Uniformly Accelerated Motion Questions in English

(C) Total distance to be covered for crossing the bridge $=$ length of train $+$ length of bridge.
Total distance $= 150 \,m + 850 \,m = 1000 \,m$.
Convert the velocity from $km/h$ to $m/s$: $45 \times \frac{5}{18} = 12.5 \,m/s$.
Time taken $= \frac{\text{Total distance}}{\text{Velocity}} = \frac{1000 \,m}{12.5 \,m/s} = 80 \,sec$.

(D) Length of the train $= 100\, m$.
Velocity of the train $= 45\, km/hr = 45 \times \frac{5}{18} = 12.5\, m/s$.
Length of the bridge $= 1\, km = 1000\, m$.
To cross the bridge completely,the train must cover a total distance equal to the sum of the length of the train and the length of the bridge.
Total distance $= 100\, m + 1000\, m = 1100\, m$.
Time taken $= \frac{\text{Total distance}}{\text{Velocity}} = \frac{1100\, m}{12.5\, m/s} = 88\, s$.

(B) Let the initial velocity of the bullet be $u$.
After penetrating $3\,cm$,its velocity becomes $u/2$.
Using the equation of motion $v^2 = u^2 - 2as$:
$(u/2)^2 = u^2 - 2a(3)$
$u^2/4 = u^2 - 6a$
$6a = 3u^2/4$
$a = u^2/8$
Let the bullet penetrate an additional distance $x$ before coming to rest.
For this motion,initial velocity is $u/2$,final velocity is $0$,and acceleration is $-a = -u^2/8$.
Using $v^2 = u^2 - 2as$:
$0^2 = (u/2)^2 - 2(u^2/8)x$
$0 = u^2/4 - (u^2/4)x$
$u^2/4 = (u^2/4)x$
$x = 1\,cm$.

(B) Given that the particle starts from rest,initial velocity $u = 0$. Let the constant acceleration be $a$.
For the first $10 \,s$ $(t_1 = 10 \,s)$:
Using the equation of motion $S = ut + \frac{1}{2}at^2$,we get:
${S_1} = 0 \times 10 + \frac{1}{2}a(10)^2 = 50a$ .....$(i)$
To find the distance in the next $10 \,s$,we first find the velocity at $t = 10 \,s$:
$v = u + at = 0 + a(10) = 10a$.
For the next $10 \,s$ $(t_2 = 10 \,s)$,the initial velocity is $10a$:
${S_2} = (10a)(10) + \frac{1}{2}a(10)^2$
${S_2} = 100a + 50a = 150a$ .....(ii)
Comparing equations $(i)$ and (ii):
${S_1} = 50a$ and ${S_2} = 150a$
Therefore,${S_2} = 3{S_1}$,which implies ${S_1} = {S_2}/3$.

(C) The displacement of the particle is given by the equation: $x = a_0 + a_1t + a_2t^2$.
To find the velocity $v$,we differentiate the displacement $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(a_0 + a_1t + a_2t^2) = 0 + a_1 + 2a_2t$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(a_1 + 2a_2t) = 0 + 2a_2 = 2a_2$.
Thus,the acceleration of the particle is $2a_2$.

(A) Given that the velocity $v$ is a function of time $t$ as $v = kt$,where $k = 2 \, m/s^2$.
To find the distance $S$ travelled in the first $3 \, s$,we integrate the velocity with respect to time from $t = 0$ to $t = 3$.
$S = \int_{0}^{3} v \, dt = \int_{0}^{3} kt \, dt$.
Substituting the value of $k = 2$,we get:
$S = \int_{0}^{3} 2t \, dt = [t^2]_{0}^{3}$.
$S = (3)^2 - (0)^2 = 9 \, m$.
Therefore,the distance travelled is $9 \, m$.

(A) Given that the displacement $S$ is proportional to the cube of time $t$,we can write $S = kt^3$,where $k$ is a constant.
To find the velocity $v$,we differentiate $S$ with respect to time $t$:
$v = \frac{dS}{dt} = \frac{d}{dt}(kt^3) = 3kt^2$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(3kt^2) = 6kt$.
Since $a = 6kt$,the acceleration $a$ is directly proportional to time $t$ $(a \propto t)$.
Therefore,the magnitude of the acceleration increases with time.

