Question

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

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

Answer 1

Answer: (B)

At $t=5s$

$S=5.5\times 5\frac{1}{2}\times 1\times (5{)}^2$

$=15m$

at $t=6s$

$S=5.5\times 6-\frac{1}{2}\times 1\times (6{)}^2$

$=15m$

at $t=5.5s$

$S=5.5\times 5.5-\frac{1}{2}\times 1\times (5.5{)}^2$

$=15.125m$



Distance covered in $6$ th second

$=-2\times 0.125=0.25m$