Question

A body travels a distance of \[20m\] in the \[7th\] second and \[24m\] in the \[9th\] second. How much distance shall it travel in the \[15th\] second?

Answer 1

Hint: For a body traveling with uniform acceleration, we can use the formula for distance traveled by the body in the $nth$ second. Simplify it by applying the rules of a quadratic equation. Use the kinematic equation to find the distance traveled by the particle in the following time intervals to obtain the required solution.

Complete answer:
If we can assume that the body is traveling with uniform acceleration $a$, having an initial velocity $u$, let the distance traveled at\[7th\] and \[9th\] second be \[{S_7}\] and \[{S_9}\]. According to the relation of the distance traveled by the body at the $nth$ second, we get the equation ${S_n} = u + \dfrac{1}{2}a(2n - 1)$. Using this relation for \[7th\] and \[9th\] second, we get
${S_7} = 20 = u + \dfrac{1}{2}a(13)$….. $1$. (Here $n = 7$ )
And,
 ${S_9} = 24 = u + \dfrac{1}{2}a(17)$(Here $n = 9$ )
Hence by subtracting equations $1\& 2$ we get,
$4 = \dfrac{1}{2}a\left( 4 \right)$
Therefore the acceleration $a = 2m{s^{ - 2}}$.
Substituting this in the equation $1$ we get,
$20 = u + \dfrac{1}{2} \times 2 \times (13)$ Which gives us the initial velocity $u = 7m/s$.
Hence the distance traveled in the \[{15^{th}}\] second is given by ${S_{15}} = 7 + \dfrac{1}{2} \times 2 \times (29) = 36cm$
Thus the body will travel \[36m\] in the \[{15^{th}}\] second.

Note:
If the body is traveling at a constant acceleration and it does change direction at the \[nth\] second, then we need to initially find the time when it varies direction which is called the turning point, then find the displacement from the position at \[n - 1\] seconds up to the turning point. The uniformly accelerated motion will create arithmetic progression because the velocity of the particle is rising every second with an equally common difference. Then obtain the displacement from the turning point to the \[nth\] second, and at the end add those two to obtain the absolute value for those two displacements, one will be positive and one will be negative.