(A) Given that the body starts from rest,so initial velocity $u = 0$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$,we have:
$S_1 = \frac{1}{2}a(p - 1)^2$
$S_2 = \frac{1}{2}ap^2$
Adding these two displacements:
$S_1 + S_2 = \frac{1}{2}a(p^2 - 2p + 1) + \frac{1}{2}ap^2 = \frac{1}{2}a(2p^2 - 2p + 1)$
Now,the displacement in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.
For $n = (p^2 - p + 1)$,the displacement is:
$S_{(p^2 - p + 1)^{th}} = 0 + \frac{a}{2}[2(p^2 - p + 1) - 1] = \frac{a}{2}[2p^2 - 2p + 2 - 1] = \frac{a}{2}(2p^2 - 2p + 1)$.
Comparing the results,we find that the displacement in the $(p^2 - p + 1)^{th}$ second is equal to $S_1 + S_2$.

(B) Given velocity $v = 4t^3 - 2t$.
Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}(4t^3 - 2t) = 12t^2 - 2$.
Position $x = \int v\,dt = \int (4t^3 - 2t)\,dt = t^4 - t^2$ (assuming $x=0$ at $t=0$).
When the particle is at $x = 2\, m$,we have $t^4 - t^2 = 2$,or $t^4 - t^2 - 2 = 0$.
Let $u = t^2$,then $u^2 - u - 2 = 0$,which factors as $(u - 2)(u + 1) = 0$.
Since $t^2$ must be positive,$t^2 = 2$,so $t = \sqrt{2}\,s$.
Substituting $t^2 = 2$ into the acceleration equation: $a = 12(2) - 2 = 24 - 2 = 22\,m/s^2$.

(B) Let $u_1, u_2, u_3$ and $u_4$ be the velocities at times $t = 0, t_1, (t_1 + t_2)$ and $(t_1 + t_2 + t_3)$ respectively,and let $a$ be the uniform acceleration.
The average velocities are given by:
$v_1 = \frac{u_1 + u_2}{2}, v_2 = \frac{u_2 + u_3}{2}, v_3 = \frac{u_3 + u_4}{2}$
Using the equation of motion $v = u + at$,we have:
$u_2 = u_1 + at_1$
$u_3 = u_1 + a(t_1 + t_2)$
$u_4 = u_1 + a(t_1 + t_2 + t_3)$
Now,calculate the differences:
$v_1 - v_2 = \frac{u_1 + u_2 - u_2 - u_3}{2} = \frac{u_1 - u_3}{2} = \frac{u_1 - (u_1 + a(t_1 + t_2))}{2} = -\frac{a(t_1 + t_2)}{2}$
$v_2 - v_3 = \frac{u_2 + u_3 - u_3 - u_4}{2} = \frac{u_2 - u_4}{2} = \frac{(u_1 + at_1) - (u_1 + a(t_1 + t_2 + t_3))}{2} = -\frac{a(t_2 + t_3)}{2}$
Taking the ratio:
$\frac{v_1 - v_2}{v_2 - v_3} = \frac{-a(t_1 + t_2) / 2}{-a(t_2 + t_3) / 2} = \frac{t_1 + t_2}{t_2 + t_3}$
Thus,$(v_1 - v_2) : (v_2 - v_3) = (t_1 + t_2) : (t_2 + t_3)$.

(B) Given that the acceleration $a(t) = at$ is a function of time.
We know that acceleration is the rate of change of velocity,$a = \frac{dv}{dt}$.
Therefore,$dv = a(t) \, dt$.
Integrating both sides from initial velocity $u$ at $t = 0$ to final velocity $v$ at time $t$:
$\int_{u}^{v} dv = \int_{0}^{t} (at) \, dt$.
$[v]_{u}^{v} = a \left[ \frac{t^2}{2} \right]_{0}^{t}$.
$v - u = \frac{at^2}{2}$.
$v = u + \frac{at^2}{2}$.
Thus,the correct relation is $v = u + \frac{at^2}{2}$.

