The initial velocity of a body moving along a straight line is $7 ext{ m/s}$. It has a uniform acceleration

Question

The initial velocity of a body moving along a straight line is $7 	ext{ m/s}$. It has a uniform acceleration of $4 	ext{ m/s}^2$. The distance covered by the body in the 5th second of its motion is:

Accepted Answer

1.  **Identify Formula:** The distance covered in the $n$-th second of uniformly accelerated motion is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$, where $u$ is initial velocity, $a$ is acceleration, and $n$ is the specific second .
2.  **Alternative Method:** Calculate the displacement at $t=5$ and $t=4$ using $S = ut + \frac{1}{2}at^2$ and find the difference.
    *   $S(5) = 7(5) + \frac{1}{2}(4)(5^2) = 35 + 2(25) = 85 	ext{ m}$.
    *   $S(4) = 7(4) + \frac{1}{2}(4)(4^2) = 28 + 2(16) = 60 	ext{ m}$.
    *   Distance in 5th second $= S(5) - S(4) = 85 - 60 = 25 	ext{ m}$.
3.  **Direct Calculation:** Using the $n$-th second formula:
    $S_5 = 7 + \frac{4}{2}(2 	imes 5 - 1)$
    $S_5 = 7 + 2(10 - 1) = 7 + 2(9) = 7 + 18 = 25 	ext{ m}$.

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