travelled $(s)$,we use the second equation of motion:
$s = ut + \frac{1}{2}at^2$
$s = 0 \times 2 + \frac{1}{2} \times 5 \times (2)^2$
$s = 0 + \frac{1}{2} \times 5 \times 4$
$s = 10 \ m$.
Thus,the acceleration is $5 \ m \ s^{-2}$ and the distance travelled is $10 \ m$.

(B) The velocity of the particle is given by $v(t) = 1 - 3t^2 + 2t^3$.
We know that velocity is the rate of change of position,$v = \frac{dx}{dt}$.
Therefore,the displacement $x$ can be found by integrating the velocity with respect to time:
$x(t) = \int v(t) \ dt = \int (1 - 3t^2 + 2t^3) \ dt$.
Performing the integration:
$x(t) = t - t^3 + \frac{2t^4}{4} + C = t - t^3 + \frac{t^4}{2} + C$.
Given that at $t = 0$,$x = 0$,we substitute these values to find the constant $C$:
$0 = 0 - 0^3 + \frac{0^4}{2} + C \Rightarrow C = 0$.
So,the position function is $x(t) = t - t^3 + \frac{t^4}{2}$.
At $t = 2 \ s$,the position is:
$x(2) = 2 - (2)^3 + \frac{(2)^4}{2} = 2 - 8 + \frac{16}{2} = 2 - 8 + 8 = 2 \ m$.

(C) The acceleration of the particle is given by $f = f_0 \left(1 - \frac{t}{T}\right)$.
Since $f = \frac{dv}{dt}$,the velocity $v$ is obtained by integrating the acceleration with respect to time:
$v = \int f dt = \int f_0 \left(1 - \frac{t}{T}\right) dt = f_0 \left(t - \frac{t^2}{2T}\right) + C$.
Given that at $t = 0$,$v = 0$,we find the constant of integration $C = 0$.
Thus,the velocity at any time $t$ is $v = f_0 \left(t - \frac{t^2}{2T}\right)$.
The acceleration becomes zero when $f = 0$,which implies $1 - \frac{t}{T} = 0$,so $t = T$.
Substituting $t = T$ into the velocity equation:
$v = f_0 \left(T - \frac{T^2}{2T}\right) = f_0 \left(T - \frac{T}{2}\right) = \frac{f_0 T}{2}$.

(A) Given,$v = \alpha \sqrt{x}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \alpha \sqrt{x}$.
Separating variables: $\frac{dx}{\sqrt{x}} = \alpha dt$.
Integrating both sides: $\int x^{-1/2} dx = \int \alpha dt \Rightarrow 2\sqrt{x} = \alpha t + C$.
At $t = 0$,$x = 0$,so $C = 0$. Thus,$2\sqrt{x} = \alpha t \Rightarrow \sqrt{x} = \frac{\alpha t}{2}$.
Squaring both sides: $x = \frac{\alpha^2 t^2}{4}$.
Velocity $v = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{\alpha^2 t^2}{4} \right) = \frac{\alpha^2}{4} (2t) = \frac{\alpha^2 t}{2}$.
Acceleration $a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{\alpha^2 t}{2} \right) = \frac{\alpha^2}{2}$.

(B) Given that,deceleration $a = -3 v^{2/3} \,m/s^2$.
At initial point $(t=0)$,velocity $u = 8 \,m/s$.
We know that acceleration $a = v \frac{dv}{ds}$.
Substituting the given expression for $a$:
$-3 v^{2/3} = v \frac{dv}{ds}$
$-3 v^{2/3} ds = v dv$
$ds = -\frac{1}{3} v^{1 - 2/3} dv = -\frac{1}{3} v^{1/3} dv$.
Integrating both sides:
$\int_{0}^{s} ds = -\frac{1}{3} \int_{8}^{0} v^{1/3} dv$
$s = -\frac{1}{3} \left[ \frac{v^{4/3}}{4/3} \right]_{8}^{0}$
$s = -\frac{1}{3} \cdot \frac{3}{4} [0 - 8^{4/3}]$
$s = -\frac{1}{4} [0 - (2^3)^{4/3}]$
$s = -\frac{1}{4} [0 - 16] = 4 \,m$.
Therefore,the distance travelled by the object before it stops is $4 \,m$.

