Question

A body covers $2.0m$ in the first $2s$ and $2.2m$ in the next $4s.$ Then the velocity of the body at the end of a $7th$ second is
A. $10m/s$
B. $1m/s$
C. $10cm/s$
D. $1cm/s$

Answer 1

Hint: Here we will use distance formula related to the initial velocity and the acceleration. Also, since the required answer is centimetre per second and given terms are in metre per second, before starting the solution convert the given distance in centimetres and then substitute in the formula and simplify.

Complete step by step answer:
Let us consider that here the body is under the uniform acceleration. Then the total distance “d” travelled by the body after the first “t” seconds can be given by –
$d = ut + \dfrac{1}{2}a{t^2}$
Where, $u = $ the initial velocity
And $a = $ the acceleration
A body covers $2.0m$ in the first $2s$ and $2.2m$ in the next $4s$ .
(Given)
Convert the above given statement.
$
  {d_1} = 2m = 200cm \\
  {t_1} = 2\operatorname{Sec} \\
  {d_2} = 2.2m = 220cm \\
  {t_2} = 4\operatorname{Sec} \\
$
Place value for first two seconds in the below equation –
$
  {d_1} = u{t_1} + \dfrac{1}{2}a{t_1}^2 \\
  200 = u \times 2 + \dfrac{1}{2}a{(2)^2} \\
$
Simplify the above equation using simple mathematical operations-
$200 = 2u + 2a$
Take common from both the sides of the equations and remove it.
 $a + u = 100$ .......(1)
Let distance be equal to ${d_2}$ travelled after six seconds.
$
  {d_2} = 200 + 220 \\
  {d_2} = 420cm \\
$
$
  {t_2} = 2 + 4 \\
  {t_2} = 6\sec \\
$
Place the above values in the below equations -
$
  {d_2} = u{t_2} + \dfrac{1}{2}a{t_2}^2 \\
  420 = u \times 6 + \dfrac{1}{2}a{(6)^2} \\
$
Simplify the above equations
$420 = 6u + 18a$
Take “$6$” common from both the sides of the equations and remove it.
$
  70 = u + 3a \\
  u + 3a = 70 \\
$ ........(2)
Subtract equation$(1)from{\text{ the equation (2)}}$
$
  2a = - 30 \\
  \Rightarrow a = \dfrac{{ - 30}}{2} \\
  \Rightarrow a = - 15cm/{s^2} \\
$
Place value of “a” to get the value of “u” in the equation $(1)$
$u + a = 100 \\$
$\implies u - 15 = 100 \\$
$\implies u = 115cm/s \\ $
Now, for the final velocity, $V = u + at$
Place values of $u,{\text{ a and t = 7s}}$ in the above equation
$V = 115 + ( - 15)(7) \\$
$\implies V = 115 - 105 \\$
$\therefore V = 10cm/s \\ $

So, the correct answer is “Option A”.

Note:
Always check the given units and the units asked in the solution. All the solutions must be in the same system of units. Also remember the conversional relation among systems of units. There are three systems of units.
- MKS unit (Metre Kilogram Second)
- CGS unit (Centimetre Gram Second)
- SI (System International)