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

By BYJU'S Exam Prep

Updated on: September 25th, 2023

A train is travelling at a speed of 90kmh-1. Brakes are applied so as to produce a uniform acceleration of -0.5 ms-2. The train will go 625 m before it comes to rest.

Steps to Calculate how far the train will go before it is brought to rest:

Step 1: It is given that

• Initial train speed u = 90 kmh-1 = 90 x 5/18 ms-1 = 25 ms-1
• Final velocity v = 0 ms-1
• Acceleration a = – 0.5 ms-1

Let the train’s prior stopping distance, s, be.

Step 2: We can make use of the formula:

v2 = u2 + 2as

Step 3: Measurement of distance:

From the third order motion equation

• v2 = u2 + 2as
• 0 = (25ms-1) + 2 x (-0.5 ms-2) x s
• s = 625 m

While distance and displacement seem to mean the same thing, their definitions and meanings are actually extremely different. While distance refers to how much land an object has traversed throughout its journey, displacement measures how far an object is out of place.
The total movement of an object, independent of direction, is its distance. The amount of space that an object has travelled, regardless of where it started or ended, is referred to as distance.
As a result, the train will travel 625 metres before stopping.

Summary:

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

A train is moving at 90 kilometres per hour. The application of brakes results in a uniform acceleration of -0.5 ms-2. The train will travel 625 metres before stopping. Distance is the space travelled by an object irrespective of where it started or ended.

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