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 $125 \mathrm{~m}$ and $50 \mathrm{~m} / \mathrm{s}$, respectively. The distance travelled by the particle in $(t+2)^{\text {th }} \mathrm{s}$ is $\mathrm{m}$.
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Answer: (175)
Solution:
Considering acceleration is constant
$v=u+a t$
$u+50=u+a \Rightarrow a=50 \mathrm{~m} / \mathrm{s}^{2}$
$125=u t+\frac{1}{2} a t^{2}$
$125=u+\frac{a}{2}$
$\Rightarrow u=100 \mathrm{~m} / \mathrm{s}$
$\therefore S_{n^{\text {in }}}=u+\frac{a}{2}[2 n-1]$
$=175 \mathbf{m}$
