The displacement and the increase in the velocity of a moving particle in the time interval of to s are 125 m and 50 m/s, respectively. The distance travelled by the particle in is __________ m

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Solution

Q.
The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t + 1)$ s are 125 m and 50 m/s, respectively. The distance travelled by the particle in ${(t + 2)}^{t h} s$ is __________ m. [2024]

Ans.
(175) $a = \frac{Δ V}{Δ t} = \frac{50}{1} = 50 {m / s}^{2}$

$displacement in {(t + 1)}^{th} s e c o n d$

$S_{1} = 125 m$

$\Rightarrow 125 = u + \frac{50}{2} [2 (t + 1) - 1]$

$\Rightarrow u = 125 - 25 [2 t + 1]$ ... (1)

$displacement in {(t + 2)}^{th} second$

$S_{2} = u + \frac{50}{2} [2 (t + 2) - 1] = u + 25 [2 t + 3]$ ...(2)

Using equation (1) in equation (2)

$S_{2} = 125 - 25 (2 t + 1) + 25 (2 t + 3)$

$= 125 - 25 + 75 = 175 m$

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