Motion In One Dimension Question 10
Question 10 - 2024 (30 Jan Shift 1)
The displacement and the increase in the velocity of a moving particle in the time interval of to $(t+1) s$ are $125 m$ and $50 m / s$, respectively. The distance travelled by the particle in $(t+2)^{\text {th }} s$ is $m$.
Show Answer
Answer: (175)
Solution:
Considering acceleration is constant
$v=u+a t$
$u+50=u+a \Rightarrow a=50 m / s^{2}$
$125=u t+\frac{1}{2} a t^{2}$
$125=u+\frac{a}{2}$
$\Rightarrow u=100 m / s$
$\therefore S _{n^{\text {in }}}=u+\frac{a}{2}[2 n-1]$
$=175 \mathbf{m}$