(A) Given deceleration $\omega = -\frac{dv}{dt} = a \sqrt{v}$.
$1$. To find time $t$:
$\frac{dv}{dt} = -a \sqrt{v} \implies \int_{v_0}^{0} v^{-1/2} dv = \int_{0}^{t} -a dt$
$[2 \sqrt{v}]_{v_0}^{0} = -at \implies -2 \sqrt{v_0} = -at \implies t = \frac{2 \sqrt{v_0}}{a}$.
$2$. To find distance $s$:
Using $\omega = v \frac{dv}{ds} = a \sqrt{v} \implies v \frac{dv}{ds} = -a \sqrt{v} \implies \sqrt{v} dv = -a ds$.
Integrating both sides: $\int_{v_0}^{0} v^{1/2} dv = \int_{0}^{s} -a ds$
$[\frac{2}{3} v^{3/2}]_{v_0}^{0} = -as \implies -\frac{2}{3} v_0^{3/2} = -as \implies s = \frac{2 v_0^{3/2}}{3 a}$.

(B) The displacement of the particle is given by $x = A t^3 + B t^2 + C t + D$.
To find the velocity $v$,we differentiate $x$ with respect to time $t$:
$v = \frac{dx}{dt} = 3At^2 + 2Bt + C$.
The initial velocity $(v_{\text{initial}})$ is the velocity at $t = 0$:
$v_{\text{initial}} = 3A(0)^2 + 2B(0) + C = C$.
To find the acceleration $a$,we differentiate $v$ with respect to time $t$:
$a = \frac{dv}{dt} = 6At + 2B$.
The initial acceleration $(a_{\text{initial}})$ is the acceleration at $t = 0$:
$a_{\text{initial}} = 6A(0) + 2B = 2B$.
The ratio between the initial acceleration and initial velocity is $\frac{a_{\text{initial}}}{v_{\text{initial}}} = \frac{2B}{C}$.

(D) The displacement of a particle in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$,where $u$ is the initial velocity and $a$ is the constant acceleration.
For the $3^{rd}$ second $(n=3)$: $10 = u + \frac{a}{2}(2(3) - 1) \implies 10 = u + 2.5a$ --- (Equation $1$)
For the $5^{th}$ second $(n=5)$: $18 = u + \frac{a}{2}(2(5) - 1) \implies 18 = u + 4.5a$ --- (Equation $2$)
Subtracting Equation $1$ from Equation $2$: $(18 - 10) = (u + 4.5a) - (u + 2.5a) \implies 8 = 2a \implies a = 4 \ ms^{-2}$.
Substituting $a = 4$ into Equation $1$: $10 = u + 2.5(4) \implies 10 = u + 10 \implies u = 0 \ ms^{-1}$.
The velocity at time $t$ is given by $v = u + at$.
At $t = 4 \ s$: $v = 0 + (4)(4) = 16 \ ms^{-1}$.

(C) Given the relation: $t = ax^2 + bx$.
Differentiating both sides with respect to $x$: $\frac{dt}{dx} = 2ax + b$.
Since velocity $v = \frac{dx}{dt}$,we have $\frac{dt}{dx} = \frac{1}{v}$.
Therefore,$\frac{1}{v} = 2ax + b$,which implies $v = (2ax + b)^{-1}$.
Acceleration $a_{acc} = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx}$.
Differentiating $v$ with respect to $x$: $\frac{dv}{dx} = -1(2ax + b)^{-2} \cdot (2a) = -2a(2ax + b)^{-2}$.
Substituting $(2ax + b) = \frac{1}{v}$: $\frac{dv}{dx} = -2a \cdot (\frac{1}{v})^{-2} = -2av^2$.
Thus,acceleration $a_{acc} = v \cdot (-2av^2) = -2av^3$.

(B) Given, resistance force $F \propto \sqrt{v}$, so $ma = -k\sqrt{v}$, which implies $a = -c\sqrt{v}$ where $c$ is a constant.
We know $a = \frac{dv}{dt} = -c\sqrt{v}$.
Rearranging, $\frac{dv}{\sqrt{v}} = -c \, dt$.
Integrating from $t=0$ $(v=u)$ to $t=T$ $(v=0)$:
$\int_{u}^{0} v^{-1/2} \, dv = \int_{0}^{T} -c \, dt \Rightarrow [2\sqrt{v}]_{u}^{0} = -cT \Rightarrow -2\sqrt{u} = -cT \Rightarrow c = \frac{2\sqrt{u}}{T}$.
Now, $v \frac{dv}{ds} = -c\sqrt{v} \Rightarrow \sqrt{v} \, dv = -c \, ds$.
Integrating from $s=0$ $(v=u)$ to $s=S$ $(v=0)$:
$\int_{u}^{0} v^{1/2} \, dv = \int_{0}^{S} -c \, ds \Rightarrow [\frac{2}{3} v^{3/2}]_{u}^{0} = -cS \Rightarrow -\frac{2}{3} u^{3/2} = -cS \Rightarrow S = \frac{2u^{3/2}}{3c}$.
Substituting $c = \frac{2\sqrt{u}}{T}$:
$S = \frac{2u^{3/2}}{3(2\sqrt{u}/T)} = \frac{uT}{3}$.
Given $u = 120 \,m/s$ and $T = 1.5 \,s$:
$S = \frac{120 \times 1.5}{3} = \frac{180}{3} = 60 \,m$.
Since $60 \,m$ is not in the options, the provided options are incorrect.

