Uniformly Accelerated Motion Questions in English

(D) Given the velocity $v = k\sqrt{x}$.
We know that acceleration $a = v \frac{dv}{dx}$.
First,find $\frac{dv}{dx}$ by differentiating $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(k x^{1/2}) = k \cdot \frac{1}{2} x^{-1/2} = \frac{k}{2\sqrt{x}}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula: $a = (k\sqrt{x}) \cdot \left(\frac{k}{2\sqrt{x}}\right)$.
Simplifying this,we get $a = \frac{k^2}{2}$.
Since $k$ is a constant,$a$ is also a constant.

(B) The correct answer is $B$.
For uniformly accelerated motion,the equation of motion is $v^2 = u^2 + 2as$,where $v$ is final velocity,$u$ is initial velocity,$a$ is acceleration,and $s$ is displacement.
Comparing this with the equation of a straight line $y = mx + c$,where $y = v^2$,$x = s$,$m = 2a$ (slope),and $c = u^2$ (intercept on the $v^2$ axis).
Since the graph is a straight line,it represents uniformly accelerated motion. This makes statement $A$ correct.
Since the graph has a non-zero intercept on the $v^2$ axis,$u^2 \neq 0$,which implies the initial velocity $u \neq 0$. Therefore,statement $B$ is wrong.
For uniformly accelerated motion,$s = ut + \frac{1}{2}at^2$,which is a quadratic equation in $t$,representing a parabola in the $s-t$ graph. Therefore,statement $C$ is correct.
Since statement $B$ is wrong,the correct choice is $B$.

(D) From the third equation of motion,we have $v^2 = u^2 + 2as$,which can be rewritten as $v^2 = 2as + u^2$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = v^2$ and $x = s$,the slope of the graph is $m = 2a$.
From the given graph,the initial value of $v^2$ at $s = 0$ is $25 \, m^2/s^2$,and at $s = 2 \, m$,the value of $v^2$ is $9 \, m^2/s^2$.
The slope of the line is calculated as:
$\text{Slope} = \frac{v_2^2 - v_1^2}{s_2 - s_1} = \frac{9 - 25}{2 - 0} = \frac{-16}{2} = -8 \, m/s^2$.
Since the slope is equal to $2a$,we have:
$2a = -8 \, m/s^2$
$a = -4 \, m/s^2$.
Therefore,the acceleration of the particle is $-4 \, m/s^2$.

(A) Given that the body starts from rest,initial velocity $u = 0$.
Using the first equation of motion,$v = u + at$,we get $v = 0 + an$,which implies $a = \frac{v}{n}$.
The total displacement in $n \ s$ is $S_n = \frac{1}{2}an^2$.
The displacement in the first $(n-3) \ s$ is $S_{n-3} = \frac{1}{2}a(n-3)^2$.
The displacement in the last $3 \ s$ is given by $\Delta S = S_n - S_{n-3}$.
Substituting the values: $\Delta S = \frac{1}{2}a[n^2 - (n-3)^2]$.
Expanding the term: $\Delta S = \frac{1}{2}a[n^2 - (n^2 - 6n + 9)] = \frac{1}{2}a(6n - 9)$.
Substituting $a = \frac{v}{n}$: $\Delta S = \frac{1}{2} \cdot \frac{v}{n} \cdot (6n - 9) = \frac{v(6n - 9)}{2n}$.

(D) The velocity $v$ decreases linearly with displacement $s$. The equation of a straight line is $v = ms + c$.
At $s=0$,$v=20 \ m/s$,so $c=20$.
At $s=30 \ m$,$v=0$,so $0 = m(30) + 20$,which gives $m = -20/30 = -2/3$.
Thus,the velocity function is $v = -\frac{2}{3}s + 20$.
The acceleration $a$ is given by $a = v \frac{dv}{ds}$.
First,find $\frac{dv}{ds} = \frac{d}{ds}(-\frac{2}{3}s + 20) = -\frac{2}{3}$.
At $s=15 \ m$,the velocity is $v = -\frac{2}{3}(15) + 20 = -10 + 20 = 10 \ m/s$.
Therefore,the acceleration $a = (10)(-\frac{2}{3}) = -\frac{20}{3} \ m/s^2$.

(A) Given: Initial velocity $u = 20 \; km/h$,final velocity $v = 60 \; km/h$,and time $t = 4 \; h$.
First,we calculate the acceleration $a$ using the equation $v = u + at$:
$60 = 20 + (a \times 4)$
$40 = 4a$
$a = 10 \; km/h^2$.
Now,we calculate the distance $d$ using the equation $d = ut + \frac{1}{2}at^2$:
$d = (20 \times 4) + \frac{1}{2} \times 10 \times (4)^2$
$d = 80 + 5 \times 16$
$d = 80 + 80 = 160 \; km$.

