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:

• A. $0$
• B. $0.5m$
• C. $2m$
• D. $0.25m$

Answer 1

The correct option is B $0.5m$

Here the particle velocity will become zero in 4.5 seconds, as

$\begin{array}{c}v=u+at\\ 0=9-2\times t\\ t=4.5\sec \end{array}$

Displacement in the same period will be

$\begin{array}{c}{v}^2-u^2=2as\\ -81=2s\\ s_{4.5}=20.25m\end{array}$

Let us call the point of velocity being zero as Point A.

now body will start moving backwards,

displacement in next 0.5 sec from point A,

$\begin{array}{c}s=ut+\frac{1}{2}a{t}^2\\ s_{0.5}=\frac{1}{2}\left(-2\right){\left(0.25\right)}^2=-0.25m\end{array}$

hence distance covered is 0.25 m

(-ve sign indicates body moved in backward direction. It is to be noted that displacement and distance are not to be confused to be same)

Distance traveled in 4 seconds

$\begin{array}{c}{s}_4=9\times 4+\frac{1}{2}\left(-2\right){\left(4\right)}^2\\ s_4=36-16=20\end{array}$

Thus,

distance covered in 5th second is

$D=\left(s_{4.5}+\left|s_{0.5}\right|-s_4\right)=20.5+0.25-20=0.5$