Question

A body covers 20m, 22m, 24m in 8th, 9th and 10th seconds respectively. The body starts
A. from rest and moves with uniform velocity.
B. from rest and moves with uniform acceleration.
C. with an initial velocity and moves with uniform acceleration.
D. with an initial velocity and moves with uniform velocity.

Answer 1

Hint: In order to solve this question, we will use the equation which gives the distance travelled by a body in nth second. The given values of time and distance can be used to obtain equations which have initial velocity and acceleration as variable. By simultaneously solving the equations, we find out which of the options is correct.

Formula used:
The distance travelled by a body in n seconds is given as
\[{S_n} = u + \dfrac{a}{2}\left( {2n - 1} \right){\text{ }}...{\text{(i)}}\]
Here n signifies the seconds elapsed for the motion of the body, S is the displacement of the body after n seconds, u is the initial velocity of the body while a is the acceleration of the body.

Complete step-by-step answer:
We are given that a body covers 20m, 22m, 24m in 8th, 9th and 10th seconds respectively.
We know the expression for displacement of the body for n seconds as given in equation (i). Now using the given information, we substitute the distances given with respective values of n.
Therefore, we can write that
$
  {S_8} = u + \dfrac{a}{2}\left( {2 \times 8 - 1} \right) \\
  \Rightarrow 20 = u + \dfrac{{15a}}{2}{\text{ }}...{\text{(A)}} \\
  {S_9} = u + \dfrac{a}{2}\left( {2 \times 9 - 1} \right) \\
  \Rightarrow 22 = u + \dfrac{{17a}}{2}{\text{ }}...{\text{(B)}} \\
  {S_{10}} = u + \dfrac{a}{2}\left( {2 \times 10 - 1} \right) \\
  \Rightarrow 24 = u + \dfrac{{19a}}{2}{\text{ }}...{\text{(C)}} \\
 $
Now we will try to obtain values of u and a by solving the first two equations A and B simultaneously. This can be done in the following way.
$
  20 = u + \dfrac{{15a}}{2} \\
   \Rightarrow u = 20 - \dfrac{{15a}}{2}{\text{ }}...{\text{(D)}} \\
 $
Using this value of u in equation B, we get
$
  22 = u + \dfrac{{17a}}{2} \\
  \Rightarrow 22 = 20 - \dfrac{{15a}}{2} + \dfrac{{17a}}{2} \\
  \Rightarrow 22 = 20 + \dfrac{{2a}}{2} = 20 + a \\
   \Rightarrow a = 22 - 20 = 2m/{s^2} \\
 $
Substituting this value of a back in equation D, we get
$u = 20 - \dfrac{{15a}}{2} = 20 - \dfrac{{15 \times 2}}{2} = 20 - 15 = 5m/s$
We also need to check if acceleration is constant or not. This can be done by substituting values of u and a in equation C. If we get the distance to be 24m then the acceleration of the body is said to be uniform.
${S_{10}} = u + \dfrac{{17a}}{2} = 5 + \dfrac{{17 \times 2}}{2} = 5 + 17 = 22m$
Hence, we can say the correct answer is option C.

So, the correct answer is “Option C”.

Note: It should be noted that in the three equations A, B and C, we have two unknown variables u and a. In order to obtain their values, we need only two equations which involve these variables. So we can use any two of the three equations to obtain the required values of u and a.