(A) The distance traveled by a particle in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
Here,the initial velocity $u = 10 \ m/s$,the acceleration $a = -2 \ m/s^2$ (since it is retardation),and $n = 5$.
Substituting these values into the formula:
$S_5 = 10 + \frac{-2}{2}(2 \times 5 - 1)$
$S_5 = 10 - 1(10 - 1)$
$S_5 = 10 - 9 = 1 \ m$.
Therefore,the distance moved by the particle in the $5^{th}$ second is $1 \ m$.

(B) Given: Initial velocity $u = 20\,m/s$,Final velocity $v = 0\,m/s$,Distance $S = 10\,m$.
Using the third equation of motion: $v^2 = u^2 + 2aS$.
Substituting the values: $0^2 = (20)^2 + 2 \times a \times 10$.
$0 = 400 + 20a$.
$20a = -400$.
$a = -20\,m/s^2$.
Thus,the acceleration is $-20\,m/s^2$.

(D) Given:
Initial velocity $(u)$ = $10\,m/s$
Uniform acceleration $(a)$ = $2\,m/s^2$
Time interval $(t)$ = $4\,s$
Using the first equation of motion:
$v = u + at$
Substituting the values:
$v = 10 + (2 \times 4)$
$v = 10 + 8$
$v = 18\,m/s$
Therefore,the velocity after $4\,s$ is $18\,m/s$.

(C) Given that the particle starts from rest,so initial velocity $u = 0$.
Let the constant acceleration be $a$.
The distance covered in time $t$ is given by $s = ut + \frac{1}{2}at^2$.
For the first $2$ seconds ($t = 2$ s),the distance $x$ is:
$x = 0(2) + \frac{1}{2}a(2)^2 = 2a$.
For the first $4$ seconds ($t = 4$ s),the total distance covered is $x + y$:
$x + y = 0(4) + \frac{1}{2}a(4)^2 = 8a$.
Substituting $x = 2a$ into the equation:
$2a + y = 8a$
$y = 6a$.
Now,comparing $x$ and $y$:
$\frac{y}{x} = \frac{6a}{2a} = 3$.
Therefore,$y = 3x$.

(A) The distance covered by a body in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
Here,the initial velocity $u = 7 \, m/s$,acceleration $a = 4 \, m/s^2$,and the time $n = 5 \, s$.
Substituting these values into the formula:
$S_5 = 7 + \frac{4}{2}(2 \times 5 - 1)$
$S_5 = 7 + 2(10 - 1)$
$S_5 = 7 + 2(9)$
$S_5 = 7 + 18 = 25 \, m$.
Therefore,the distance covered in the $5^{th}$ second is $25 \, m$.

(A) The formula for the distance travelled by a body in the $n^{th}$ second is given by: $S_n = u + \frac{a}{2}(2n - 1)$.
Given:
Initial velocity $u = 0\,m/s$.
Acceleration $a = 8\,m/s^2$.
Time $n = 5^{th}$ second.
Substituting the values into the formula:
$S_5 = 0 + \frac{8}{2}(2 \times 5 - 1)$
$S_5 = 4(10 - 1)$
$S_5 = 4 \times 9$
$S_5 = 36\,m$.
Therefore,the distance travelled in the fifth second is $36\,m$.

(D) Given: Initial velocity $u = 1 \, km/s = 1000 \, m/s$,Final velocity $v = 9 \, km/s = 9000 \, m/s$,Distance $s = 4 \, m$.
Using the equation of motion $v^2 = u^2 + 2as$:
$(9000)^2 = (1000)^2 + 2 \times a \times 4$
$81,000,000 - 1,000,000 = 8a$
$80,000,000 = 8a$
$a = 10^7 \, m/s^2$.
Now,using $v = u + at$:
$9000 = 1000 + 10^7 \times t$
$8000 = 10^7 \times t$
$t = \frac{8000}{10^7} = 8 \times 10^{-4} \, s$.

(B) Given:
Initial velocity $u = 10 \, m/s$
Final velocity $v = -2 \, m/s$ (since it is in the opposite direction)
Time $t = 4 \, s$
Using the first equation of motion: $v = u + at$
Substituting the values: $-2 = 10 + a \times 4$
$-2 - 10 = 4a$
$-12 = 4a$
$a = -3 \, m/s^2$
Therefore,the acceleration produced is $-3 \, m/s^2$.

