Question

Consider a particle initially moving with a velocity of $5\text{m/s}$ starts decelerating at a constant rate of $2m/s^2$. Find the distance travelled in the $2^{nd}$ second.

• A. $1\text{m}$
• B. $5\text{m}$
• C. $2\text{m}$
• D. $8\text{m}$

Answer 1

The correct option is C $2\text{m}$

Given, $u=5\text{m/s},a=-2{\text{m/s}}^2$
As we know that the distance travelled by the particle in $n^{th}$ second is given as
$S_n=u+\frac{1}{2}a(2n-1)$
$\Rightarrow S_{2^{nd}}=5+\frac{1}{2}\times (-2)\left[2\right(2)-1]=2\text{m}$