Question

Initially, the velocity of a particle is 11 m/s towards positive x-axis and a constant acceleration towards negative x-axis of magnitude is acting on the particle. The distance traveled by the particle during 6th second is

• A. 0.25 m
• B. 0.5 m
• C. 1.0 m
• D. zero

Answer 1

The correct option is B 0.5 m

$u=11m/s$
$a=-2m/s^2$
They will become zero when $0=11-2t\Rightarrow t=5.5s$
At t=5.5 s, it will change direction of motion.
Velocity at the end of 5s $=11-2\times 5=1m/s$
From 5.5 s to 6s,
$s^{\prime }=\frac{1}{2}\times 2\times {\left(\frac{1}{2}\right)}^2=\frac{1}{4}m$
$\therefore$ Total distance, s+s' $=\frac{1}{2}m=0.5m$.