(C) The displacement is given by $x = at + bt^2 - ct^3$.
Velocity $(v)$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = a + 2bt - 3ct^2$.
Acceleration $(a_{acc})$ is the derivative of velocity with respect to time: $a_{acc} = \frac{dv}{dt} = 2b - 6ct$.
Setting acceleration to zero: $2b - 6ct = 0 \implies t = \frac{2b}{6c} = \frac{b}{3c}$.
Now,substitute $t = \frac{b}{3c}$ into the velocity equation:
$v = a + 2b(\frac{b}{3c}) - 3c(\frac{b}{3c})^2$
$v = a + \frac{2b^2}{3c} - 3c(\frac{b^2}{9c^2})$
$v = a + \frac{2b^2}{3c} - \frac{b^2}{3c}$
$v = a + \frac{b^2}{3c}$.

(B) The distance between the two particles is given by $s = s_1 - s_2$,where $s_1 = ut$ and $s_2 = \frac{1}{2}ft^2$.
$s = ut - \frac{1}{2}ft^2$.
To find the maximum distance,we differentiate $s$ with respect to $t$ and set it to zero:
$\frac{ds}{dt} = u - ft = 0 \implies t = \frac{u}{f}$.
At this time $t = \frac{u}{f}$,the velocity of the second particle equals the velocity of the first particle.
Substituting $t = \frac{u}{f}$ into the expression for $s$:
$s_{\max} = u(\frac{u}{f}) - \frac{1}{2}f(\frac{u}{f})^2$
$s_{\max} = \frac{u^2}{f} - \frac{u^2}{2f} = \frac{u^2}{2f}$.

(A) The position of a particle moving with constant acceleration is given by the equation $x(t) = ut + \frac{1}{2}at^2$.
For particle $A$ at $t=2\,s$:
$x_A = (10)(2) + \frac{1}{2}(-4)(2)^2 = 20 - 8 = 12\,m$.
For particle $B$ at $t=2\,s$:
$x_B = (20)(2) + \frac{1}{2}(-2)(2)^2 = 40 - 4 = 36\,m$.
The distance between the two particles is $|x_B - x_A| = |36 - 12| = 24\,m$.

(C) Given the relation $u^2 = kr$,we have $u = \sqrt{k} \cdot r^{1/2}$.
Since $u = \frac{dr}{dt}$,we can write $\frac{dr}{dt} = \sqrt{k} \cdot r^{1/2}$.
Separating the variables,we get $r^{-1/2} \, dr = \sqrt{k} \, dt$.
Integrating both sides with the initial condition $r = 0$ at $t = 0$: $\int_{0}^{r} r^{-1/2} \, dr = \int_{0}^{t} \sqrt{k} \, dt$.
This gives $[2r^{1/2}]_{0}^{r} = \sqrt{k} \cdot t$,so $2\sqrt{r} = \sqrt{k} \cdot t$,which implies $\sqrt{r} = \frac{\sqrt{k}}{2} t$.
Substituting $\sqrt{r}$ back into the velocity equation $u = \sqrt{k} \cdot \sqrt{r}$:
$u = \sqrt{k} \cdot (\frac{\sqrt{k}}{2} t) = \frac{k}{2} t$.
At $t = 1 \, s$,the velocity $u = \frac{k}{2} (1) = \frac{k}{2}$.

(B) Given: mass $m = 1\,kg$,initial velocity $u = 60\,ms^{-1}$,force $F = -10\,N$ (directed towards $O$,opposite to motion).
Acceleration $a = \frac{F}{m} = \frac{-10}{1} = -10\,ms^{-2}$.
To find the time taken to return to point $O$,the displacement $s$ must be $0$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$:
$0 = 60t + \frac{1}{2}(-10)t^2$
$0 = 60t - 5t^2$
$5t^2 = 60t$
Since $t \neq 0$,we divide by $5t$:
$t = \frac{60}{5} = 12\,s$.
Thus,the body will again cross $O$ in $12\,s$.

(A) Let the initial velocity be $u = 10\,m/s$ and the acceleration be $a$. Let the displacement when the velocity becomes $v_1 = 50\,m/s$ be $s$.
Using the equation of motion $v^2 = u^2 + 2as$:
$(50)^2 = (10)^2 + 2as$
$2500 = 100 + 2as$
$2as = 2400$
Now,the acceleration is reversed,so the new acceleration is $-a$. The particle travels from the point where $v = 50\,m/s$ back to the starting point,covering a displacement of $-s$.
Using the equation of motion $v_f^2 = v_i^2 + 2a's'$:
$v_f^2 = (50)^2 + 2(-a)(-s)$
$v_f^2 = 2500 + 2as$
Substitute $2as = 2400$ into the equation:
$v_f^2 = 2500 + 2400 = 4900$
$v_f = \sqrt{4900} = 70\,m/s$.

