Uniformly Accelerated Motion Questions in English

(D) Given the deceleration equation: $\frac{dv}{dt} = -kv^3$.
Rearranging the terms to integrate: $\frac{dv}{v^3} = -k dt$.
Integrating both sides with limits from $v_0$ to $v$ for velocity and $0$ to $t$ for time:
$\int_{v_0}^{v} v^{-3} dv = \int_{0}^{t} -k dt$.
Evaluating the integrals:
$\left[ \frac{v^{-2}}{-2} \right]_{v_0}^{v} = -kt$.
$-\frac{1}{2} \left( \frac{1}{v^2} - \frac{1}{v_0^2} \right) = -kt$.
$\frac{1}{v^2} - \frac{1}{v_0^2} = 2kt$.
$\frac{1}{v^2} = \frac{1}{v_0^2} + 2kt = \frac{1 + 2v_0^2kt}{v_0^2}$.
Taking the reciprocal and square root:
$v^2 = \frac{v_0^2}{1 + 2v_0^2kt}$.
$v = \frac{v_0}{\sqrt{1 + 2v_0^2kt}}$.

(C) Let the initial velocity of the bullet be $u$ and the thickness of one plank be $s$. The velocity after passing through one plank is $v = u - u/20 = 19u/20$.
Using the equation of motion $v^2 = u^2 + 2as$,where $a$ is the constant retardation:
$(19u/20)^2 = u^2 + 2as \implies 2as = (19/20)^2 u^2 - u^2 = u^2 (361/400 - 1) = -39u^2/400$.
Let $n$ be the number of planks required to stop the bullet. For the final velocity to be $0$ after passing through $n$ planks of total thickness $ns$:
$0^2 = u^2 + 2a(ns) = u^2 + n(2as)$.
Substituting $2as = -39u^2/400$:
$0 = u^2 + n(-39u^2/400) \implies 1 = n(39/400) \implies n = 400/39 \approx 10.25$.
Since the number of planks must be an integer,we need at least $11$ planks to stop the bullet.

(C) Let the uniform acceleration of the particle be $a$ and initial velocity $u = 0 \ m/s$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,the distance covered in the first $5 \ s$ is:
$x = 0(5) + \frac{1}{2}a(5)^2 = 12.5a \implies a = \frac{x}{12.5} = 0.08x \ m/s^2$.
Now,the distance covered in the first $10 \ s$ (total time) is:
$s_{10} = 0(10) + \frac{1}{2}a(10)^2 = 50a$.
Substituting $a = 0.08x$:
$s_{10} = 50(0.08x) = 4x \ m$.
The distance covered in the next $5 \ s$ is the difference between the distance covered in $10 \ s$ and the distance covered in $5 \ s$:
$\text{Distance} = s_{10} - x = 4x - x = 3x \ m$.

(A) Let the initial acceleration be $a_1$. Since the car starts from rest,$u = 0$. After $t_1 = 9 \ s$,the velocity becomes $v = a_1 t_1 = 9a_1$. Thus,$a_1 = v/9$.
Next,the motorist maintains the speed $v$ for $t_2 = 50 \ s$.
Finally,the motorist decelerates to rest. Let the deceleration be $a_2$. We are given $a_2 = 3a_1 = 3(v/9) = v/3$.
Using the equation of motion $v_f = u_f - a_2 t_3$,where $v_f = 0$ (final rest),$u_f = v$ (initial speed for this phase),and $t_3$ is the time taken to stop:
$0 = v - (v/3) t_3$
$v = (v/3) t_3$
$t_3 = 3 \ s$.

(C) Given that the particle starts from rest,initial velocity $u = 0$. The acceleration $a$ is constant.
For the first $10\, s$ $(t_1 = 10\, s)$:
$s_1 = ut_1 + \frac{1}{2}at_1^2 = 0(10) + \frac{1}{2}a(10)^2 = 50a$.
For the total time of $20\, s$ $(t_{total} = 20\, s)$:
$s_{total} = s_1 + s_2 = u(t_{total}) + \frac{1}{2}a(t_{total})^2 = 0(20) + \frac{1}{2}a(20)^2 = 200a$.
Substituting $s_1 = 50a$ into the total distance equation:
$50a + s_2 = 200a$.
Therefore,$s_2 = 200a - 50a = 150a$.
Comparing $s_1$ and $s_2$:
$\frac{s_2}{s_1} = \frac{150a}{50a} = 3$.
Thus,$s_2 = 3s_1$.

(C) Given: Initial velocity $u = 20\ m/s$,final velocity $v = 0\ m/s$,and retardation $a = -5\ m/s^2$.
Using the third equation of motion: $v^2 = u^2 + 2as$.
Substituting the values: $0^2 = (20)^2 + 2(-5)s$.
$0 = 400 - 10s$.
$10s = 400$.
$s = 40\ m$.
Therefore,the distance travelled by the car until it stops is $40\ m$.

(A) Let the initial velocity be $u$ and the uniform acceleration be $a$.
For the first interval of $t = 4 \ sec$,the distance covered is $s_1 = 24 \ m$.
Using the equation $s = ut + \frac{1}{2}at^2$:
$24 = u(4) + \frac{1}{2}a(4)^2$
$24 = 4u + 8a$
Dividing by $4$,we get: $6 = u + 2a$ --- $(1)$
For the total time of $t = 8 \ sec$,the total distance covered is $s_{total} = 24 + 64 = 88 \ m$.
Using the equation $s = ut + \frac{1}{2}at^2$:
$88 = u(8) + \frac{1}{2}a(8)^2$
$88 = 8u + 32a$
Dividing by $8$,we get: $11 = u + 4a$ --- $(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(u + 4a) - (u + 2a) = 11 - 6$
$2a = 5 \Rightarrow a = 2.5 \ m/s^2$
Substituting $a = 2.5$ into equation $(1)$:
$6 = u + 2(2.5)$
$6 = u + 5$
$u = 1 \ m/s$.

