Question

A body covers $200\,\,cm$in the first $2\,\,s$ and $220\,\,cm$ in the next two seconds. What will be its velocity at the end of $7\,s$? Also, find the displacement in $7\,s$.

Answer 1

Hint:The given problem can be solved using one of the three equations of motion, that is the formula for the second equation of motion. The equation of motion consists of velocity of initial and final, acceleration, acceleration due to gravity, time taken and even the displacement.

Formulae Used:
The distance travelled by the body is given by the formula of;
$d = ut + \dfrac{1}{2}a{t^2}$
Where, $d$ denotes the displacement of the body, $u$denotes the initial velocity of the body, $a$ is the acceleration of the body, $t$ denotes the time taken by the body for displacement.

Complete step-by-step solution:
The data given in the problem is;
Distance travelled by the body, ${d_1} = 200\,\,cm$
Distance travelled by the body, ${d_2} = 220\,\,cm$
Time taken by the body, $t = 2\,\,s$
Assume that the body is under uniform acceleration here. Then, the total distance d travelled after the first t seconds is given by;
$d = ut + \dfrac{1}{2}a{t^2}$
At first two seconds:
The distance travelled by the body is given by the formula of;
${d_1} = ut + \dfrac{1}{2}a{t^2}$
Substitute the values of ${d_1}$ and $t$ in the above equation;
$200 = 2u + \dfrac{1}{2}a \times 4$
$200 = 2u + 2a\,\,..........\left( 1 \right)$
At the end of six seconds:
$d = ut + \dfrac{1}{2}a{t^2}$
$d = {d_1} + {d_2}$
$
  d = 200 + 220 \\
  d = 420\,\,cm \\
 $
Substitute the values of $d$ and $t$ is the above equation;
$420 = 6u + \dfrac{1}{2}a \times 36$
$420 = 6u + 18a\,\,..........\left( 2 \right)$
On equating the equation (1) and equation (2) we get;
\[2u + 2a = 200\]
$6u + 18a = 420$
$a = - 15\,\,cm\,{s^{ - 2}}$, $u = 115\,\,cm\,{s^{ - 2}}$
Therefore, the velocity at the end of seven seconds is;
$v = u + at$
Substitute the values of $u$, $a$ and $t$,
$v = 115 + \left( { - 15 \times 7} \right)$
$
  v = 115 - 105 \\
  \Rightarrow v = 10\,\,cm\,{s^{ - 1}} \\
 $
The displacement at the end of seven seconds is;
$d = ut + \dfrac{1}{2}a{t^2}$
Substitute the values of $u$, $a$ and $t$,
$d = \left( {115 \times 7} \right) + \dfrac{1}{2}\left( { - 115 \times {7^2}} \right)$
On simplifying the above equation, we get;
$
  d = 805 - 368 \\
  \Rightarrow d = 437\,\,cm \\
 $
Therefore, the velocity of the body at the end of seven seconds is $v = 10\,\,cm\,{s^{ - 1}}$ and the displacement of the body at the end of seven seconds is $d = 437\,\,cm$.

Note:- There are three equations of motion, in which two of the equations of motion can be used in finding the displacement of the body, that is if we have both initial and final velocity, we can use formula of equation of motion that is ${v^2} = {u^2} + 2as$.