(D) Given that the distance $s$ is proportional to $t^{1/2}$,we can write $s = k t^{1/2}$,where $k$ is a constant.
The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(k t^{1/2}) = \frac{1}{2} k t^{-1/2}$.
The acceleration $a$ is the second derivative of displacement with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(\frac{1}{2} k t^{-1/2}) = -\frac{1}{4} k t^{-3/2}$.
The negative sign indicates that the acceleration is in the opposite direction of motion,which represents retardation (deceleration).
As time $t$ increases,the magnitude of acceleration $|a| = \frac{k}{4 t^{3/2}}$ decreases.
Therefore,the motion is characterized by decreasing retardation.

(B) Let $t_1$ be the time of acceleration and $t_2$ be the time of retardation. The maximum velocity reached is $v = a t_1$.
Since the lift comes to rest,$v = (2a) t_2$. Thus,$a t_1 = 2a t_2$,which implies $t_1 = 2 t_2$.
Given the total time $t = t_1 + t_2$,we substitute $t_1 = 2 t_2$ to get $t = 2 t_2 + t_2 = 3 t_2$. Therefore,$t_2 = \frac{t}{3}$ and $t_1 = \frac{2t}{3}$.
The total height $h$ is the sum of the distances covered during acceleration and retardation:
$h = \frac{1}{2} a t_1^2 + (v t_2 - \frac{1}{2} (2a) t_2^2)$
Substituting $v = a t_1 = 2a t_2$:
$h = \frac{1}{2} a (2 t_2)^2 + (2a t_2 \cdot t_2 - a t_2^2) = 2 a t_2^2 + a t_2^2 = 3 a t_2^2$.
Substituting $t_2 = \frac{t}{3}$:
$h = 3 a (\frac{t}{3})^2 = 3 a (\frac{t^2}{9}) = \frac{a t^2}{3}$.

(B) Let the initial position be $x=0$. The particle starts from rest $(u=0)$ with acceleration $a$ for time $t_0$.
At $t=t_0$,the displacement is $s_1 = \frac{1}{2} a t_0^2$ and the velocity is $v_1 = a t_0$.
After $t_0$,the acceleration becomes $-a$. Let the particle return to the initial position $(x=0)$ at a total time $T = t_0 + t$,where $t$ is the time elapsed after the acceleration change.
The displacement equation for the second phase is $s = u t + \frac{1}{2} a t^2$. Here,the initial velocity is $v_1 = a t_0$ and acceleration is $-a$. We want the total displacement from the start to be zero,so the displacement in the second phase must be $-s_1 = -\frac{1}{2} a t_0^2$.
Thus,$-\frac{1}{2} a t_0^2 = (a t_0) t - \frac{1}{2} a t^2$.
Dividing by $-\frac{1}{2} a$,we get $t_0^2 = -2 t_0 t + t^2$,which rearranges to $t^2 - 2 t_0 t - t_0^2 = 0$.
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $t = \frac{2 t_0 \pm \sqrt{4 t_0^2 + 4 t_0^2}}{2} = t_0 \pm \sqrt{2} t_0$.
Since $t > 0$,we take $t = (1 + \sqrt{2}) t_0$.
The total time is $T = t_0 + t = t_0 + (1 + \sqrt{2}) t_0 = (2 + \sqrt{2}) t_0$.

(B) The displacement of the particle is given by $s = t^3 - 6t^2 + 3t + 4$.
Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = 3t^2 - 12t + 3$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = 6t - 12$.
To find the time when acceleration is zero,set $a = 0$: $6t - 12 = 0 \implies t = 2 \ s$.
Now,substitute $t = 2 \ s$ into the velocity equation: $v = 3(2)^2 - 12(2) + 3 = 3(4) - 24 + 3 = 12 - 24 + 3 = -9 \ m/s$.

(B) Let the acceleration of the particle be $x$ and the initial velocity be $u=0$.
For the first interval of time $p$,the distance covered is $a = \frac{1}{2} x p^2$.
For the total time $(p+q)$,the total distance covered is $a+b = \frac{1}{2} x (p+q)^2$.
From the first equation,$x = \frac{2a}{p^2}$.
Substituting this into the second equation: $a+b = \frac{1}{2} (\frac{2a}{p^2}) (p+q)^2 = a \frac{(p+q)^2}{p^2}$.
This approach is complex. Let's use the average velocity method.
Velocity at time $p$ is $v_1 = xp$.
Distance $a = \frac{1}{2} xp^2$.
Distance $b$ is covered in time $q$ starting with velocity $v_1$. So,$b = v_1 q + \frac{1}{2} x q^2 = (xp)q + \frac{1}{2} x q^2 = xpq + \frac{1}{2} x q^2$.
We have $a = \frac{1}{2} xp^2 \implies x = \frac{2a}{p^2}$.
Substitute $x$ in $b$: $b = (\frac{2a}{p^2})pq + \frac{1}{2} (\frac{2a}{p^2}) q^2 = \frac{2aq}{p} + \frac{aq^2}{p^2} = \frac{2aqp + aq^2}{p^2} = \frac{aq(2p+q)}{p^2}$.
Thus,$x = \frac{2(bp - aq)}{pq(p+q)}$ is not the standard form. Let's re-evaluate: $b = \frac{1}{2}x(p+q)^2 - \frac{1}{2}xp^2 = \frac{1}{2}x(q^2+2pq)$.
$x = \frac{2b}{q^2+2pq} = \frac{2b}{q(q+2p)}$.

