Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

$20 \text{ m/s}^2$

$10 \text{ m/s}^2$

$2 \text{ m/s}^2$

$1 \text{ m/s}^2$

Convert Units: First, convert the initial velocity from km/h to m/s. $u = 72 \text{ km/h} = 72 \times \frac{5}{18} \text{ m/s} = 20 \text{ m/s}$ .
Identify Given Values: Initial velocity, $u = 20 \text{ m/s}$ . Final velocity, $v = 0 \text{ m/s}$ (brought to rest).

Select Formula: Use the kinematic equation relating velocity, acceleration, and displacement: $v^2 = u^2 + 2as$ .
Calculation: $0^2 = (20)^2 + 2a(200)$ $0 = 400 + 400a$ $400a = -400$ $a = -1 \text{ m/s}^2$ .
Conclusion: The acceleration is $-1 \text{ m/s}^2$ . Retardation is the magnitude of negative acceleration, which is $1 \text{ m/s}^2$ .

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