(D) Given acceleration $a = v \frac{dv}{dx} = -\alpha x^2$.
Separating the variables,we get $v dv = -\alpha x^2 dx$.
Integrating both sides from the initial state $(x=0, v=v_0)$ to the final state $(x=d, v=0)$:
$\int_{v_0}^{0} v dv = \int_{0}^{d} -\alpha x^2 dx$.
Evaluating the integrals:
$[\frac{v^2}{2}]_{v_0}^{0} = -\alpha [\frac{x^3}{3}]_{0}^{d}$.
$0 - \frac{v_0^2}{2} = -\alpha (\frac{d^3}{3})$.
$\frac{v_0^2}{2} = \frac{\alpha d^3}{3}$.
Solving for $d$:
$d^3 = \frac{3v_0^2}{2\alpha}$.
$d = (\frac{3v_0^2}{2\alpha})^{1/3}$.

(A) Let $u_1, u_2, u_3, u_4$ be the velocities at $t=0, t_1, (t_1+t_2)$ and $(t_1+t_2+t_3)$ respectively. Let $a$ be the constant acceleration.
The average velocity in an interval is the arithmetic mean of initial and final velocities:
$v_1 = \frac{u_1 + u_2}{2}$,$v_2 = \frac{u_2 + u_3}{2}$,$v_3 = \frac{u_3 + u_4}{2}$.
Using $v = u + at$:
$u_2 = u_1 + at_1$
$u_3 = u_2 + at_2 = u_1 + a(t_1 + t_2)$
$u_4 = u_3 + at_3 = u_1 + a(t_1 + t_2 + t_3)$
Now,$v_1 - v_2 = \frac{u_1 + u_2}{2} - \frac{u_2 + u_3}{2} = \frac{u_1 - u_3}{2} = \frac{u_1 - (u_1 + a(t_1 + t_2))}{2} = -\frac{a(t_1 + t_2)}{2}$.
Similarly,$v_2 - v_3 = \frac{u_2 + u_3}{2} - \frac{u_3 + u_4}{2} = \frac{u_2 - u_4}{2} = \frac{(u_1 + at_1) - (u_1 + a(t_1 + t_2 + t_3))}{2} = -\frac{a(t_2 + t_3)}{2}$.
Dividing the two expressions:
$\frac{v_1 - v_2}{v_2 - v_3} = \frac{t_1 + t_2}{t_2 + t_3}$.

(D) The formula for distance traveled in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.
Given that the body starts from rest,the initial velocity $u = 0 \, m/s$.
The distance traveled in the $6^{th}$ second is $S_6 = 120 \, cm = 1.2 \, m$.
Substituting the values into the formula:
$1.2 = 0 + \frac{a}{2}(2 \times 6 - 1)$
$1.2 = \frac{a}{2}(11)$
$a = \frac{1.2 \times 2}{11} = \frac{2.4}{11} \, m/s^2 \approx 0.218 \, m/s^2$.
Converting back to $cm/s^2$:
$a = 0.218 \times 100 \, cm/s^2 = 21.8 \, cm/s^2$.

(A) Given the velocity equation: $v = \sqrt{2x + 9}$.
Squaring both sides,we get: $v^2 = 2x + 9$.
Comparing this with the kinematic equation $v^2 = u^2 + 2ax$,where $u$ is the initial velocity and $a$ is the acceleration:
At the origin,$x = 0$. Substituting $x = 0$ into the velocity equation: $v = \sqrt{2(0) + 9} = \sqrt{9} = 3\, \text{unit}$. Thus,the initial velocity $u = 3\, \text{unit}$.
Comparing $v^2 = 2x + 9$ with $v^2 = u^2 + 2ax$,we identify $2a = 2$,which gives $a = 1\, \text{unit}$.
Therefore,the initial acceleration is $1\, \text{unit}$ and the initial velocity is $3\, \text{unit}$.

(B) Given,$v = 2\sqrt{x}$.
Acceleration $a = v \frac{dv}{dx}$.
First,find $\frac{dv}{dx}$:
$\frac{dv}{dx} = \frac{d}{dx}(2x^{1/2}) = 2 \cdot \frac{1}{2} x^{-1/2} = \frac{1}{\sqrt{x}}$.
Now,calculate acceleration:
$a = (2\sqrt{x}) \cdot \left(\frac{1}{\sqrt{x}}\right) = 2\,m/s^2$.
Since the acceleration is constant and independent of time and position,the net force $F$ acting on the particle is given by Newton's second law:
$F = m \cdot a = 1\,kg \cdot 2\,m/s^2 = 2\,N$.
Thus,the net force at any instant,including at $t = 2\,s$,is $2\,N$.

