Question

The acceleration of the particle moving in $x-$direction with an initial velocity $3\text{m/s}$ is $2{m/s}^2$. Then its displacement at $t=3s$ will be

• A. $36\text{m}$
• B. $26\text{m}$
• C. $8\text{m}$
• D. $18\text{m}$

Answer 1

The correct option is D $18\text{m}$

Given, $u=3\text{m/s},a=2{\text{m/s}}^2,t=3s$
using $2^{nd}$ equation of motion i.e.,
$s=ut+\frac{1}{2}a{t}^2$
$s=3\times 3+\frac{1}{2}\times 2\times 3^2$
$s=9+9=18\text{m}$