Question

A stone is thrown up with speed $55\,\,m{s^{ - 1}}$. distance travelled by the stone in the time interval between $G = 5\,\,s$and $t = 7\,\,s$ .
(A) $2.50\,\,m$
(B) $6.50\,\,m$
(C) $10\,\,m$
(D) $12.50\,\,m$

Answer 1

Hintthe above equation of motion can be solved using one of the three equations of motion. There are three equations of motion that can be used to derive components such as displacement $\left( s \right)$, velocity (initial and final), time $\left( u \right)$ and acceleration $\left( a \right)$.

Useful formula
The total distance travelled by the stone in the given time interval is found by:
$S = ut - \dfrac{1}{2}g{t^2}$
Where, $S$ denotes the distance travelled by the stone in the given time interval, $u$ denotes the initial velocity of the stone, $t$ denotes the time interval, $g$ denotes the acceleration due to gravity acting on the stone.

Complete step by step solution
The data given in the problem are as follows:
Initial velocity by which the stone is thrown is, $u = 55\,\,m{s^{ - 1}}$,
Time interval is given as, $G = 5\,\,s$,
Time interval is given as, $t = 7\,\,s$.
The time taken by the stone that is thrown to reach the maximum height is;
$T = \dfrac{u}{g}$
Where, $T$ denotes the time taken by the stone that is thrown to reach the maximum height, $u$ denotes the initial velocity of the stone, $g$ denotes the acceleration due to gravity acting on the stone.
$
  T = \dfrac{{55}}{{10}} \\
  T = 5.5\,\,s \\
 $

From the time interval $G = 5\,\,s$ to $T = 5.5\,\,s$ the stone goes up to a distance of ${S_1}$.
And from the time $T = 5.5\,\,s$ to $t = 7\,\,s$ the stone goes down to a distance of ${S_2}$.

For the time interval $G = 5\,\,s$ to $T = 5.5\,\,s$ the stone goes up to a distance of;
${S_1} = \left[ {uT - \dfrac{1}{2}g{T^2}} \right] - \left[ {uG - \dfrac{1}{2}g{G^2}} \right]$
Substitute the values of initial velocity, acceleration due to gravity and the time intervals $G$and $T$ in the above equation.
\[
  \Rightarrow {S_1} = \left[ {55 \times 5.5 - \dfrac{1}{2}10 \times {{5.5}^2}} \right] - \left[ {55 \times 5 - \dfrac{1}{2} \times 10 \times {5^2}} \right] \\
  \Rightarrow {S_1} = \left[ {302.5 - 151.25} \right] - \left[ {275 - 125} \right] \\
  \Rightarrow {S_1} = 151.25 - 150 \\
  \Rightarrow {S_1} = 1.25\,\,m \\
 \]
For the time interval $G = 5\,\,s$ to $T = 5.5\,\,s$ the stone goes up to a distance of \[{S_1} = 1.25\,\,m\].
For the time interval $t = 7\,\,s$ to $t = 7\,\,s$ the stone goes down to a distance of;
${S_2} = \left[ {ut - \dfrac{1}{2}g{t^2}} \right] - \left[ {uT - \dfrac{1}{2}g{T^2}} \right]$
Substitute the values of initial velocity, acceleration due to gravity and the time intervals $t$and $T$ in the above equation.
$
  \Rightarrow {S_2} = \left[ {55 \times 7 - \dfrac{1}{2} \times 10 \times {7^2}} \right] - \left[ {55 \times 5.5 - \dfrac{1}{2} \times 10 \times {{5.5}^2}} \right] \\
  \Rightarrow {S_2} = \left[ {385 - 245} \right] - \left[ {302.5 - 151.25} \right] \\
  \Rightarrow {S_2} = 140 - 151.25 \\
  \Rightarrow {S_2} = 11.25\,\,m \\
 $
For the time interval $G = 5\,\,s$ to $T = 5.5\,\,s$ the stone goes down to a distance of \[{S_2} = 11.25\,\,m\].

The total distance travelled by the stone in the given time interval is found as;
$S = {S_1} + {S_2}$
Substitute the values of ${S_1}$ and ${S_2}$ in the above equation;
$
  \Rightarrow S = 1.25 + 11.25 \\
  \Rightarrow S = 12.50\,\,m \\
 $
Therefore, the total distance travelled by the stone in the given time interval is found as $S = 12.50\,\,m$.

Hence the option (D) $S = 12.50\,\,m$ is the correct answer.

Note The equations of motion of kinematics denotes the most basic concepts of motion of an object. These equations control the motion of an object in $1D$, $2D$ and $3D$. They can easily be used to calculate expressions such as the distance, velocity, or acceleration of a body at various times.