(B) Let the acceleration of the car be $a$ and the distance between $P$ and $Q$ be $s$.
Using the equation of motion $v^2 = u^2 + 2as$ for the path $PQ$:
$40^2 = 30^2 + 2as$
$1600 = 900 + 2as$
$2as = 700$
$as = 350$
Let $V$ be the velocity at the midpoint of $PQ$. The distance from $P$ to the midpoint is $s/2$.
Using the equation of motion $V^2 = u^2 + 2a(s/2)$:
$V^2 = 30^2 + as$
$V^2 = 900 + 350$
$V^2 = 1250$
$V = \sqrt{1250} = \sqrt{625 \times 2} = 25\sqrt{2}\; km/h$.
Alternatively,for uniform acceleration,the velocity at the midpoint $V_{mid}$ is given by:
$V_{mid} = \sqrt{\frac{v_P^2 + v_Q^2}{2}}$
$V_{mid} = \sqrt{\frac{30^2 + 40^2}{2}} = \sqrt{\frac{900 + 1600}{2}} = \sqrt{\frac{2500}{2}} = \sqrt{1250} = 25\sqrt{2}\; km/h$.

(B) Let the initial velocity be $v$ and the constant retardation be $a$.
After penetrating $s_1 = 3\,cm$,the velocity becomes $v/2$.
Using the equation of motion $v_f^2 = v_i^2 + 2as$:
$(v/2)^2 = v^2 - 2a(3) \implies v^2/4 = v^2 - 6a \implies 6a = 3v^2/4 \implies a = v^2/8$.
Now,let the total distance penetrated be $s_2$ until the final velocity becomes $0$.
$0^2 = v^2 - 2a(s_2) \implies v^2 = 2(v^2/8)s_2 \implies v^2 = (v^2/4)s_2 \implies s_2 = 4\,cm$.
The additional distance covered is $s_2 - s_1 = 4\,cm - 3\,cm = 1\,cm$.

(D) For a particle starting from rest $(u = 0)$ under constant acceleration $(a)$:
The displacement in $n$ seconds is given by the kinematic equation:
$s_n = ut + \frac{1}{2}an^2 = 0 + \frac{1}{2}an^2 = \frac{1}{2}an^2$
The displacement in the $n^{th}$ second is given by:
$s_{nth} = u + \frac{a}{2}(2n - 1) = 0 + \frac{a}{2}(2n - 1) = \frac{a}{2}(2n - 1)$
Taking the ratio of displacement in $n$ seconds to the displacement in the $n^{th}$ second:
$\frac{s_n}{s_{nth}} = \frac{\frac{1}{2}an^2}{\frac{a}{2}(2n - 1)} = \frac{n^2}{2n - 1}$
Thus,the correct option is $D$.

(C) Let the motion be divided into three parts: $A$ to $B$,$B$ to $C$,and $C$ to $D$.
For part $A$ to $B$: Initial velocity $u = 0$,acceleration $a = \alpha$,distance $d_1 = d$. Using $v^2 = u^2 + 2ad$,we get $v^2 = 2\alpha d$.
For part $C$ to $D$: Final velocity $v' = 0$,acceleration $a = -\alpha/2$,distance $d_3$. Using $v'^2 = v^2 + 2ad_3$,we get $0 = v^2 - 2(\alpha/2)d_3$,which implies $v^2 = \alpha d_3$. Since $v^2 = 2\alpha d$,we have $2\alpha d = \alpha d_3$,so $d_3 = 2d$.
Given total distance $d_1 + d_2 + d_3 = 15d$. Substituting $d_1 = d$ and $d_3 = 2d$,we get $d + d_2 + 2d = 15d$,which gives $d_2 = 12d$.
For part $B$ to $C$: The car moves with constant speed $v$ for time $t$. So,$d_2 = v \cdot t$,which means $12d = vt$,or $v = 12d/t$.
Substituting $v$ into the equation $v^2 = 2\alpha d$: $(12d/t)^2 = 2\alpha d$.
$144d^2 / t^2 = 2\alpha d$.
$d = (2\alpha t^2) / 144 = \frac{1}{72} \alpha t^2$.

(D) Let the acceleration be $\alpha$ and the retardation be $\beta$. Given $\beta = 3\alpha$.
Let $t_1$ be the time of acceleration and $t_2$ be the time of retardation.
The total time is $t_1 + t_2 = 8 \, s$.
Since the particle starts from rest and comes to rest, the maximum velocity $v$ reached is $v = \alpha t_1 = \beta t_2$.
Substituting $\beta = 3\alpha$, we get $\alpha t_1 = 3\alpha t_2$, which implies $t_1 = 3t_2$.
Substituting $t_1 = 3t_2$ into the total time equation: $3t_2 + t_2 = 8 \, s$.
$4t_2 = 8 \, s$, so $t_2 = 2 \, s$.
Therefore, the time of acceleration $t_1 = 3 \times 2 = 6 \, s$.

(B) Let $t_1$ be the time taken to accelerate to maximum speed $v$ and $t_2$ be the time taken to retard to rest.
Using $v = u + at$,we get $v = 0 + at_1 \implies t_1 = \frac{v}{a}$.
For retardation,$0 = v - at_2 \implies t_2 = \frac{v}{a}$.
Total time $T = t_1 + t_2 = \frac{2v}{a}$.
Distance covered during acceleration $s_1 = \frac{1}{2}at_1^2 = \frac{1}{2}a(\frac{v}{a})^2 = \frac{v^2}{2a}$.
Distance covered during retardation $s_2 = \frac{v^2}{2a}$.
Total distance $s = s_1 + s_2 = \frac{v^2}{a} \implies v^2 = as \implies v = \sqrt{as}$.
Substituting $v$ in $T = \frac{2v}{a}$,we get $T = \frac{2\sqrt{as}}{a} = 2\sqrt{\frac{s}{a}}$.
Thus,the total time is $2\sqrt{\frac{s}{a}}$ and maximum speed is $\sqrt{as}$.

