a train traveling at 72km/hr is to brought to rest in a distance of 200m then its retadetion should be what.
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72 km 200 m is a train travelling at distance
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✇ A train is traveling at 72km/hr is to rest in a distance of 200m.
- ♳ Initial Velocity (u) = 72 km/hr
- ♴ Final Velocity (v) = 0 m/s
- ♵ Distance (s) = 200 m
✇ We need to find the retardation of the train.
☉ First of all we need to convert the unit of initial velocity (u) from km/hr to m/s.
☉After making units same just simply plug in the known values in third kinematical equation of motion.
✪ CALCULATION ✪
↠ Initial Velocity (u) = 72 km/hr
↠ Initial Velocity (u) = 72 × (5/18)
↠ Initial Velocity (u) = 360/18
↠ Initial Velocity (u) = 20 m/s
∴ Initial Velocity (u) of the train is 20 m/s.
⇢v² - u² = 2as
⇢(0)² - (20)² = 2 × a × 200
⇢ 0 - 400 = 400a
⇢-400 = 400a
⇢a = -400/400
⇢a = -1 m/s²
∴ The retardation of the train is 1 m/s².
✣ Note :- Retardation is nothing but negative Acceleration.
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