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

(A) The stopping distance $S$ of a vehicle is given by the formula $S = \frac{u^2}{2a}$,where $u$ is the initial velocity and $a$ is the magnitude of deceleration.
Since $a$ is constant for the same car and same braking force,we have $S \propto u^2$.
Given $u_1 = 40 \, km/h$ and $S_1 = 2 \, m$.
For $u_2 = 80 \, km/h$,we have the ratio $\frac{S_2}{S_1} = \left( \frac{u_2}{u_1} \right)^2$.
Substituting the values: $\frac{S_2}{2} = \left( \frac{80}{40} \right)^2 = (2)^2 = 4$.
Therefore,$S_2 = 2 \times 4 = 8 \, m$.

(A) Given:
Initial velocity,$u = 0\,m/s$ (since the body starts from rest).
Acceleration,$a = 5\,m/s^2$.
Time,$t = 10\,s$.
Using the first equation of motion:
$v = u + at$
Substituting the values:
$v = 0 + (5\,m/s^2 \times 10\,s)$
$v = 50\,m/s$.
Therefore,the instantaneous speed at the end of $10\,s$ is $50\,m/s$.

(A) The distance travelled by a body in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
Since the body starts from rest,the initial velocity $u = 0$.
Therefore,the distance travelled in the $n^{th}$ second is $S_n = \frac{a}{2}(2n - 1)$.
For the $4^{th}$ second $(n = 4)$: $S_4 = \frac{a}{2}(2 \times 4 - 1) = \frac{a}{2}(7) = \frac{7a}{2}$.
For the $3^{rd}$ second $(n = 3)$: $S_3 = \frac{a}{2}(2 \times 3 - 1) = \frac{a}{2}(5) = \frac{5a}{2}$.
The ratio of the distance travelled in the $4^{th}$ second to the $3^{rd}$ second is: $\frac{S_4}{S_3} = \frac{7a/2}{5a/2} = \frac{7}{5}$.

(B) The relationship between acceleration $a$ and velocity $v$ is given by $a = \frac{dv}{dt}$,which implies $dv = a\,dt$.
Integrating both sides,we get $v = \int a\,dt + C$.
Given $a = 3t^2 + 2t + 2$,we integrate with respect to $t$:
$v = \int (3t^2 + 2t + 2) dt = t^3 + t^2 + 2t + C$.
At $t = 0$,the initial velocity $u = 2\,m/s$,so $2 = (0)^3 + (0)^2 + 2(0) + C$,which gives $C = 2$.
Thus,the velocity function is $v(t) = t^3 + t^2 + 2t + 2$.
At $t = 2\,s$,the velocity is $v(2) = (2)^3 + (2)^2 + 2(2) + 2 = 8 + 4 + 4 + 2 = 18\,m/s$.

(D) The displacement 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 $t = 2$ into the velocity equation:
$v = 3(2)^2 - 12(2) + 3 = 3(4) - 24 + 3 = 12 - 24 + 3 = -9 \ m s^{-1}$.

(D) is incorrect: Consider a ball thrown vertically upward. At the highest point,it is at momentary rest $(v = 0)$,but the acceleration due to gravity $(g)$ still acts downwards.
$B$ is incorrect: $A$ slowing body has positive retardation (deceleration).
$C$ is incorrect: Distance traveled is always a positive scalar quantity,regardless of the direction of motion.
$D$ is correct: From the second equation of kinematics,$x = x_0 + ut + \frac{1}{2}at^2$. By knowing the initial position $(x_0)$,initial velocity $(u)$,and constant acceleration $(a)$ at instant $t = 0$,we can determine the displacement $(x)$ at any future time $t$.

(C) The distance traveled by an object in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
Given that the body starts from rest,the initial velocity $u = 0 \,m/s$.
The distance traveled in the $6^{th}$ second is $S_6 = 120 \,cm = 1.2 \,m$.
Substituting the values into the formula:
$1.2 = 0 + \frac{a}{2}(2 \times 6 - 1)$
$1.2 = \frac{a}{2}(11)$
$a = \frac{1.2 \times 2}{11} = \frac{2.4}{11} \approx 0.218 \,m/s^2$.
Rounding to two decimal places,we get $a \approx 0.22 \,m/s^2$.