(C) For the first part of the journey: $u = 0, a = 4 \,m/s^2$, time $= t_1$. The velocity attained at the end of time $t_1$ is $v_1 = u + at_1 = 4t_1$.
The displacement in the first $t_1$ seconds is $s_1 = \frac{1}{2}at_1^2 = \frac{1}{2} \times 4 \times t_1^2 = 2t_1^2$.
For the second part of the journey: initial velocity $v_1 = 4t_1$, time $= t_2$, and acceleration $= 0$.
The displacement in the next $t_2$ seconds is $s_2 = v_1 t_2 = 4t_1 t_2$.
For the third part of the journey: initial velocity $v_1 = 4t_1$, final velocity $= 0$, time interval $= t_1$.
The displacement in the third part is $s_3 = \frac{v_1 + v_f}{2} \times t_1 = \frac{4t_1 + 0}{2} \times t_1 = 2t_1^2$.
Total time $T = t_1 + t_2 + t_1 = 2t_1 + t_2 = 10 \,s$, so $t_2 = 10 - 2t_1$.
Average velocity $v_{\text{avg}} = \frac{\text{Total displacement}}{\text{Total time}} = \frac{s_1 + s_2 + s_3}{10} = 5.1$.
Substituting the values: $5.1 = \frac{2t_1^2 + 4t_1 t_2 + 2t_1^2}{10} = \frac{4t_1^2 + 4t_1(10 - 2t_1)}{10}$.
$51 = 4t_1^2 + 40t_1 - 8t_1^2$.
$4t_1^2 - 40t_1 + 51 = 0$.
Solving the quadratic equation: $t_1 = \frac{40 \pm \sqrt{1600 - 4(4)(51)}}{2(4)} = \frac{40 \pm \sqrt{1600 - 816}}{8} = \frac{40 \pm \sqrt{784}}{8} = \frac{40 \pm 28}{8}$.
$t_1 = \frac{68}{8} = 8.5 \,s$ (not possible) or $t_1 = \frac{12}{8} = 1.5 \,s$.
Therefore, $t_1 = 1.5 \,s$.

(C) $1$. For the first $3 \,s$, the particle starts from rest $(u=0)$ with acceleration $a_1 = 2 \,m/s^2$.
$2$. Velocity at $t=3 \,s$ is $v = u + a_1 t = 0 + 2(3) = 6 \,m/s$.
$3$. Displacement at $t=3 \,s$ is $s_1 = ut + 0.5 a_1 t^2 = 0 + 0.5(2)(3^2) = 9 \,m$.
$4$. After $t=3 \,s$, the acceleration reverses, so $a_2 = -2 \,m/s^2$. Let the additional time be $t'$.
$5$. The position at time $t'$ is $s = s_1 + vt' + 0.5 a_2 (t')^2$.
$6$. For the particle to return to its initial position, $s=0$.
$7$. $0 = 9 + 6t' - 0.5(2)(t')^2 \Rightarrow 0 = 9 + 6t' - (t')^2 \Rightarrow (t')^2 - 6t' - 9 = 0$.
$8$. Using the quadratic formula $t' = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 - 4(1)(-9)}}{2} = \frac{6 \pm \sqrt{72}}{2} = 3 \pm 3\sqrt{2}$.
$9$. Since $t' > 0$, $t' = 3 + 3\sqrt{2} = 3(1+\sqrt{2}) \,s$.
$10$. Total time $T = 3 + t' = 3 + 3 + 3\sqrt{2} = 6 + 3\sqrt{2} = 3(2+\sqrt{2}) \,s$.

(D) For uniformly accelerated motion,the distance covered in the $n^{th}$ second is given by $S_n = u + (2n - 1) \frac{a}{2}$.
For the fourth second $(n=4)$: $25 = u + (2 \times 4 - 1) \frac{a}{2} = u + 3.5a$ ...$(i)$
For the sixth second $(n=6)$: $37 = u + (2 \times 6 - 1) \frac{a}{2} = u + 5.5a$ ...(ii)
Subtracting $(i)$ from (ii): $12 = 2a \implies a = 6 \ m/s^2$.
Substituting $a$ in $(i)$: $25 = u + 3.5(6) = u + 21 \implies u = 4 \ m/s$.
The distance covered in the next two seconds (i.e.,from $t=6 \ s$ to $t=8 \ s$) is calculated using $S = ut + \frac{1}{2}at^2$ with $u=4 \ m/s$,$a=6 \ m/s^2$,and $t=2 \ s$:
$S = (4 \times 2) + \frac{1}{2} \times 6 \times (2)^2 = 8 + 12 = 20 \ m$ is incorrect based on the provided solution logic. Re-evaluating: The velocity at $t=6 \ s$ is $v = u + at = 4 + (6 \times 6) = 40 \ m/s$.
Distance covered in the next $2 \ s$ starting from $t=6 \ s$ is $S = vt + \frac{1}{2}at^2 = (40 \times 2) + \frac{1}{2} \times 6 \times (2)^2 = 80 + 12 = 92 \ m$.