(A) The ground floor is the origin $(x = 0)$,and the upward direction is positive. Since the lift is between the ground floor and the $8^{th}$ floor,its position $x$ is positive $(x > 0)$.
As the lift is moving downwards,its velocity $v$ is in the negative direction,so $v < 0$.
Since the lift is about to stop at the $4^{th}$ floor,it must be decelerating (retarding). For a downward-moving object,deceleration means the acceleration $a$ must be in the opposite direction to the velocity,i.e.,in the upward (positive) direction. Therefore,$a > 0$.
Thus,the correct signs are $x > 0, v < 0, a > 0$.

(D) Given $x^{2} = at^{2} + 2bt + c$.
Differentiating with respect to $t$:
$2x \frac{dx}{dt} = 2at + 2b$
$x v = at + b$,where $v = \frac{dx}{dt}$.
Differentiating again with respect to $t$:
$v \frac{dx}{dt} + x \frac{dv}{dt} = a$
$v^{2} + x a' = a$,where $a' = \frac{dv}{dt}$ is the acceleration.
From $x v = at + b$,we have $v = \frac{at+b}{x}$.
Substituting $v$ into the equation:
$x a' = a - v^{2} = a - \left(\frac{at+b}{x}\right)^{2}$
$x a' = \frac{ax^{2} - (at+b)^{2}}{x^{2}}$
Substitute $x^{2} = at^{2} + 2bt + c$:
$x a' = \frac{a(at^{2} + 2bt + c) - (a^{2}t^{2} + 2abt + b^{2})}{x^{2}}$
$x a' = \frac{a^{2}t^{2} + 2abt + ac - a^{2}t^{2} - 2abt - b^{2}}{x^{2}}$
$x a' = \frac{ac - b^{2}}{x^{2}}$
$a' = \frac{ac - b^{2}}{x^{3}}$
Thus,$a' \propto x^{-3}$. Comparing with $x^{-n}$,we get $n = 3$.

(N/A) By definition,acceleration is $a = \frac{dv}{dt}$.
Rearranging gives $dv = a dt$.
Integrating both sides from initial velocity $v_0$ at $t=0$ to $v$ at time $t$:
$\int_{v_0}^{v} dv = \int_{0}^{t} a dt = a \int_{0}^{t} dt$
$v - v_0 = at \implies v = v_0 + at$.
Since $v = \frac{dx}{dt}$,we have $dx = v dt = (v_0 + at) dt$.
Integrating both sides from initial position $x_0$ at $t=0$ to $x$ at time $t$:
$\int_{x_0}^{x} dx = \int_{0}^{t} (v_0 + at) dt$
$x - x_0 = v_0 t + \frac{1}{2} at^2 \implies x = x_0 + v_0 t + \frac{1}{2} at^2$.
Using the chain rule,$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$.
Thus,$v dv = a dx$.
Integrating both sides from $v_0$ to $v$ and $x_0$ to $x$:
$\int_{v_0}^{v} v dv = \int_{x_0}^{x} a dx$
$\frac{v^2 - v_0^2}{2} = a(x - x_0)$
$v^2 = v_0^2 + 2a(x - x_0)$.

(N/A) Let the distance travelled by the vehicle before it stops be $d_{s}$.
Using the third equation of motion,$v^{2} = v_{0}^{2} + 2ax$,where $v$ is the final velocity,$v_{0}$ is the initial velocity,$a$ is the acceleration (which is negative in this case,i.e.,$-a$),and $x$ is the displacement.
At the point where the vehicle stops,the final velocity $v = 0$.
Substituting these values into the equation: $0^{2} = v_{0}^{2} + 2(-a)d_{s}$.
Rearranging the terms to solve for $d_{s}$:
$2ad_{s} = v_{0}^{2}$
$d_{s} = \frac{v_{0}^{2}}{2a}$.
Thus,the stopping distance is directly proportional to the square of the initial velocity $(d_{s} \propto v_{0}^{2})$.

(D) First,convert the initial velocity from $km/h$ to $m/s$:
$u = 126 \times \frac{5}{18} = 35 \; m/s$.
Since the car comes to a stop,the final velocity $v = 0 \; m/s$ and the distance covered $s = 200 \; m$.
Using the third equation of motion,$v^2 - u^2 = 2as$:
$0^2 - (35)^2 = 2 \times a \times 200$.
$-1225 = 400a$.
$a = -3.0625 \; m/s^2$.
Now,use the first equation of motion,$v = u + at$,to find the time $t$:
$0 = 35 + (-3.0625)t$.
$t = \frac{35}{3.0625} \approx 11.4 \; s$.

