If a train travelling at $72 ext{ km/h}$ is to be brought to rest in a distance of $200 ext{ metres}$, the

Question

If a train travelling at $72 	ext{ km/h}$ is to be brought to rest in a distance of $200 	ext{ metres}$, then its retardation should be:

Accepted Answer

1.  **Convert Units:** First, convert the initial velocity from km/h to m/s.
    $u = 72 	ext{ km/h} = 72 	imes \frac{5}{18} 	ext{ m/s} = 20 	ext{ m/s}$ .
2.  **Identify Given Values:**
    *   Initial velocity, $u = 20 	ext{ m/s}$.
    *   Final velocity, $v = 0 	ext{ m/s}$ (brought to rest).
    *   Displacement, $s = 200 	ext{ m}$.
3.  **Select Formula:** Use the kinematic equation relating velocity, acceleration, and displacement: $v^2 = u^2 + 2as$ .
4.  **Calculation:**
    $0^2 = (20)^2 + 2a(200)$
    $0 = 400 + 400a$
    $400a = -400$
    $a = -1 	ext{ m/s}^2$.
5.  **Conclusion:** The acceleration is $-1 	ext{ m/s}^2$. Retardation is the magnitude of negative acceleration, which is $1 	ext{ m/s}^2$.

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