Question

The distance travelled by a body in the ${n^{th}}$ second is given by the expression $\left( {2 + 3n} \right)$. Find the initial velocity and acceleration. Also, find its velocity at the end of $2\,\sec$ ?

Answer 1

Hint
By using the equation of motion formula which shows the relation between the distance, initial velocity, acceleration and time, and substituting the given distance value in that formula, the acceleration and the initial velocity is determined. And by using the acceleration formula with two velocities, the velocity at the end of $2\,\sec$ can be determined.
From the equation of motion, the distance is given by,
$\Rightarrow d = u + \dfrac{1}{2}at^2$
Where, $d$ is the distance, ${u}$ is the initial velocity, $a$ is the acceleration and $t$ is the time.
Acceleration of the object with two velocities,
$\Rightarrow a = \dfrac{{v - u}}{t}$
Where, $a$ is the acceleration, $v$ is the final velocity, $u$ is the initial velocity and $t$ is the time.

Complete step by step answer
Given that,
The distance travelled by a body is given in the expression, $d = \left( {2 + 3n} \right)$
The time taken by the body to reach the given distance is given by, $t = 2n - 1$
Now,
From the equation of motion, the distance is given by,
$\Rightarrow d = u + \dfrac{1}{2}at^2 … (1)$
By substituting the distance and the time in the above equation (1), then the above equation (1) is written as,
$\Rightarrow 2 + 3n = u + \dfrac{1}{2}a\left( {2n - 1} \right)$
On multiplying the terms in RHS, then the above equation is written as,
$\Rightarrow 2 + 3n = u + \dfrac{{2an}}{2} - \dfrac{a}{2}$
Now cancelling the same terms in numerator and in denominator, then
$\Rightarrow 2 + 3n = u + an - \dfrac{a}{2}$
By rearranging the terms, then the above equation is written as,
$\Rightarrow 2 + 3n = u - \dfrac{a}{2} + an$
Now, by equating the constant terms and the coefficient of $n$, then
$\Rightarrow 2 = u - \dfrac{a}{2}\,.........\left( 2 \right)$ and
$3n = an\,.............\left( 3 \right)$
Now, in the equation (3), by cancelling the same terms on both the side, then
$\Rightarrow a = 3\,m{s^{ - 2}}$
And substituting the $a$ value in the equation (2), then
$\Rightarrow 2 = u - \dfrac{3}{2}$
By keeping the initial velocity in one side and the other terms in other side, then
$\Rightarrow u = 2 + \dfrac{3}{2}$
By simplifying the terms, then
$\Rightarrow u = 2 + 1.5$
By adding the terms in the above equation, then
$\Rightarrow u = 3.5\,m{s^{ - 1}}$
Now, the velocity after $2\,\sec$ is given by,
$\Rightarrow a = \dfrac{{v - u}}{t}\,...............\left( 4 \right)$
By rearranging the terms, then
$\Rightarrow v - u = at$
By keeping the final velocity in one side and the other terms in other side, then
$\Rightarrow v = u + at$
By substituting the initial velocity, acceleration and time in the above equation, then
$\Rightarrow v = 3.5 + \left( {3 \times 2} \right)$
By multiplying the terms, then
$\Rightarrow v = 3.5 + 6$
By adding the terms, then
$\Rightarrow v = 9.5\,m{s^{ - 1}}$.
The initial velocity is, $u = 3.5\,m{s^{ - 1}}$
The acceleration is, $a = 3\,m{s^{ - 1}}$
The velocity after $2\,\sec$ is, $v = 9.5\,m{s^{ - 1}}$.

Note
By comparing the given distance in the question with the distance formula, and by comparing the coefficients the acceleration of the body and the initial velocity of the body is determined. And then by using the first law of motion equation, the final velocity of the body is determined.