(B) First,convert the initial velocity from $km/h$ to $m/s$:
$u = 126 \; km/h = 126 \times \frac{5}{18} \; m/s = 35 \; m/s$.
Since the car comes to a stop,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: $0^2 - (35)^2 = 2 \times a \times 200$.
$-1225 = 400a$.
$a = -\frac{1225}{400} = -3.0625 \; m/s^2$.
The retardation is the magnitude of negative acceleration,which is $3.0625 \; m/s^2$.

(A) The plot is a straight line.
The distance covered by a body in the $n^{\text{th}}$ second is given by the formula:
$D_{n} = u + \frac{a}{2}(2n - 1) \quad \dots(i)$
Where:
$u = \text{Initial velocity}$
$a = \text{Acceleration}$
$n = \text{Time interval in seconds}$
In the given case,$u = 0$ and $a = 1\; m/s^{2}$. Substituting these values into equation $(i)$:
$D_{n} = 0 + \frac{1}{2}(2n - 1) = n - 0.5$
This relation $D_{n} = n - 0.5$ shows that $D_{n}$ is a linear function of $n$. Therefore,the plot of $D_{n}$ versus $n$ is a straight line.
Substituting different values of $n$ from $1$ to $10$:

$n$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$D_{n}$	$0.5$	$1.5$	$2.5$	$3.5$	$4.5$	$5.5$	$6.5$	$7.5$	$8.5$	$9.5$

Since the three-wheeler moves with uniform velocity after $10\; s$,the acceleration becomes zero. For $n > 10$,the distance covered in each subsequent second remains constant,so the plot becomes a horizontal line parallel to the $n$-axis.

(C) If the decrease in velocity is the same in equal intervals of time,the acceleration $a$ is constant and negative $(a < 0)$.
Since $v = u + at$,the velocity decreases linearly with time.
The position $x$ is given by the kinematic equation $x(t) = ut + \frac{1}{2}at^2$.
Since $a$ is negative,the $x-t$ graph is a parabola that opens downwards.
This represents a particle undergoing uniform retardation.

(A) The average velocity is defined as the total displacement divided by the total time interval,$v_{avg} = \frac{\Delta x}{\Delta t}$.
The instantaneous velocity is defined as the velocity at a specific instant of time,$v = \frac{dx}{dt}$.
If the velocity of an object remains constant over a time interval,then the rate of change of position is constant.
In this case,the displacement $\Delta x$ in time $\Delta t$ is given by $\Delta x = v \cdot \Delta t$.
Therefore,$v_{avg} = \frac{v \cdot \Delta t}{\Delta t} = v$.
This condition implies that the velocity does not change with time,which is the definition of uniform motion.

(N/A) The velocity-time $(v-t)$ graphs for motion with constant acceleration are as follows:
$(a)$ An object is moving in a positive direction with a positive acceleration.
$(b)$ An object is moving in a positive direction with a negative acceleration (deceleration).
$(c)$ An object is moving in a negative direction with a negative acceleration.
$(d)$ An object is moving in a positive direction until time $t_{1}$,then it turns back and moves with the same negative acceleration.
An interesting feature of a velocity-time graph for any moving object is that the area under the curve represents the displacement over a given time interval.

(C) For motion with constant negative acceleration $(a < 0)$,the position-time equation is given by the kinematic formula: $x(t) = x_0 + v_0t + \frac{1}{2}at^2$.
Since the acceleration $a$ is negative,the coefficient of the $t^2$ term is negative.
This equation represents a quadratic function in $t$,which graphically corresponds to a parabola.
Because the coefficient of $t^2$ is negative,the parabola opens downwards.
Therefore,the $x-t$ graph for negative acceleration is a downward-opening parabola.

(N/A) Let the velocity of the particle at $t=0$ be $v_{0}$ and at time $t$ be $v$.
$1$. First Equation of Motion $(v = v_{0} + at)$:
The acceleration $a$ is the slope of the velocity-time graph line $AB$.
$a = \text{slope} = \frac{v - v_{0}}{t - 0} = \frac{v - v_{0}}{t}$
$\therefore v = v_{0} + at$.
$2$. Second Equation of Motion $(x = v_{0}t + \frac{1}{2}at^{2})$:
The displacement $x$ is the area under the $v-t$ graph (trapezium $OABD$).
$x = \text{Area of rectangle } OACD + \text{Area of } \Delta ACB$
$x = (v_{0} \times t) + \frac{1}{2} \times (t) \times (v - v_{0})$
Since $(v - v_{0}) = at$,we get:
$x = v_{0}t + \frac{1}{2}at^{2}$.
$3$. Third Equation of Motion $(v^{2} - v_{0}^{2} = 2ax)$:
The displacement $x$ is the area of the trapezium $OABD$.
$x = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$x = \frac{1}{2} (v_{0} + v) \times t$
Since $t = \frac{v - v_{0}}{a}$,we substitute $t$:
$x = \frac{1}{2} (v + v_{0}) \times \frac{(v - v_{0})}{a}$
$2ax = v^{2} - v_{0}^{2}$
$\therefore v^{2} = v_{0}^{2} + 2ax$.

