Question

A particle has an initial velocity of $5.5m/s$ due east and a constant acceleration of $1m/s^2$ due west. The distance covered by the particle in $6^{th}$ second of its motion is

• A. $0.25m$
• B. $0.5m$
• C. $0$
• D. $0.75m$

Answer 1

The correct option is A $0.25m$

At $t=5s$
$S_1=5.5\times 5-\frac{1}{2}\times 1\times (5{)}^2$
$=15m$

At $t=5.5s$
$S_2=5.5\times 5.5-\frac{1}{2}\times 1\times (5.5{)}^2$
$=15.125m$

At $0.5s$
$S=S_2-S_1=0.125m$

Distance covered in $6^{th}$ second $=2\times 0.125=0.25m$

Final Answer: (b)