(B) Given: Initial velocity $u = 0 \, m/s$,final velocity $v = 144 \, km/h = 144 \times \frac{5}{18} \, m/s = 40 \, m/s$,and time $t = 20 \, s$.
Using the first equation of motion,$v = u + at$:
$40 = 0 + a \times 20 \Rightarrow a = 2 \, m/s^2$.
Now,using the second equation of motion,$s = ut + \frac{1}{2}at^2$:
$s = 0 \times 20 + \frac{1}{2} \times 2 \times (20)^2 = 400 \, m$.
Therefore,the distance covered is $400 \, m$.

(D) Given: Initial velocity $u = 72 \text{ km/h} = 72 \times \frac{5}{18} \text{ m/s} = 20 \text{ m/s}$.
Final velocity $v = 0 \text{ m/s}$ (as the train comes to rest).
Distance $s = 200 \text{ m}$.
Using the third equation of motion: $v^2 = u^2 - 2as$,where $a$ is the retardation.
$0^2 = (20)^2 - 2 \times a \times 200$.
$0 = 400 - 400a$.
$400a = 400$.
$a = 1 \text{ m/s}^2$.
Therefore,the retardation is $1 \text{ m/s}^2$.

(A) For a body moving with uniform acceleration $a$,the displacement $S$ after time $t$ is given by the second equation of motion.
According to the kinematic equations for uniform acceleration:
$v = u + at$
$S = ut + \frac{1}{2}at^2$
$v^2 = u^2 + 2aS$
Where $u$ is the initial velocity,$v$ is the final velocity,$a$ is the acceleration,$t$ is the time,and $S$ is the displacement.
Comparing this with the given options,the correct relation is $S = ut + \frac{1}{2}at^2$.

(B) Let the cars $A$ and $B$ meet after time $t$ seconds. At the point of meeting,the distance traveled by both cars must be equal.
For car $A$ moving with uniform velocity: $S_A = v_A \times t = 40t$.
For car $B$ starting from rest with constant acceleration: $S_B = u_B t + \frac{1}{2} a_B t^2 = 0 + \frac{1}{2} \times 4 \times t^2 = 2t^2$.
Since $S_A = S_B$,we have $40t = 2t^2$.
Dividing both sides by $2t$ (assuming $t \neq 0$),we get $t = 20\, s$.

(C) For a body starting from rest $(u = 0)$ with constant acceleration $(a)$,the distance traveled in time $t$ is given by $S = \frac{1}{2}at^2$.
Let the time interval be $T = 5\, s$.
Distance in the first $5\, s$ $(t=T)$: ${S_1} = \frac{1}{2}aT^2$.
Distance in the first $10\, s$ $(t=2T)$: ${S_1} + {S_2} = \frac{1}{2}a(2T)^2 = 4 \times (\frac{1}{2}aT^2) = 4{S_1}$. Thus,${S_2} = 3{S_1}$.
Distance in the first $15\, s$ $(t=3T)$: ${S_1} + {S_2} + {S_3} = \frac{1}{2}a(3T)^2 = 9 \times (\frac{1}{2}aT^2) = 9{S_1}$. Thus,${S_3} = 9{S_1} - ({S_1} + {S_2}) = 9{S_1} - 4{S_1} = 5{S_1}$.
Therefore,the ratio is ${S_1}:{S_2}:{S_3} = 1:3:5$.
This implies ${S_1} = \frac{1}{3}{S_2} = \frac{1}{5}{S_3}$.