(A) The distance travelled in the $n^{th}$ second is defined as the difference between the total distance covered in $n$ seconds and the total distance covered in $(n-1)$ seconds.
Using the kinematic equation $s = v_0 t + \frac{1}{2} a t^2$:
$d_n = s_n - s_{n-1}$
$d_n = (v_0 n + \frac{1}{2} a n^2) - [v_0 (n-1) + \frac{1}{2} a (n-1)^2]$
$d_n = v_0 n + \frac{1}{2} a n^2 - [v_0 n - v_0 + \frac{1}{2} a (n^2 - 2n + 1)]$
$d_n = v_0 n + \frac{1}{2} a n^2 - v_0 n + v_0 - \frac{1}{2} a n^2 + a n - \frac{1}{2} a$
$d_n = v_0 + a n - \frac{1}{2} a$
$d_n = v_0 + \frac{a}{2} (2n - 1)$

(N/A) When brakes are applied to a moving vehicle,the distance it travels before coming to a complete stop is called the stopping distance.
Suppose the initial velocity of the moving vehicle is $v_{0}$. After applying brakes,the retardation is $-a$.
Let the distance covered be $d_{s}$ (stopping distance) and the final velocity be $v = 0$.
Using the kinematic equation $v^{2} - v_{0}^{2} = 2as$:
$0 - v_{0}^{2} = 2(-a)(d_{s})$
$v_{0}^{2} = 2ad_{s}$
$d_{s} = \frac{v_{0}^{2}}{2a}$
Here,$a$ is the magnitude of retardation,which is constant.
Thus,the stopping distance is proportional to the square of the initial velocity: $d_{s} \propto v_{0}^{2}$.
If the initial velocity is doubled,i.e.,$(v_{0})_{2} = 2(v_{0})_{1}$,then:
$\frac{(d_{s})_{2}}{(d_{s})_{1}} = \frac{(v_{0})_{2}^{2}}{(v_{0})_{1}^{2}} = (2)^{2} = 4$.
Therefore,the stopping distance becomes $4$ times the original distance.

(N/A) The stopping distance of a vehicle is the distance it travels before coming to a complete rest after the brakes are applied. It depends on the following factors:
$1$. Initial velocity $(v_0)$ of the vehicle: The stopping distance is directly proportional to the square of the initial velocity $(d \propto v_0^2)$.
$2$. Deceleration capacity: It depends on the braking efficiency or the maximum deceleration $(a)$ that the vehicle can achieve,which is determined by the friction between the tires and the road surface.

(B) The average acceleration $a_{avg}$ over a time interval $\Delta t$ is defined as the change in velocity divided by the time interval: $a_{avg} = \frac{\Delta v}{\Delta t}$.
The instantaneous acceleration $a$ is defined as the limit of the average acceleration as $\Delta t$ approaches zero: $a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t}$.
For the instantaneous acceleration to be equal to the average acceleration at any time,the acceleration must be constant (uniform acceleration).
If the acceleration is constant,the rate of change of velocity is constant,making the average acceleration over any interval equal to the instantaneous acceleration at any point within that interval.

(N/A) Stopping distance is the distance a vehicle travels after the brakes are applied until it comes to a complete stop.
It depends on the initial velocity $(v_0)$ of the vehicle and the deceleration $(a)$ produced by the brakes.
Using the third equation of motion,$v^2 = v_0^2 + 2as$,where $v$ is the final velocity $(0 \ m/s)$,$v_0$ is the initial velocity,$a$ is the negative acceleration (deceleration),and $s$ is the stopping distance.
Setting $v = 0$,we get $0 = v_0^2 - 2as$,which simplifies to $s = \frac{v_0^2}{2a}$.

(A) Given that the car starts from rest $(u = 0)$ and moves with uniform acceleration $(a = \text{constant})$.
$(i)$ For $x-t$ graph: Using the equation of motion $x = ut + \frac{1}{2}at^2$, since $u=0$, we get $x = \frac{1}{2}at^2$. This represents a parabola passing through the origin.
$(ii)$ For $v-t$ graph: Using the equation $v = u + at$, since $u=0$, we get $v = at$. This represents a straight line passing through the origin with a positive slope.
$(iii)$ For $a-t$ graph: Since the acceleration is uniform, $a$ remains constant with respect to time. This represents a straight line parallel to the time axis.

(A) The average acceleration over a time interval $\Delta t$ is defined as $a_{avg} = \frac{\Delta v}{\Delta t}$.
The instantaneous acceleration is defined as $a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt}$.
For the average acceleration to be equal to the instantaneous acceleration at any time,the rate of change of velocity must be constant.
Therefore,the acceleration must be constant.