(B) Let $u$ be the initial velocity and $a$ be the constant acceleration.
Using the first equation of motion,$v = u + at$,at $t = 10 \, s$,$v = 20 \, m/s$:
$20 = u + 10a \quad \dots(i)$
Using the formula for distance traveled in the $n^{th}$ second,$s_n = u + \frac{a}{2}(2n - 1)$,for $n = 10$ and $s_{10} = 10 \, m$:
$10 = u + \frac{a}{2}(2 \times 10 - 1)$
$10 = u + \frac{19a}{2} \quad \dots(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(20 - 10) = (u - u) + (10a - 9.5a)$
$10 = 0.5a$
$a = \frac{10}{0.5} = 20 \, m/s^2$.

(B) Let the time interval between the departures of the two cars be $\Delta t$ seconds.
Let the first car travel for $t_1 = 120\, s$ ($2$ minutes).
The distance covered by the first car is $S_1 = \frac{1}{2} a t_1^2 = \frac{1}{2} \times 0.4 \times (120)^2 = 0.2 \times 14400 = 2880\, m$.
The second car starts $\Delta t$ seconds later,so its travel time is $t_2 = (120 - \Delta t)\, s$.
The distance covered by the second car is $S_2 = \frac{1}{2} a t_2^2 = \frac{1}{2} \times 0.4 \times (120 - \Delta t)^2 = 0.2(120 - \Delta t)^2$.
The distance between the cars is $S_1 - S_2 = 1.9\, km = 1900\, m$.
Substituting the values: $2880 - 0.2(120 - \Delta t)^2 = 1900$.
$0.2(120 - \Delta t)^2 = 2880 - 1900 = 980$.
$(120 - \Delta t)^2 = \frac{980}{0.2} = 4900$.
Taking the square root: $120 - \Delta t = 70$.
$\Delta t = 120 - 70 = 50\, s$.

(A) Given that the body starts from rest,initial velocity $u = 0$.
After $n$ seconds,the velocity is $v$.
Using the first equation of motion,$v = u + an$,we get $v = 0 + an$,which implies $a = \frac{v}{n}$.
The displacement in the last $2 \ s$ is the difference between the total displacement in $n$ seconds and the displacement in $(n - 2)$ seconds.
$S_{last 2s} = S_n - S_{n-2} = \frac{1}{2}an^2 - \frac{1}{2}a(n-2)^2$.
$S_{last 2s} = \frac{1}{2}a [n^2 - (n^2 - 4n + 4)] = \frac{1}{2}a [4n - 4] = 2a(n - 1)$.
Substituting $a = \frac{v}{n}$,we get $S_{last 2s} = 2(\frac{v}{n})(n - 1) = \frac{2v(n - 1)}{n}$.

(D) The particle starts from rest at $x=-A$. The acceleration is $a = F/m$ towards the right.
Let $t_1$ be the time taken to travel from $x=-A$ to $x=0$. Using $s = ut + \frac{1}{2}at^2$ with $u=0$ and $s=A$:
$A = 0 + \frac{1}{2} (F/m) t_1^2 \implies t_1 = \sqrt{\frac{2mA}{F}}$.
Let $t_2$ be the time taken to travel from $x=-A$ to $x=-A/2$. Using $s = A/2$:
$A/2 = 0 + \frac{1}{2} (F/m) t_2^2 \implies t_2 = \sqrt{\frac{mA}{F}}$.
The time taken to travel from $x=-A/2$ to $x=0$ is $\Delta t = t_1 - t_2$.
$\Delta t = \sqrt{\frac{2mA}{F}} - \sqrt{\frac{mA}{F}} = \sqrt{\frac{mA}{F}}(\sqrt{2}-1)$.

(B) Given the velocity function: $v = \sqrt{c_1 - c_2 x}$.
Squaring both sides,we get: $v^2 = c_1 - c_2 x$.
Differentiating both sides with respect to time $t$:
$\frac{d}{dt}(v^2) = \frac{d}{dt}(c_1 - c_2 x)$.
Using the chain rule: $2v \frac{dv}{dt} = -c_2 \frac{dx}{dt}$.
Since acceleration $a = \frac{dv}{dt}$ and velocity $v = \frac{dx}{dt}$,we have:
$2v a = -c_2 v$.
Assuming $v \neq 0$,we divide both sides by $2v$:
$a = -\frac{c_2}{2}$.
Thus,the acceleration of the particle is constant and equal to $-\frac{c_2}{2}$.

(C) Given:
Initial velocity $u = 0\,m/s$ (starting from rest).
Time $t = 8\,s$.
Acceleration $a = 20\,cm/s^2 = 0.2\,m/s^2$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$s = (0)(8) + \frac{1}{2} \times (0.2) \times (8)^2$
$s = 0 + 0.1 \times 64$
$s = 6.4\,m$
Since $1\,m = 100\,cm$,then $s = 6.4 \times 100 = 640\,cm$.
Therefore,the correct option is $C$.

(B) Using the equation of motion $v^2 - u^2 = 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the acceleration (deceleration),and $s$ is the stopping distance.
For the first case: $u_1 = 40\, km/h$,$v_1 = 0$,$s_1 = 40\, m$.
$0^2 - u_1^2 = 2a(40) \implies a = -u_1^2 / 80$.
Since the deceleration $a$ is constant for the same braking force,we have $s \propto u^2$.
Therefore,$s_2 / s_1 = (u_2 / u_1)^2$.
Given $u_2 = 80\, km/h$,$u_1 = 40\, km/h$,and $s_1 = 40\, m$.
$s_2 = 40 \times (80 / 40)^2 = 40 \times 2^2 = 40 \times 4 = 160\, m$.
The minimum stopping distance is $160\, m$.