(A) Let the initial velocity at $t = 0$ be $u$ and constant acceleration be $a$.
For the first $5\,s$,distance $s_5 = 10\,m$:
$s = ut + \frac{1}{2}at^2 \Rightarrow 10 = 5u + \frac{1}{2}a(5)^2 \Rightarrow 10 = 5u + 12.5a \Rightarrow 2u + 5a = 4$ ...$(i)$
For the first $8\,s$ $(5\,s + 3\,s)$,distance $s_8 = 10\,m + 10\,m = 20\,m$:
$20 = 8u + \frac{1}{2}a(8)^2 \Rightarrow 20 = 8u + 32a \Rightarrow 2u + 8a = 5$ ...(ii)
Subtracting $(i)$ from (ii): $(2u + 8a) - (2u + 5a) = 5 - 4 \Rightarrow 3a = 1 \Rightarrow a = \frac{1}{3}\,m/s^2$.
Substituting $a$ in $(i)$: $2u + 5(\frac{1}{3}) = 4 \Rightarrow 2u = 4 - \frac{5}{3} = \frac{7}{3} \Rightarrow u = \frac{7}{6}\,m/s$.
Now,distance in total $10\,s$ $(s_{10})$:
$s_{10} = u(10) + \frac{1}{2}a(10)^2 = (\frac{7}{6})(10) + \frac{1}{2}(\frac{1}{3})(100) = \frac{70}{6} + \frac{50}{3} = \frac{70 + 100}{6} = \frac{170}{6} \approx 28.33\,m$.
Distance in the next $2\,s$ is $s_{10} - s_8 = 28.33 - 20 = 8.33\,m \approx 8.3\,m$.

(A) Given that the distance $s$ is proportional to the square of time $t$,we have $s \propto t^2$.
This can be written as $s = kt^2$,where $k$ is a constant.
The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(kt^2) = 2kt$.
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,the acceleration $a = 2k$ is also a constant. Therefore,the particle travels with uniform acceleration.

(D) Given the velocity equation: $v = at$.
We know that distance $x$ is the integral of velocity with respect to time: $x = \int v \ dt$.
Substituting the given equation: $x = \int (at) \ dt = \frac{1}{2}at^2$.
To find the distance travelled in the first $4 \ s$,we substitute $t = 4 \ s$ into the expression:
$x = \frac{1}{2} \times a \times (4)^2 = \frac{1}{2} \times a \times 16 = 8a$.
Therefore,the distance travelled is $8a$.

(B) Given: Distance $(s)$ = $3.06 \ m$,Average velocity $(v_{avg})$ = $0.34 \ ms^{-1}$,Change in velocity $(\Delta v)$ = $0.18 \ ms^{-1}$.
First,calculate the time taken $(t)$ using the formula: $t = \frac{s}{v_{avg}} = \frac{3.06}{0.34} = 9 \ s$.
Now,calculate the uniform acceleration $(a)$ using the formula: $a = \frac{\Delta v}{t} = \frac{0.18}{9} = 0.02 \ ms^{-2}$.

(A) The distance traveled by an object in the $n$th second is given by $S_n = u + \frac{a}{2}(2n - 1)$.
For body $A$,the distance traveled in the $5$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 $5$th second after the start of $A$ corresponds to the $(5 - 2) = 3$rd second of motion for body $B$.
The distance traveled by body $B$ in its $3$rd second is $S_{B,3} = 0 + \frac{a_2}{2}(2 \times 3 - 1) = \frac{5a_2}{2}$.
Given that these distances are equal: $\frac{9a_1}{2} = \frac{5a_2}{2}$.
Simplifying this,we get $9a_1 = 5a_2$,which implies $\frac{a_1}{a_2} = \frac{5}{9}$.

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

(A) According to Newton's second law of motion,$F = ma$. If the force $F$ is fixed in both magnitude and direction,the acceleration $a = F/m$ is also constant in magnitude and direction.
If the particle starts from rest or moves in the direction of the force,the velocity will change only in magnitude along the same line.
If the particle has an initial velocity not parallel to the force,the path is a parabola (like projectile motion).
However,in the context of a force fixed in magnitude and direction acting on a particle,if we consider the simplest case where the force is the only influence,the particle moves in a straight line if the initial velocity is zero or parallel to the force. Given the standard interpretation of this physics problem,the motion is a straight line.

