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A train is travelling at a speed of 90 km h^(−1). Brakes are applied so as to produce a uniform acceleration of −0.5 m s^(−2) - CBSE Class 9 - Science

ConceptEquations of Motion by Graphical Method - Equation for Position-time Relation

Question

A train is travelling at a speed of 90 km h−1. Brakes are applied so as to produce a uniform acceleration of −0.5 m s−2. Find how far the train will go before it is brought to rest.

Solution

Initial speed of the train, u = 90 km/h = 25 m/s

Final speed of the train, = 0 (finally the train comes to rest)

Acceleration = −0.5 m s−2

According to third equation of motion:

v2 = u2 + 2 as

(0)2 = (25)2 + 2 (−0.5) s

Where, s is the distance covered by the train

s=25^2/(2(0.5))=625m

The train will cover a distance of 625 m before it comes to rest.

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Solution A train is travelling at a speed of 90 km h^(−1). Brakes are applied so as to produce a uniform acceleration of −0.5 m s^(−2) Concept: Equations of Motion by Graphical Method - Equation for Position-time Relation.
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