Correct option is (B)

Let's analyze the given information before determining the distance between the two cars when they come to rest. Each car is traveling towards the other at a speed of $20{\text{m s}}^{-1}$ and they both start braking when they are $300\text{m}$ apart. The deceleration (negative acceleration) of each car is given as $2{\text{m s}}^{-2}.$

To find the distance each car travels before coming to rest, we can use the kinematic equation that relates initial velocity ($v_i$), final velocity ($v_f$), acceleration ($a$), and distance ($d$), which is:

$v_f^2=v_i^2+2ad$

Since the final velocity $v_f=0$ (they come to rest), we can rearrange the equation to solve for $d$ (the distance each car travels before stopping):

$0=v_i^2+2ad\Rightarrow d=-\frac{v_i^2}{2a}$

Plugging in the values for each car (noting that acceleration $a$ is negative because it is deceleration, so $a=-2{\text{m s}}^{-2}$):

$d=-\frac{(20{)}^2}{2(-2)}=-\frac{400}{-4}=100\text{m}$

Each car travels $100\text{m}$ before coming to rest. Since they both start $300\text{m}$ apart and each travels $100\text{m}$ towards the other, the total distance covered by both cars before stopping is $2\times 100\text{m}=200\text{m}$.

To find the distance between them when they come to rest, we subtract the total distance covered by both cars from the original distance between them:

$300\text{m}-200\text{m}=100\text{m}$

So, the distance between the cars when they come to rest is $100\text{m}$. Therefore, the correct option is:

Option B 100 m

Correct option is (a)

The distance covered by a body in the $n^{\text{th}}$ second can be found using the equation:

$S_n=u+{\displaystyle \frac{1}{2}}a(2n-1)$

where,

• $S_n$ is the distance covered in the $n^{\text{th}}$ second,
• $u$ is the initial velocity,
• $a$ is the acceleration, and
• $n$ is the nth second.

The distance covered in the $n^{\text{th}}$ second is given as $102.5\text{m}$, so we have:

$102.5=u+{\displaystyle \frac{1}{2}}a(2n-1)$ ---- (1)

For the $(n+2{)}^{\text{th}}$ second, the distance covered is:

$S_{n+2}=u+{\displaystyle \frac{1}{2}}a(2(n+2)-1)$

Substituting $n+2$ in place of $n$, we get:

$115.0=u+{\displaystyle \frac{1}{2}}a(2n+3)$ ---- (2)

Subtracting equation (1) from equation (2), we get:

$115.0-102.5={\displaystyle \frac{1}{2}}a(2n+3-2n+1)$

$12.5={\displaystyle \frac{1}{2}}a(4)$

So, solving for $a$ gives:

$a={\displaystyle \frac{12.5\times 2}{4}}={\displaystyle \frac{25}{4}}=6.25{\text{m/s}}^2$

Therefore, the acceleration of the body is:

$6.25{\text{m/s}}^2$

Which corresponds to Option A: $6.25{\text{m/s}}^2$.

Correct option is (d)

Let's denote the constant acceleration with which the body moves as $a$. We know that the displacement $S$ covered by a body starting from rest under a constant acceleration $a$ in time $t$ is given by the equation of motion: $S=\frac{1}{2}a{t}^2$.

Considering the displacement ${\mathrm{S}}_1$ in first $(p-1)$ seconds, we apply the equation of motion:

${\mathrm{S}}_1=\frac{1}{2}a(p-1{)}^2$

Similarly, for the displacement ${\mathrm{S}}_2$ in first $p$ seconds:

${\mathrm{S}}_2=\frac{1}{2}a{p}^2$

To find out the total time it will take to cover the displacement ${\mathrm{S}}_1+{\mathrm{S}}_2$, we first find the sum of these two displacements:

${\mathrm{S}}_1+{\mathrm{S}}_2=\frac{1}{2}a(p-1{)}^2+\frac{1}{2}a{p}^2$

Let's simplify this:

${\mathrm{S}}_1+{\mathrm{S}}_2=\frac{1}{2}a((p-1{)}^2+p^2)$ ${\mathrm{S}}_1+{\mathrm{S}}_2=\frac{1}{2}a(p^2-2p+1+p^2)$ ${\mathrm{S}}_1+{\mathrm{S}}_2=\frac{1}{2}a(2p^2-2p+1)$ ${\mathrm{S}}_1+{\mathrm{S}}_2=a\left(\frac{2p^2-2p+1}{2}\right)$

If we consider the total displacement ${\mathrm{S}}_1+{\mathrm{S}}_2$ is to be covered in a time $t$ seconds from rest, we should set this equal to the equation of motion:

${\mathrm{S}}_1+{\mathrm{S}}_2=\frac{1}{2}a{t}^2$

Equating the two equations:

$a\left(\frac{2p^2-2p+1}{2}\right)=\frac{1}{2}a{t}^2$

Since $a\ne 0$, we can simplify by dividing both sides by $\frac{1}{2}a$:

$\frac{2p^2-2p+1}{2}=\frac{t^2}{2}$ $2p^2-2p+1=t^2$

To find $t$, we take the square root of both sides:

$t=\sqrt{2p^2-2p+1}$

Therefore, the time taken to cover the displacement ${\mathrm{S}}_1+{\mathrm{S}}_2$ will be:

$t=\sqrt{(2p^2-2p+1)}s$

Hence, the correct option would be:

Option D: $\sqrt{(2p^2-2p+1)}s$

Correct answer is 19

Given that a body is moving on a frictionless plane and starts from rest, the motion can be assumed to be uniformly accelerated motion. The formula for the distance covered in uniformly accelerated motion from rest is given by $s=ut+\frac{1}{2}a{t}^2$, where:

• $s$ is the distance covered,


• $u$ is the initial velocity (which is 0 since the body starts from rest),


• $a$ is the acceleration, and


• $t$ is the time.

Since the body starts from rest ($u=0$), the formula simplifies to $s=\frac{1}{2}a{t}^2$.

The distance moved between $t=n-1$ and $t=n$, denoted as $S_n$, can be found by calculating the distance covered by the end of time $n$ and subtracting the distance covered by the end of time $n-1$. Let's denote the total distance covered by time $n$ as $S(n)$, which according to the formula is $S(n)=\frac{1}{2}a{n}^2$. Thus, $S_n=S(n)-S(n-1)$.

Therefore, we have:

$S_n=\frac{1}{2}a{n}^2-\frac{1}{2}a(n-1{)}^2$

Simplifying this, we get:

$S_n=\frac{1}{2}a(n^2-(n^2-2n+1))$

This simplifies to:

$S_n=\frac{1}{2}a(2n-1)$

Similarly,

$S_{n-1}=\frac{1}{2}a((n-1{)}^2-(n-2{)}^2)$

Simplifying:

$S_{n-1}=\frac{1}{2}a((n-1{)}^2-(n^2-4n+4))$

Which further simplifies to:

$S_{n-1}=\frac{1}{2}a(2n-3)$

Hence, the ratio $\frac{S_{n-1}}{S_n}$ can be calculated:

$\frac{S_{n-1}}{S_n}=\frac{\frac{1}{2}a(2n-3)}{\frac{1}{2}a(2n-1)}=\frac{2n-3}{2n-1}$

For $n=10$,

$\frac{S_{n-1}}{S_n}=\frac{2(10)-3}{2(10)-1}=\frac{20-3}{20-1}=\frac{17}{19}$

Given that the ratio is represented as $(1-\frac{2}{x})$, we have:

$\frac{17}{19}=1-\frac{2}{x}$

Solving this equation for $x$ gives:

$1-\frac{17}{19}=\frac{2}{x}$

$\frac{2}{19}=\frac{2}{x}$

Thus:

$x=19$

Correct answer is 40

$\begin{array}{rl}& u=72\times \frac{5}{18}=20\mathrm{m}/\mathrm{s}\\ & v=0\\ & t=4\\ & \Rightarrow 0=20+4a\\ & \Rightarrow a=-5\mathrm{m}/{\mathrm{s}}^2\\ & \begin{array}{rl}\therefore S& =20\times 4-\frac{1}{2}\times 5\times 16\\ & =40\mathrm{m}\end{array}\end{array}$