(D) Using the equation of motion $v^2 - u^2 = 2as$,where $v = 0$ (final velocity),$u$ is the initial velocity,$a$ is the retardation,and $s$ is the stopping distance.
Since $v = 0$,we have $-u^2 = 2as$,which implies $s = \frac{u^2}{2|a|}$.
This shows that the stopping distance $s \propto u^2$.
Given $u_1 = 50 \,km/hr$ and $s_1 = 6 \,m$.
For $u_2 = 100 \,km/hr$,the ratio is $\frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^2$.
$\frac{s_2}{6} = \left(\frac{100}{50}\right)^2 = (2)^2 = 4$.
Therefore,$s_2 = 4 \times 6 = 24 \,m$.

(A) For body $A$,the distance covered in time $t$ with initial velocity $u = 0$ and acceleration $a$ is given by the equation of motion: $s_A = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}at^2 = \frac{1}{2}at^2$.
For body $B$,the distance covered in time $t$ with constant velocity $v$ is: $s_B = vt$.
Since both bodies start from the same point and meet after time $t$,their displacements must be equal: $s_A = s_B$.
Equating the two expressions: $\frac{1}{2}at^2 = vt$.
Dividing both sides by $t$ (assuming $t \neq 0$): $\frac{1}{2}at = v$.
Solving for $t$,we get: $t = \frac{2v}{a}$.

(C) Given: Initial velocity $u = 0$,final velocity $v = 27.5 \,m/s$ at time $t = 10 \,s$.
First,calculate the acceleration $a$:
$a = \frac{v - u}{t} = \frac{27.5 - 0}{10} = 2.75 \,m/s^2$.
For the next $10 \,s$,the initial velocity is the velocity at the end of the first $10 \,s$,which is $u' = 27.5 \,m/s$.
Using the equation of motion $S = u't + \frac{1}{2}at^2$ for the next $10 \,s$:
$S = (27.5 \times 10) + \frac{1}{2} \times 2.75 \times (10)^2$
$S = 275 + \frac{1}{2} \times 2.75 \times 100$
$S = 275 + 137.5 = 412.5 \,m$.

(C) Given the velocity of the particle as $v = (180 - 16x)^{1/2}$.
We know that acceleration $a$ is defined as $a = \frac{dv}{dt}$.
Using the chain rule,we can write $a = \frac{dv}{dx} \cdot \frac{dx}{dt}$.
Since $\frac{dx}{dt} = v$,we have $a = v \cdot \frac{dv}{dx}$.
First,differentiate $v$ with respect to $x$:
$\frac{dv}{dx} = \frac{1}{2}(180 - 16x)^{-1/2} \cdot (-16) = -8(180 - 16x)^{-1/2}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (180 - 16x)^{1/2} \cdot [-8(180 - 16x)^{-1/2}]$.
$a = -8 \cdot (180 - 16x)^{1/2 - 1/2} = -8 \cdot (180 - 16x)^0$.
$a = -8 \cdot 1 = -8 \text{ m/s}^2$.

(D) Given that the displacement $x$ is proportional to the cube of time $t$,we can write:
$x = K t^3$,where $K$ is a constant.
To find the velocity $v$,we differentiate the displacement with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt}(K t^3) = 3 K t^2$.
To find the acceleration $a$,we differentiate the velocity with respect to time:
$a = \frac{dv}{dt} = \frac{d}{dt}(3 K t^2) = 6 K t$.
Since $6$ and $K$ are constants,we have $a \propto t$.
Therefore,the acceleration of the particle is directly proportional to time.

(A) Given that the particle starts from rest,the initial velocity $u = 0$ at $t = 0$.
The acceleration is $a = \frac{dv}{dt} = 2(t - 1)$.
To find the velocity $v$,we integrate the acceleration with respect to time:
$dv = 2(t - 1) \, dt$
Integrating both sides from $t = 0$ to $t = 5$:
$v = \int_{0}^{5} 2(t - 1) \, dt$
$v = 2 \left[ \frac{t^2}{2} - t \right]_{0}^{5}$
$v = 2 \left[ \left( \frac{5^2}{2} - 5 \right) - (0 - 0) \right]$
$v = 2 \left( \frac{25}{2} - 5 \right) = 2 \left( 12.5 - 5 \right) = 2(7.5) = 15 \, m/s$.

