Question

A body is moving in a straight line with initial velocity $u$ and uniform acceleration $a$. Sum of the distances travelled in $n^{\text{th}}$ and $(n+1{)}^{\text{th}}\text{seconds}$ is $50\text{m}$. If $v$ is its velocity after $n\text{seconds}$ and $u=5\text{m/s}$, then $\frac{v}{u}$ is

• A. $5$
• B. $10$
• C. $15$
• D. $20$

Answer 1

Answer: (A)

Step 1: Given that:

Initial velocity of the body = $u$

Uniform acceleration in the body = $a$

Sum of the distances travelled in $n^{th}$ and ${(n+1)}^{th}$ seconds = $50m$

After $nseconds$ velocity of the body = $v$

$u=5m{s}^{-1}$

Step 2: Formula used:

The distance covered by a body in $n^{th}$ second is given by;

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

Where u = initial velocity of the body

a = Acceleration in the body

The first equation of motion is given as;

$v=u+at$

Where v= final velocity of the body

u= Initial velocity of the body

a= Acceleration in the body

t= Time taken by the body

Step 3: Calculation of final velocity of the body:

The distance covered by the body in $n^{th}$ second will be

$s_n=u+\frac{1}{2}a(2n-1)$ ............(1)

The distance covered by the body in ${(n+1)}^{th}$ second is given as;

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

$s_{n+1}=u+\frac{1}{2}a\{2n+2-1\}$

$s_{n+1}=u+\frac{1}{2}a\{2n+1\}$ ............(2)

Since, $s_n+s_{n+1}=50m$

Thus,

$u+\frac{1}{2}a\{2n-1\}+u+\frac{1}{2}a\{2n+1\}=50$

$2u+\frac{1}{2}2na-\frac{1}{2}a+\frac{1}{2}2na+\frac{1}{2}a=50$

$2u+2na=50$

$u+na=25$

$u+an=25$

According to the first equation of motion for time n second

$v=u+an$

that is

$u+an=v$

Now, comparing both we get

$v=25m{s}^{-1}$

Step 4: calculation of the value of $\frac{v}{u}$ :

Since $u=5m{s}^{-1}$

and, $v=25m{s}^{-1}$

Now,

$\frac{v}{u}=\frac{25m{s}^{-1}}{5m{s}^{-1}}$

$\frac{v}{u}=5$

Thus,

The value of $\frac{v}{u}$ will be 5.