Question

If the distance travelled by a body in the ${{n}^{th}}$ second is given by $\left( 4+6n \right)m$, then find the initial velocity and acceleration of the body respectively.

Answer 1

Hint: The formula for distance travelled by a body in $n$ seconds is ${{d}_{n}}=un+\dfrac{1}{2}a{{n}^{2}}$. Now when we know the formula, we can find the distance travelled by the body in the ${{n}^{th}}$ second and equate it with the value given in the question and eventually we can find out the initial velocity and the acceleration of the body.
Formula used: ${{d}_{n}}=un+\dfrac{1}{2}a{{n}^{2}}$

Complete answer:
Let us write down the formula of distance travelled by a body in $n$ seconds:
${{d}_{n}}=un+\dfrac{1}{2}a{{n}^{2}}$
Here, $u$ is the initial velocity of the particle and $a$ is the acceleration of the body.
We are given that the distance travelled by a body in the ${{n}^{th}}$ second is $\left( 4+6n \right)m$ which can be written as the distance travelled by the body in $n$ seconds and subtract it from the distance travelled by the body in $\left( n-1 \right)$ seconds. This can be written in the form of an equation in the following way:
\[\begin{align}
  & {{d}_{{{n}^{th}}}}={{d}_{n}}-{{d}_{n-1}} \\
 & \Rightarrow 4+6n=\left( un+\dfrac{1}{2}a{{n}^{2}} \right)-\left( u\left( n-1 \right)+\dfrac{1}{2}a{{\left( n-1 \right)}^{2}} \right) \\
 & \Rightarrow 4+6n=\left( un+\dfrac{1}{2}a{{n}^{2}} \right)-\left( un-u+\dfrac{1}{2}a\left( {{n}^{2}}+1-2n \right) \right) \\
 & \Rightarrow 4+6n=\left( un+\dfrac{1}{2}a{{n}^{2}} \right)-\left( un-u+\dfrac{1}{2}a{{n}^{2}}+\dfrac{1}{2}a-an \right) \\
 & \Rightarrow 4+6n=u-\dfrac{1}{2}a+an \\
 & \Rightarrow 4+6n=\dfrac{2u-1}{2}a+an \\
\end{align}\]
On comparing the values, we get:
$\begin{align}
  & 6n=an \\
 & \therefore a=6\text{ m/}{{\text{s}}^{2}} \\
\end{align}$
And, now as we got the value of acceleration of the body, we will compare the other value to calculate the initial velocity of the body:
$\begin{align}
  & 4=\dfrac{2u-a}{2} \\
 & \Rightarrow 8=2u-a \\
 & \Rightarrow 8=2u-6 \\
 & \Rightarrow 8+6=2u \\
 & \Rightarrow 14=2u \\
 & \therefore u=7\text{ m/s} \\
\end{align}$
Thus, we get the initial velocity of the body as $u=7\text{ m/s}$ and the acceleration of the body came out to be $a=6\text{ m/}{{\text{s}}^{2}}$.

Note:
The question could have been made simpler if we memorize the formula of distance covered by a body in ${{n}^{th}}$ which is $d=u+a\left( n-\dfrac{1}{2} \right)$. We can compare the values of the equation given in the question to this formula to get our answer. But in case we forget this formula then the above solution would help us to arrive at the same result.