(D) Let $S$ be the total length of the train and $a$ be the uniform acceleration.
Using the equation of motion $v^2 - u^2 = 2aS$,we have:
$v^2 - u^2 = 2aS \implies aS = \frac{v^2 - u^2}{2}$.
Let $V_c$ be the velocity of the middle wagon as it passes the pole. The distance covered by the train from the engine to the middle wagon is $S/2$.
Applying the equation of motion for this distance:
$V_c^2 - u^2 = 2a(S/2) = aS$.
Substituting the value of $aS$:
$V_c^2 - u^2 = \frac{v^2 - u^2}{2}$.
$V_c^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{2u^2 + v^2 - u^2}{2} = \frac{u^2 + v^2}{2}$.
Therefore,$V_c = \sqrt{\frac{u^2 + v^2}{2}}$.

(C) Let the distance of the race be $L$. For car $A$,$L = \frac{1}{2}a_1 t_1^2$,so $t_1 = \sqrt{\frac{2L}{a_1}}$.
For car $B$,$L = \frac{1}{2}a_2 t_2^2$,so $t_2 = \sqrt{\frac{2L}{a_2}}$.
Given $t_2 - t_1 = t$,so $\sqrt{2L} \left( \frac{1}{\sqrt{a_2}} - \frac{1}{\sqrt{a_1}} \right) = t \Rightarrow \sqrt{2L} \left( \frac{\sqrt{a_1} - \sqrt{a_2}}{\sqrt{a_1 a_2}} \right) = t$.
Final speeds are $v_A = a_1 t_1 = a_1 \sqrt{\frac{2L}{a_1}} = \sqrt{2a_1 L}$ and $v_B = a_2 t_2 = a_2 \sqrt{\frac{2L}{a_2}} = \sqrt{2a_2 L}$.
Given $v_A - v_B = v$,so $\sqrt{2L} (\sqrt{a_1} - \sqrt{a_2}) = v$.
Dividing the two equations: $\frac{v}{t} = \frac{\sqrt{2L}(\sqrt{a_1} - \sqrt{a_2})}{\sqrt{2L} \frac{(\sqrt{a_1} - \sqrt{a_2})}{\sqrt{a_1 a_2}}} = \sqrt{a_1 a_2}$.
Therefore,$v = \sqrt{a_1 a_2} t$.

(D) Given that the particle starts from rest,the initial velocity $u = 0$.
Since the particle moves with uniform acceleration,the acceleration $a$ is constant over time.
$1$. For acceleration-time graph: Since $a$ is constant,the graph $(a)$ is a horizontal line parallel to the $t$-axis,which is correct.
$2$. For velocity-time graph: Using the equation $v = u + at$,we get $v = 0 + at = at$. This represents a straight line passing through the origin,so graph $(b)$ is correct.
$3$. For displacement-time graph: Using the equation $x = ut + \frac{1}{2}at^2$,we get $x = 0(t) + \frac{1}{2}at^2 = \frac{1}{2}at^2$. This represents a parabola opening upwards starting from the origin,so graph $(d)$ is correct.
Therefore,the figures $(a)$,$(b)$,and $(d)$ correctly represent the motion.

(D) Given the position function: $x(t) = at + bt^2 - ct^3$.
The velocity $v(t)$ is the first derivative of position with respect to time:
$v(t) = \frac{dx}{dt} = a + 2bt - 3ct^2$.
The acceleration $a(t)$ is the derivative of velocity with respect to time:
$a(t) = \frac{dv}{dt} = 2b - 6ct$.
Set the acceleration to zero to find the time $t$ when this occurs:
$0 = 2b - 6ct \implies 6ct = 2b \implies t = \frac{b}{3c}$.
Now,substitute $t = \frac{b}{3c}$ into the velocity equation:
$v = a + 2b\left(\frac{b}{3c}\right) - 3c\left(\frac{b}{3c}\right)^2$
$v = a + \frac{2b^2}{3c} - 3c\left(\frac{b^2}{9c^2}\right)$
$v = a + \frac{2b^2}{3c} - \frac{b^2}{3c}$
$v = a + \frac{b^2}{3c}$.

(B) Given the speed $v = b\sqrt{x}$.
We know that $v = \frac{dx}{dt}$,so $\frac{dx}{dt} = b\sqrt{x}$.
Separating the variables,we get $\frac{dx}{\sqrt{x}} = b dt$.
Integrating both sides from $t = 0$ (where $x = 0$) to $t = \tau$ (where $x = x$):
$\int_{0}^{x} x^{-1/2} dx = \int_{0}^{\tau} b dt$
$[2\sqrt{x}]_{0}^{x} = b\tau$
$2\sqrt{x} = b\tau$,which implies $\sqrt{x} = \frac{b\tau}{2}$.
Substituting this into the expression for speed $v = b\sqrt{x}$:
$v = b \left( \frac{b\tau}{2} \right) = \frac{b^2\tau}{2}$.

