Question

A particle has an initial velocity of $9m/s$ due east and a constant acceleration of $2m/s^2$ due west. The distance covered by the particle in the fifth second of its motion is : ( in metres)

• A. $0.5$
  
• B. $1$
  
• C. $2$
  
• D. $3$
  

Answer 1

The correct option is A $0.5$

Given, initial velocity, $\overrightarrow{u}=9\hat{i}m/s$ and acceleration, $\overrightarrow{a}=-2\hat{i}m/s^2$

Using formula, $S_n=u+\frac{1}{2}(2n-1)a$, the particle displacement in 5 th sec is

$S_5=9\hat{i}+0.5(10-1)(-2\hat{i})=0$

Thus, the particle will be reached in the initial position in 5 th sec.

Let, the velocity will be zero at time $t$.

So using, $v=u+at\Rightarrow 0=9-2t$ or $t=4.5sec$

The velocity at 4 sec is $v=9-2\left(4\right)=1m/s$

Using formula, $S=ut+\left(1/2\right)a{t}^2$, distance covered in time 4 sec to 4.5 sec is $S_1=1\left(1/2\right)+\left(1/2\right)(-2)(1/2{)}^2=0.25m$ and distance covered in time 4.5 sec to 5 sec is $S_2=\left(1/2\right)\left(2\right)(1/2{)}^2=0.25m$ (as velocity at 4.5 sec is zero)

Thus, total distance covered in 5 th sec $S_1+S_2=0.25+0.25=0.5m$