(A) Using the third equation of motion: $v^2 - u^2 = 2as$.
Here,the final velocity $v = 0$ (as the vehicle stops).
Let the initial velocity be $u$ and the acceleration be $a = -|a|$ (retardation).
Substituting these values: $0^2 - u^2 = 2(-|a|)s$.
$-u^2 = -2|a|s$.
$s = \frac{u^2}{2|a|}$.
Since $2|a|$ is constant for a given acceleration,the stopping distance $s$ is proportional to the square of the initial speed $(s \propto u^2)$.

(B) The stopping distance $d$ of a vehicle is given by the formula $d = \frac{v_0^2}{2|a|}$, where $v_0$ is the initial velocity and $a$ is the constant deceleration.
Since $d \propto v_0^2$, the ratio of the stopping distances for two vehicles with initial velocities $v_1 = 80 \, km/hr$ and $v_2 = 40 \, km/hr$ is:
$\frac{d_1}{d_2} = \left(\frac{v_1}{v_2}\right)^2 = \left(\frac{80}{40}\right)^2 = (2)^2 = 4$.
Therefore, the stopping distance for the vehicle moving at $80 \, km/hr$ is $4$ times that of the vehicle moving at $40 \, km/hr$.

(B) The average velocity of an object is defined as the total displacement divided by the total time interval. The instantaneous velocity is the velocity at a specific instant of time. When an object moves with uniform velocity (constant velocity) in a straight line,its velocity does not change with time. Therefore,the velocity at any instant is equal to the average velocity over any time interval.

(A) The equation $\Delta S = v \Delta t$ represents motion with a constant velocity $(v)$.
Since the velocity $v$ is constant and does not change with time,the displacement $\Delta S$ is directly proportional to the time interval $\Delta t$.
This is the characteristic equation for uniform motion in a straight line.

(B) Given that the displacement $x$ is proportional to the square of time $t$,we have $x = k t^{2}$,where $k$ is a constant.
The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(k t^{2}) = 2kt$.
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. Therefore,the object has a constant acceleration.

(N/A) To represent periodic motion,we use trigonometric functions like $\sin t$ or $\cos t$.
$(a)$ Consider the motion $x(t) = t - \sin t$.
Velocity $v(t) = \frac{dx}{dt} = 1 - \cos t$.
At $t = 2n\pi$,$v = 1 - 1 = 0$,so the particle comes to rest. Since $v \ge 0$ for all $t$,the particle always moves in the positive $x$-direction.
$(b)$ Consider the motion $x(t) = \sin t$.
Velocity $v(t) = \frac{dx}{dt} = \cos t$.
At $t = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$,the velocity $v = 0$,so the particle comes to rest. Since $\cos t$ alternates between positive and negative values,the particle moves forward and backward periodically.

(N/A) Consider a particle moving in one dimension whose position is given by the function $x(t) = x_0 + A e^{-kt}$,where $x_0 > 0$,$A > 0$,and $k > 0$.
$1$. Position: $x(t) = x_0 + A e^{-kt}$. Since $x_0, A, k, t > 0$,the term $A e^{-kt}$ is positive,so $x(t) > 0$.
$2$. Velocity: $v(t) = \frac{dx}{dt} = -Ak e^{-kt}$. Since $A, k, e^{-kt} > 0$,the product $-Ak e^{-kt}$ is negative,so $v(t) < 0$.
$3$. Acceleration: $a(t) = \frac{dv}{dt} = Ak^2 e^{-kt}$. Since $A, k^2, e^{-kt} > 0$,the acceleration $a(t) > 0$.
An example of such motion is a particle moving towards the origin from the positive side of the $x$-axis,slowing down as it approaches a limiting position $x_0$.

(N/A) Given,$x(t) = x_0 (1 - e^{-\gamma t})$.
Velocity $v(t) = \frac{dx}{dt} = x_0 \gamma e^{-\gamma t}$.
Acceleration $a(t) = \frac{dv}{dt} = -x_0 \gamma^2 e^{-\gamma t}$.
$(a)$ At $t = 0$,$x(0) = x_0(1 - e^0) = 0$. The particle starts at the origin.
Velocity at $t = 0$ is $v(0) = x_0 \gamma e^0 = x_0 \gamma$.
$(b)$
$x(t)$: As $t \to \infty$,$x(t) \to x_0$ (maximum). At $t = 0$,$x(t) = 0$ (minimum). Since $\frac{dx}{dt} > 0$,$x(t)$ increases with time.
$v(t)$: At $t = 0$,$v(0) = x_0 \gamma$ (maximum). As $t \to \infty$,$v(t) \to 0$ (minimum). Since $\frac{dv}{dt} < 0$,$v(t)$ decreases with time.
$a(t)$: At $t = 0$,$a(0) = -x_0 \gamma^2$ (minimum). As $t \to \infty$,$a(t) \to 0$ (maximum). Since $\frac{da}{dt} = x_0 \gamma^3 e^{-\gamma t} > 0$,$a(t)$ increases with time.

