Question

Particle has initial velocity $9m{s}^{-1}$ due east and constant acceleration of $2m{s}^{-2}$ due west. If the distance covered by it in fifth second of its motion is $\frac{n}{10}m$, then the value of ${}^{\prime }{n}^{\prime }$ is:

• A. $5$
• B. $1$
• C. $3$
• D. $2$

Answer 1

The correct option is A $5$

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

$v=u+at$

$0=9-2t$

$t=4.5sec$

Displacement in the same period will be

$v^2-u^2=2as$

$-81=2\times (-2)s$

$s_{4.5}=20.25m$

Let us call the point of velocity being 0 as Point A

Now body will start moving backwards,

Displacement in next 0.5 sec from point A,

$s=ut+\frac{1}{2}a{t}^2$

$s_{0.5}=\frac{1}{2}\times (-2)\times 0.25=-0.25m$

Hence distance covered is 0.25m

(negative 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

$s_4=36-16=20m$

Distance covered in 5th second is

$D=s_{4.5}+s_{0.5}-s_4$

$D=20.25+0.25-20$

$D=0.5m$

which means the distance covered by particle in 5th second of its motion is $\frac{n}{10}=\frac{5}{10}=0.5$

Therefore value of n is 5