(A) Let the initial velocity be $u$ and acceleration be $a$.
For the first $4\,s$ (from $A$ to $B$):
Using the equation of motion $s = ut + \frac{1}{2}at^2$,we have:
$40 = u(4) + \frac{1}{2}a(4)^2$
$40 = 4u + 8a$
Dividing by $4$,we get: $u + 2a = 10$ --- $(i)$
For the total time of $8\,s$ (from $A$ to $C$):
The total distance covered is $40\,m + 120\,m = 160\,m$.
$160 = u(8) + \frac{1}{2}a(8)^2$
$160 = 8u + 32a$
Dividing by $8$,we get: $u + 4a = 20$ --- $(ii)$
Subtracting equation $(i)$ from $(ii)$:
$(u + 4a) - (u + 2a) = 20 - 10$
$2a = 10 \implies a = 5\,m/s^2$
Substituting $a = 5$ in equation $(i)$:
$u + 2(5) = 10$
$u + 10 = 10 \implies u = 0\,m/s$
Thus,the initial velocity is $0\,m/s$ and acceleration is $5\,m/s^2$.

(A) Given the velocity $v = 5 \sqrt{x}$.
We know that acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.
First,find $\frac{dv}{dx}$ by differentiating $v = 5x^{1/2}$ with respect to $x$:
$\frac{dv}{dx} = 5 \cdot \frac{1}{2} x^{-1/2} = \frac{2.5}{\sqrt{x}}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (5 \sqrt{x}) \cdot \left( \frac{2.5}{\sqrt{x}} \right) = 5 \cdot 2.5 = 12.5 \, m/s^2$.
Thus,the acceleration of the particle is $12.5 \, m/s^2$.

(C) The particle undergoes uniformly accelerated motion. Let the starting point be $x = 0$ at $t = 0$. The position of the particle at any time $t$ is given by $x(t) = ut + \frac{1}{2}at^2$.
Since the particle returns to the starting point at $t = 10\,s$,we have $x(10) = 0$.
$10u + \frac{1}{2}a(10)^2 = 0 \implies 10u + 50a = 0 \implies u = -5a$.
Now,the position at time $t$ is $x(t) = (-5a)t + \frac{1}{2}at^2 = a(\frac{t^2}{2} - 5t)$.
We want to find $t'$ such that $x(7) = x(t')$.
$x(7) = a(\frac{49}{2} - 35) = a(24.5 - 35) = -10.5a$.
Setting $x(t') = -10.5a$,we get $a(\frac{t'^2}{2} - 5t') = -10.5a$.
$\frac{t'^2}{2} - 5t' + 10.5 = 0 \implies t'^2 - 10t' + 21 = 0$.
Solving the quadratic equation: $(t' - 7)(t' - 3) = 0$.
Thus,$t' = 7\,s$ or $t' = 3\,s$.
Since we are looking for a time other than $7\,s$,the location is the same at $3\,s$.

(B) Let the time after which the two bodies meet be $t$ seconds.
For the first body moving with uniform velocity $v = 8\,m/s$,the displacement $s_1$ in time $t$ is given by $s_1 = v \times t = 8t$.
For the second body starting from rest $(u = 0)$ with uniform acceleration $a = 4\,m/s^2$,the displacement $s_2$ in time $t$ is given by $s_2 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 4 \times t^2 = 2t^2$.
Since the two bodies meet,their displacements must be equal: $s_1 = s_2$.
Therefore,$8t = 2t^2$.
Rearranging the equation: $2t^2 - 8t = 0$,which gives $2t(t - 4) = 0$.
Since $t$ cannot be $0$ (as they meet after starting),we have $t = 4\,s$.

(D) Using the equation of motion $v^{2} = u^{2} + 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is the acceleration (retardation),and $s$ is the stopping distance.
Since the train stops,$v = 0$,so $0 = u^{2} - 2as$,which gives $s = \frac{u^{2}}{2a}$.
Since the retarding force $F = ma$ is constant,the retardation $a$ is constant.
Therefore,$s \propto u^{2}$.
Given $s_{1} = 80 \ m$ for initial speed $u_{1} = u$.
If the speed is doubled,$u_{2} = 2u$.
Then,$\frac{s_{2}}{s_{1}} = \left(\frac{u_{2}}{u_{1}}\right)^{2} = \left(\frac{2u}{u}\right)^{2} = 4$.
$s_{2} = 4 \times s_{1} = 4 \times 80 \ m = 320 \ m$.
Thus,the stopping distance becomes four times the original distance.

(C) Given:
Acceleration $a = 2\,m/s^2$.
Body $A$ starts at $t = 0$ and travels for $t_A = 2\,s$ (one second after $B$ starts,which is $1+1=2\,s$ total).
Body $B$ starts at $t = 1\,s$ and travels for $t_B = 1\,s$ (at the end of the next second).
Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$:
Distance traveled by $A$ in $2\,s$ is $s_A = \frac{1}{2} \times 2 \times (2)^2 = 4\,m$.
Distance traveled by $B$ in $1\,s$ is $s_B = \frac{1}{2} \times 2 \times (1)^2 = 1\,m$.
The separation between the two bodies is $s_A - s_B = 4\,m - 1\,m = 3\,m$.

(A) Let the initial velocity be $u$. The velocity reduces by $\frac{3}{4}u$ in time $t_0$,so the final velocity $v$ at time $t_0$ is $v = u - \frac{3}{4}u = \frac{1}{4}u$.
Using the first equation of motion,$v = u + at$,where acceleration is negative (retardation),we have $\frac{1}{4}u = u - at_0$.
Rearranging gives $at_0 = u - \frac{1}{4}u = \frac{3}{4}u$,which implies $a = \frac{3u}{4t_0}$.
Let $T$ be the total time taken for the velocity to become zero. Using $v = u - aT$ with $v = 0$,we get $0 = u - aT$,so $T = \frac{u}{a}$.
Substituting the value of $a$,we get $T = \frac{u}{(3u / 4t_0)} = \frac{4}{3}t_0$.

