A body A starts from rest with an acceleration $a_1$. After $2$ seconds, another body B starts from rest with

Question

A body A starts from rest with an acceleration $a_1$. After $2$ seconds, another body B starts from rest with an acceleration $a_2$. If they travel equal distances in the $5^{	ext{th}}$ second, after the start of A, then the ratio $a_1 : a_2$ is equal to:

Accepted Answer

1.  **Formula for Distance in n-th Second:** The distance traveled by a body in the $n$-th second of its motion is given by $S_n = u + \frac{a}{2}(2n - 1)$, where $u$ is the initial velocity and $a$ is the acceleration .
2.  **Motion of Body A:**
    *   Starts from rest: $u_A = 0$.
    *   Acceleration: $a_1$.
    *   Time interval: $5^{	ext{th}}$ second (from $t=4$ to $t=5$s).
    *   Distance $S_{A} = 0 + \frac{a_1}{2}(2 	imes 5 - 1) = \frac{9a_1}{2}$.
3.  **Motion of Body B:**
    *   Starts from rest: $u_B = 0$.
    *   Acceleration: $a_2$.
    *   Body B starts 2 seconds *after* A. Therefore, the $5^{	ext{th}}$ second from the start of A corresponds to the interval between $t=4$ and $t=5$ seconds on A's clock. Since B started at $t=2$, this interval corresponds to the time from $t' = (4-2) = 2$s to $t' = (5-2) = 3$s on B's clock. This is the **$3^{	ext{rd}}$ second** of B's motion.
    *   Distance $S_{B} = 0 + \frac{a_2}{2}(2 	imes 3 - 1) = \frac{5a_2}{2}$.
4.  **Equating Distances:**
    Given $S_{A} = S_{B}$.
    $\frac{9a_1}{2} = \frac{5a_2}{2}$
    $9a_1 = 5a_2$
    $\frac{a_1}{a_2} = \frac{5}{9}$.

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