(D) Given that the body starts from rest,so initial velocity $u = 0$.
Let the uniform acceleration be $a$.
The velocity after $n$ seconds is given by $v = u + at$.
Substituting the values,$V = 0 + a(n)$,which gives $a = \frac{V}{n}$.
The displacement $S$ in the last $t = 2 \text{ s}$ can be calculated using the formula $S = v_{final}t - \frac{1}{2}at^2$,where $v_{final} = V$.
$S = V(2) - \frac{1}{2} \left(\frac{V}{n}\right)(2)^2$.
$S = 2V - \frac{1}{2} \left(\frac{V}{n}\right)(4)$.
$S = 2V - \frac{2V}{n}$.
$S = 2V \left(1 - \frac{1}{n}\right) = \frac{2V(n-1)}{n}$.

(B) The initial velocity $u = 10 \,m/s$ and acceleration $a = 2 \,m/s^2$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$ for a time interval $\Delta t = 1 \,s$:
For $S_1$ (interval $t=3 \,s$ to $t=4 \,s$):
The velocity at $t=3 \,s$ is $v_3 = u + at = 10 + 2(3) = 16 \,m/s$.
$S_1 = v_3(1) + \frac{1}{2}a(1)^2 = 16(1) + \frac{1}{2}(2)(1)^2 = 16 + 1 = 17 \,m$.
For $S_2$ (interval $t=4 \,s$ to $t=5 \,s$):
The velocity at $t=4 \,s$ is $v_4 = u + at = 10 + 2(4) = 18 \,m/s$.
$S_2 = v_4(1) + \frac{1}{2}a(1)^2 = 18(1) + \frac{1}{2}(2)(1)^2 = 18 + 1 = 19 \,m$.
Therefore,the ratio $\frac{S_2}{S_1} = \frac{19}{17}$.

(B) Given: Acceleration $a = 6t$,Initial velocity $u = 10 \ m/s$,Initial position $x_0 = 0$.
Step $1$: Find velocity $v(t)$ by integrating acceleration.
$v(t) = \int a \ dt = \int 6t \ dt = 3t^2 + C$.
At $t = 0$,$v = 10 \ m/s$,so $C = 10$. Thus,$v(t) = 3t^2 + 10$.
Step $2$: Find position $x(t)$ by integrating velocity.
$x(t) = \int v(t) \ dt = \int (3t^2 + 10) \ dt = t^3 + 10t + C'$.
At $t = 0$,$x = 0$,so $C' = 0$. Thus,$x(t) = t^3 + 10t$.
Step $3$: Calculate distance at $t = 2 \ s$.
$x(2) = (2)^3 + 10(2) = 8 + 20 = 28 \ m$.
Since the velocity $v(t) = 3t^2 + 10$ is always positive for $t \ge 0$,the distance travelled is equal to the displacement.
The correct answer is $28 \ m$.

(B) Given,initial velocity of the truck,$u = 54 \,km/h = 54 \times \frac{5}{18} = 15 \,m/s$.
Final velocity,$v = 0$.
Deceleration,$a = -10 \,m/s^2$.
Using the equation of motion,$v^2 = u^2 + 2as$.
Substituting the values,$(0)^2 = (15)^2 + 2 \times (-10) \times s$.
$0 = 225 - 20s$.
$20s = 225$.
$s = \frac{225}{20} = 11.25 \,m$.

(A) Given that the velocity $v$ is a function of time: $v(t) = 10t \text{ cm/s}$.
Since the particle starts from the origin at $t=0$, the initial velocity $u = v(0) = 0 \text{ cm/s}$.
The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(10t) = 10 \text{ cm/s}^2$.
Using the second equation of motion for constant acceleration, the distance $s$ covered in time $t$ is given by:
$s = ut + \frac{1}{2}at^2$
Substituting the values $u = 0 \text{ cm/s}$, $a = 10 \text{ cm/s}^2$, and $t = 8 \text{ s}$:
$s = (0)(8) + \frac{1}{2} \times 10 \times (8)^2$
$s = 0 + 5 \times 64$
$s = 320 \text{ cm}$.
Thus, the distance covered by the particle in $8 \text{ s}$ is $320 \text{ cm}$.