(B) Given the position equation: $x = 4(t-2) + a(t-2)^2$.
To find the velocity $v$,we differentiate $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}[4(t-2) + a(t-2)^2] = 4 + 2a(t-2)$.
To find the acceleration $a_{acc}$,we differentiate the velocity $v$ with respect to time $t$:
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}[4 + 2a(t-2)] = 2a$.
Checking the options:
$(a)$ Initial velocity at $t=0$ is $v = 4 + 2a(0-2) = 4 - 4a$. This is not $4$ unless $a=0$.
$(b)$ The acceleration is $2a$,which matches our derivation.
$(c)$ At $t=0$,$x = 4(0-2) + a(0-2)^2 = -8 + 4a$. This is not $0$ unless $a=2$.
Therefore,the correct statement is that the acceleration of the particle is $2a$.

(C) For uniformly accelerated motion,the position $x$ must be a quadratic function of time $t$,i.e.,$x = x_0 + u t + \frac{1}{2} a t^2$.
Given option $(C)$: $t = \sqrt{\frac{x+a}{b}}$
Squaring both sides:
$t^2 = \frac{x+a}{b}$
$b t^2 = x + a$
$x = b t^2 - a$
Comparing this with the standard kinematic equation $x = x_0 + u t + \frac{1}{2} a t^2$,we see that the acceleration is constant $(a_{accel} = 2b)$ and initial velocity is zero. Thus,this represents uniformly accelerated motion.

(B) Let the total time be $T$. The first half time is $t = T/2$.
For particle $A$:
Velocity at $t = T/2$ is $v_A = 0 + 2(T/2) = T$.
Distance in first half $s_{A1} = 0 + 1/2(2)(T/2)^2 = T^2/4$.
Distance in second half $s_{A2} = v_A(T/2) + 1/2(4)(T/2)^2 = T(T/2) + 2(T^2/4) = T^2/2 + T^2/2 = T^2$.
Total distance $S_A = T^2/4 + T^2 = 5T^2/4$.
For particle $B$:
Velocity at $t = T/2$ is $v_B = 0 + 4(T/2) = 2T$.
Distance in first half $s_{B1} = 0 + 1/2(4)(T/2)^2 = T^2/2$.
Distance in second half $s_{B2} = v_B(T/2) + 1/2(2)(T/2)^2 = 2T(T/2) + 1(T^2/4) = T^2 + T^2/4 = 5T^2/4$.
Wait,let's re-evaluate the graph. The area under the $v-t$ graph represents displacement. Since both particles start from rest and reach the same final velocity at time $T$ $(v_{final} = a_1(T/2) + a_2(T/2) = (2+4)(T/2) = 3T)$,the area under the curve for $B$ is clearly larger because it gains velocity faster in the first half.
Thus,particle $B$ covers a larger distance.

(B) Given the velocity $v = \frac{dx}{dt} = \alpha \sqrt{x}$.
Separating the variables,we get $\frac{dx}{\sqrt{x}} = \alpha dt$.
Integrating both sides,$\int_{0}^{x} x^{-1/2} dx = \int_{0}^{t} \alpha dt$.
This yields $[2x^{1/2}]_{0}^{x} = \alpha t$.
So,$2\sqrt{x} = \alpha t$,which implies $\sqrt{x} = \frac{\alpha}{2} t$.
Squaring both sides,$x = \frac{\alpha^2}{4} t^2$.
Therefore,the displacement $x$ is proportional to $t^2$.

(B) Given the position-time relation: $x^2 = 2 + t$.
Differentiating both sides with respect to $t$:
$2x \frac{dx}{dt} = 1$.
Since velocity $v = \frac{dx}{dt}$,we have $2xv = 1$,which implies $v = \frac{1}{2x}$.
Acceleration $a$ is given by $a = v \frac{dv}{dx}$.
First,find $\frac{dv}{dx}$ by differentiating $v = \frac{1}{2} x^{-1}$ with respect to $x$:
$\frac{dv}{dx} = \frac{1}{2} (-1) x^{-2} = -\frac{1}{2x^2}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (\frac{1}{2x}) \times (-\frac{1}{2x^2}) = -\frac{1}{4x^3}$.

(B) For the acceleration phase starting from rest: $v_{\max} = \alpha t_1$ and $x_1 = \frac{1}{2} \alpha t_1^2$.
For the retardation phase coming to rest: $v_{\max} = \beta t_2$ and $x_2 = \frac{1}{2} \beta t_2^2$.
Equating the maximum velocities: $\alpha t_1 = \beta t_2$,which implies $\frac{t_1}{t_2} = \frac{\beta}{\alpha}$.
Now,taking the ratio of distances: $\frac{x_1}{x_2} = \frac{\frac{1}{2} \alpha t_1^2}{\frac{1}{2} \beta t_2^2} = \frac{\alpha}{\beta} \left( \frac{t_1}{t_2} \right)^2$.
Substituting $\frac{\alpha}{\beta} = \frac{t_2}{t_1}$ into the distance ratio: $\frac{x_1}{x_2} = \left( \frac{t_2}{t_1} \right) \left( \frac{t_1}{t_2} \right)^2 = \frac{t_1}{t_2}$.
Since $\frac{t_1}{t_2} = \frac{\beta}{\alpha}$,we get $\frac{x_1}{x_2} = \frac{t_1}{t_2} = \frac{\beta}{\alpha}$.

