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Motion in Straight Line

Average velocity

A train is targeted to run from Delhi to Pune at an average sp

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A train is targeted to run from Delhi to Pune at an average speed of $80 kph$ but due to repairs of track looses $2 hrs$ in the first part of the journey. If then accelerates at a rate of $20 kph^{2}$ till the speed reaches $100 kph$ . Its speed is now maintained till the end of the journey. If the train now reaches station in time, find the distance from when it started accelerating?

Name: Video solution 1: A train is targeted to run from Delhi to Pune at an average speed of 80kph but due to repairs of track looses 2hrs in the first part of the journey. If then accelerates at a rate of 20kph2 till the speed reaches 100kph. Its speed is now maintained till the end of the journey. If the train now reaches station in time, find the distance from when it started accelerating?
Uploaded: 2023-05-23T04:28:17.000Z
Duration: 7 min
Description: Video solution 1: A train is targeted to run from Delhi to Pune at an average speed of 80kph but due to repairs of track looses 2hrs in the first part of the journey. If then accelerates at a rate of 20kph2 till the speed reaches 100kph. Its speed is now maintained till the end of the journey. If the train now reaches station in time, find the distance from when it started accelerating?

Views: 6,596 students

Updated on: May 23, 2023

Book:

Physics Galaxy Mechanics Vol 1 (Ashish Arora)

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Chloe Discussed

A train is targeted to run from Delhi to Pune at an average speed of

80 kph

but due to repairs of track looses

2 hrs

in the first part of the journey. If then accelerates at a rate of

20 kph^{2}

till the speed reaches

100 kph

. Its speed is now maintained till the end of the journey. If the train now reaches station in time, find the distance from when it started accelerating?

5 mins ago

Discuss this question LIVE

5 mins ago

Text solutionVerified

t

is normal runing time then
Solution Image

Here

t_{1} = \frac{x _{1}}{80} + 2 t_{2} = \frac{100 - 80}{20} = 1 hr t_{3} = \frac{l - x _{1} - x _{2}}{100}

Here

x_{2} = 80 (1) + \frac{1}{2} (20) (1)^{2} = 90 km

Thus total time

t = t_{1} + t_{2} + t_{3} = \frac{l}{80}

(\frac{x _{1}}{80} + 2) + 1 + (\frac{l - x _{1} - 90}{100}) = \frac{l}{80}

Soving it gives

l - x_{1} = 840 km

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Question Text	A train is targeted to run from Delhi to Pune at an average speed of $80 kph$ but due to repairs of track looses $2 hrs$ in the first part of the journey. If then accelerates at a rate of $20 kph^{2}$ till the speed reaches $100 kph$ . Its speed is now maintained till the end of the journey. If the train now reaches station in time, find the distance from when it started accelerating?
Updated On	May 23, 2023
Topic	Motion in Straight Line
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 1
Upvotes	173
Avg. Video Duration	7 min

Text solutionVerified

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