(C) Given: Distance in first $5 \, s$ $(S_1)$ = $40 \, m$,time $(t)$ = $5 \, s$. Distance in next $5 \, s$ $(S_2)$ = $65 \, m$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$:
For the first $5 \, s$: $40 = 5u + \frac{1}{2}a(5)^2$ $\Rightarrow 40 = 5u + 12.5a$ ... $(i)$
For the total $10 \, s$: $S_1 + S_2 = u(10) + \frac{1}{2}a(10)^2$ $\Rightarrow 40 + 65 = 10u + 50a$ $\Rightarrow 105 = 10u + 50a$ ... (ii)
Multiply equation $(i)$ by $2$: $80 = 10u + 25a$ ... (iii)
Subtracting (iii) from (ii): $(105 - 80) = (10u - 10u) + (50a - 25a)$ $\Rightarrow 25 = 25a$ $\Rightarrow a = 1 \, m/s^2$.
Substitute $a = 1$ in equation $(i)$: $40 = 5u + 12.5(1)$ $\Rightarrow 5u = 40 - 12.5 = 27.5$ $\Rightarrow u = 5.5 \, m/s$.

(D) Using the third equation of motion,$v^2 = u^2 + 2as$. Since the cars are brought to a stop,the final velocity $v = 0$.
Therefore,$0 = u^2 - 2as$,which gives the stopping distance $s = \frac{u^2}{2a}$.
Since both cars are identical and are stopped under the same braking force,the deceleration $a$ is the same for both.
Thus,$s \propto u^2$.
The ratio of the distances is $\frac{s_1}{s_2} = \left( \frac{u_1}{u_2} \right)^2 = \left( \frac{u}{4u} \right)^2 = \left( \frac{1}{4} \right)^2 = \frac{1}{16}$.
Hence,the ratio is $1:16$.

(C) Let the car start from point $A$ from rest and move up to point $B$ with acceleration $f$ over distance $S$.
The velocity of the car at point $B$ is $v = \sqrt{2fS}$ (using $v^2 = u^2 + 2as$).
The car moves distance $BC = x$ with this constant velocity $v$ in time $t$,so $x = vt = \sqrt{2fS} \cdot t$ ... $(i)$.
At point $C$,the car begins to decelerate at rate $a' = \frac{f}{2}$ until it comes to rest at point $D$. Let the distance $CD = y$.
Using $v^2 = u^2 - 2a'y$,where final velocity is $0$: $0 = v^2 - 2(\frac{f}{2})y \implies y = \frac{v^2}{f} = \frac{2fS}{f} = 2S$ ... (ii).
The total distance $AD = AB + BC + CD = S + x + 2S = 15S$.
$3S + x = 15S \implies x = 12S$.
Substituting $x = 12S$ into equation $(i)$: $12S = \sqrt{2fS} \cdot t$.
Squaring both sides: $144S^2 = 2fS \cdot t^2$.
Dividing by $2S$: $72S = f \cdot t^2 \implies S = \frac{1}{72}ft^2$.

(B) Given the position equation: $x = 4(t - 2) + a(t - 2)^2$.
To find the velocity,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,we differentiate the velocity 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 $v = 4 + 2a(0 - 2) = 4 - 4a$,which is not $4$ unless $a = 0$.
$(b)$ The acceleration is $2a$,which is constant and matches the derived result.
$(c)$ At $t = 0$,$x = 4(0 - 2) + a(0 - 2)^2 = -8 + 4a$,which is not $0$ unless $a = 2$.
Therefore,option $(b)$ is correct.

(A) The distance covered by an object in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.
Since the body starts from rest,$u = 0$.
For the $5^{th}$ second $(n = 5)$:
$S_{5^{th}} = 0 + \frac{a}{2}(2 \times 5 - 1) = \frac{9a}{2}$.
The distance covered in $t$ seconds is given by $S = ut + \frac{1}{2}at^2$.
For $t = 5$ seconds:
$S_5 = 0 + \frac{1}{2} \times a \times (5)^2 = \frac{25a}{2}$.
The ratio is $\frac{S_{5^{th}}}{S_5} = \frac{9a/2}{25a/2} = \frac{9}{25}$.