(A) The equations of motion for constant acceleration are:
$1$. Velocity-time relation: $v = v_0 + at$. Thus,$(1) \to (a)$.
$2$. Position-time relation: $S = v_0t + \frac{1}{2}at^2$.
$3$. Velocity-displacement relation: $v^2 = v_0^2 + 2as$. Thus,$(2) \to (c)$.
Therefore,the correct matching is $(1-a, 2-c)$.

(C) The distance covered by a body in the $n^{\text{th}}$ second is given by $S_{n} = u + \frac{a}{2}(2n - 1)$.
For body $A$,starting from rest $(u=0)$ at $t=0$,the distance covered in the $5^{\text{th}}$ second is:
$S_{A, 5} = 0 + \frac{a_{1}}{2}(2 \times 5 - 1) = \frac{9a_{1}}{2}$.
Body $B$ starts $2$ seconds after $A$. Therefore,the $5^{\text{th}}$ second after the start of $A$ corresponds to the $(5-2) = 3^{\text{rd}}$ second of motion for body $B$.
For body $B$,starting from rest $(u=0)$,the distance covered in its $3^{\text{rd}}$ second is:
$S_{B, 3} = 0 + \frac{a_{2}}{2}(2 \times 3 - 1) = \frac{5a_{2}}{2}$.
Given that $S_{A, 5} = S_{B, 3}$,we have:
$\frac{9a_{1}}{2} = \frac{5a_{2}}{2}$.
Thus,$\frac{a_{1}}{a_{2}} = \frac{5}{9}$.

(B) The velocity is given by $v = \frac{dx}{dt} = v_{0} + gt + Ft^{2}$.
To find the displacement $s$ at $t = 1$,we integrate the velocity with respect to time from $t = 0$ to $t = 1$:
$s = \int_{0}^{1} v dt = \int_{0}^{1} (v_{0} + gt + Ft^{2}) dt$
$s = [v_{0}t + \frac{gt^{2}}{2} + \frac{Ft^{3}}{3}]_{0}^{1}$
Substituting the limits $t = 0$ and $t = 1$:
$s = (v_{0}(1) + \frac{g(1)^{2}}{2} + \frac{F(1)^{3}}{3}) - (0)$
$s = v_{0} + \frac{g}{2} + \frac{F}{3}$

(A) Let the total length of the train be $L = 2d$,where $d$ is the distance from the engine to the middle point,and also from the middle point to the last compartment.
Let $a$ be the uniform acceleration of the train.
When the engine passes the signal post,its velocity is $u$. When the middle point passes the signal post,let its velocity be $v^{\prime}$.
Using the equation of motion $v_f^2 = v_i^2 + 2as$:
For the first half of the train (from engine to middle point),we have:
$(v^{\prime})^2 = u^2 + 2ad$ --- $(1)$
For the second half of the train (from middle point to last compartment),we have:
$v^2 = (v^{\prime})^2 + 2ad$ --- $(2)$
From equation $(1)$,$2ad = (v^{\prime})^2 - u^2$.
Substituting this into equation $(2)$:
$v^2 = (v^{\prime})^2 + ((v^{\prime})^2 - u^2)$
$v^2 = 2(v^{\prime})^2 - u^2$
$2(v^{\prime})^2 = v^2 + u^2$
$(v^{\prime})^2 = \frac{v^2 + u^2}{2}$
$v^{\prime} = \sqrt{\frac{v^2 + u^2}{2}}$

(A) Given the velocity function: $v = \sqrt{5000 + 24x}$.
We know that acceleration $a$ is given by $a = v \frac{dv}{dx}$.
First,differentiate $v$ with respect to $x$:
$\frac{dv}{dx} = \frac{d}{dx} (5000 + 24x)^{1/2} = \frac{1}{2} (5000 + 24x)^{-1/2} \times 24 = \frac{12}{\sqrt{5000 + 24x}}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = \sqrt{5000 + 24x} \times \frac{12}{\sqrt{5000 + 24x}}$.
$a = 12 \; \text{m/s}^2$.

(D) From the equation of motion,$v^{2} = u^{2} + 2ax$,which is of the form $y = mx + c$,where $y = v^{2}$ and $x$ is the displacement.
Comparing this with the graph,the slope of the $v^{2}$ versus $x$ line is $2a$.
From the graph,we can calculate the slope $m$ using two points,for example,$(10, 40)$ and $(20, 60)$:
$m = \frac{v_{2}^{2} - v_{1}^{2}}{x_{2} - x_{1}} = \frac{60 - 40}{20 - 10} = \frac{20}{10} = 2 \text{ m/s}^{2}$.
Since the slope $m = 2a$,we have $2a = 2$.
Therefore,the acceleration $a = 1 \text{ m/s}^{2}$.