The initial velocity of a particle is $10 ext{ m/sec}$ and its retardation is $2 ext{ m/sec}^2$. The dista

Question

The initial velocity of a particle is $10 	ext{ m/sec}$ and its retardation is $2 	ext{ m/sec}^2$. The distance moved by the particle in the 5th second of its motion is:

Accepted Answer

1.  **Analyze Motion:** The particle has an initial velocity $u = 10 	ext{ m/s}$ and acceleration $a = -2 	ext{ m/s}^2$ (retardation). First, check when the particle comes to rest using $v = u + at$. 
    $0 = 10 - 2t \implies t = 5 	ext{ s}$.
    Since the particle stops exactly at the end of the 5th second, it does not change direction during the 5th second (interval $t=4$ to $t=5$). Thus, distance equals displacement.
2.  **Formula for $n$-th Second:** The displacement in the $n$-th second is given by $S_n = u + \frac{a}{2}(2n - 1)$.
3.  **Calculation:** Substitute $u=10$, $a=-2$, and $n=5$:
    $S_5 = 10 + \frac{-2}{2}(2 	imes 5 - 1)$
    $S_5 = 10 - 1(9)$
    $S_5 = 10 - 9 = 1 	ext{ m}$.
4.  **Alternative Method:** Calculate position at $t=4$ and $t=5$ using $x = ut + \frac{1}{2}at^2$ .
    $x_5 = 10(5) - \frac{1}{2}(2)(25) = 50 - 25 = 25 	ext{ m}$.
    $x_4 = 10(4) - \frac{1}{2}(2)(16) = 40 - 16 = 24 	ext{ m}$.
    Distance in 5th second $= x_5 - x_4 = 25 - 24 = 1 	ext{ m}$.

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