(C) Given $x \propto t^{5/2}$,so $x = K t^{5/2}$ where $K$ is a constant.
Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(K t^{5/2}) = K \cdot \frac{5}{2} t^{3/2}$.
Thus,$v \propto t^{3/2}$.
Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}(\frac{5}{2} K t^{3/2}) = \frac{5}{2} K \cdot \frac{3}{2} t^{1/2} = \frac{15}{4} K t^{1/2}$.
Thus,$a \propto t^{1/2}$ or $a \propto \sqrt{t}$.
Therefore,both $(A)$ and $(B)$ are correct.

(A) Let the initial velocity be $u$ and acceleration be $a$.
For the first interval of $t = 4\, s$,the distance $s_1 = 24\, m$:
$s = ut + \frac{1}{2}at^2$
$24 = u(4) + \frac{1}{2}a(4)^2$
$24 = 4u + 8a$
Dividing by $4$,we get: $6 = u + 2a$ $...(i)$
For the total time $t = 8\, s$,the total distance $s_{total} = 24 + 64 = 88\, m$:
$88 = u(8) + \frac{1}{2}a(8)^2$
$88 = 8u + 32a$
Dividing by $8$,we get: $11 = u + 4a$ $...(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(u + 4a) - (u + 2a) = 11 - 6$
$2a = 5 \implies a = 2.5\, m/s^2$
Substituting $a = 2.5$ in equation $(i)$:
$u + 2(2.5) = 6$
$u + 5 = 6$
$u = 1\, m/s$.

(D) Given acceleration $a = \frac{dv}{dt} = 3t^2 + 2t + 2$.
To find the velocity $v$,we integrate the acceleration with respect to time: $dv = (3t^2 + 2t + 2)dt$.
Integrating both sides with the given limits: $\int_{v_0}^{v} dv = \int_{0}^{t} (3t^2 + 2t + 2) dt$.
Given $v_0 = 2\,m/s$ at $t = 0$,we have: $v - 2 = [t^3 + t^2 + 2t]_0^2$.
Substituting the upper limit $t = 2$: $v - 2 = (2^3 + 2^2 + 2(2)) - 0$.
$v - 2 = 8 + 4 + 4 = 16$.
$v = 16 + 2 = 18\,m/s$.

(B) Given,acceleration $a = 4t$.
Since the particle starts from rest,the initial velocity $u = 0$ at $t = 0$.
We know that $a = \frac{dv}{dt}$,so $dv = a \ dt = 4t \ dt$.
Integrating both sides,we get $v = \int 4t \ dt = 2t^2 + C$.
At $t = 0$,$v = 0$,so $C = 0$. Thus,$v = 2t^2$.
We know that $v = \frac{dx}{dt}$,so $dx = v \ dt = 2t^2 \ dt$.
Integrating both sides to find the distance $d$ covered from the origin ($x=0$ at $t=0$):
$d = \int_{0}^{3} 2t^2 \ dt = \left[ \frac{2t^3}{3} \right]_{0}^{3}$.
$d = \frac{2(3)^3}{3} = \frac{2 \times 27}{3} = 2 \times 9 = 18 \ m$.

(C) Given that the body starts from rest,the initial velocity $u = 0$ at $t = 0$.
We know that acceleration $a = \frac{dv}{dt}$,so $dv = a \ dt$.
Integrating both sides from $t = 0$ to $t = 2 \ s$:
$v = \int_{0}^{2} a \ dt = \int_{0}^{2} (3t + 4) \ dt$.
$v = \left[ \frac{3t^2}{2} + 4t \right]_{0}^{2}$.
$v = \left( \frac{3(2)^2}{2} + 4(2) \right) - (0)$.
$v = \left( \frac{3 \times 4}{2} + 8 \right) = 6 + 8 = 14 \ m/s$.

(D) The displacement is given by $x = \alpha t^2 - \beta t^3$.
$1$. To check if the particle returns to its starting point,set $x = 0$:
$\alpha t^2 - \beta t^3 = 0 \implies t^2(\alpha - \beta t) = 0$.
This gives $t = 0$ and $t = \frac{\alpha}{\beta}$. Since the particle returns to $x = 0$ at $t = \frac{\alpha}{\beta}$,statement $(a)$ is incorrect.
$2$. Velocity $v = \frac{dx}{dt} = 2\alpha t - 3\beta t^2$. Setting $v = 0$ for rest:
$t(2\alpha - 3\beta t) = 0 \implies t = 0$ or $t = \frac{2\alpha}{3\beta}$. Thus,statement $(b)$ is correct.
$3$. Acceleration $a = \frac{dv}{dt} = 2\alpha - 6\beta t$. At $t = 0$,$a = 2\alpha$. Since $2\alpha \neq 0$,the initial acceleration is not zero. Thus,statement $(c)$ is incorrect.
Since both $(a)$ and $(c)$ are incorrect,the correct choice is $(d)$.

(B) The position of the particle is given by $x = 32t - \frac{8t^3}{3}$.
To find the velocity $v$,we differentiate $x$ with respect to $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(32t - \frac{8t^3}{3}) = 32 - 8t^2$.
The particle is at rest when $v = 0$:
$32 - 8t^2 = 0 \implies 8t^2 = 32 \implies t^2 = 4 \implies t = 2 \, s$ (since $t > 0$).
To find the acceleration $a$,we differentiate $v$ with respect to $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(32 - 8t^2) = -16t$.
At the instant $t = 2 \, s$,the acceleration is:
$a = -16(2) = -32 \, m/s^2$.