Physics on road.

Classical Mechanics Level pending

A man on a scooter has a speed of 9m/s. He is about to go down an inclined road of inclination 30°. He switches off his engine before going down. The incline has a length of 10m and coefficient of friction (3^0.5)/5. While just entering the incline, he sees another car with some speed 'v', engine switched off, about to go down a Similar incline. The road after the incline is frictionless. The scooter guy, after traveling 11m after the incline, hears a thud indicating the car is at the end of the incline. Using the data, findthe carscar's initial speed 'v'.


The answer is 3.

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2 solutions

Ashutosh Mishra
May 16, 2015

First calculate the time taken by scooter guy to reach the bottom of incline. That will come as 1 sec. Next find the scooter guy's speed at the end of incline, that will come as 11m/s. Hence time taken for car to come down = (1 + (11/11)) = 2 sec. Now using 10=2v+4,find v, which is 3m/s.

Aditya Kumar
May 16, 2015

There are a few typos in your problem.

1. 1. cars is typed as carscar's.

2. 2. the answer is in decimals, do mention to round of the answer to the nearest integer.

+1 for the ques

Well i typed in my phone thats why the typos... second of all... the answer is a perfect '3'.... did u make a mistake somewhere?

Ashutosh Mishra - 6 years ago

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I took g=10, maybe that created problem. It's better if you type problems through you pc. The website is better than app as u can like, report and reshare problems. Also do some LaTeXing in your problem. That'll make the presentation of the problem better .

Aditya Kumar - 6 years ago

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Look at the soln i posted, tell me if i made any errors... i also took g=10.

Ashutosh Mishra - 6 years ago

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@Ashutosh Mishra 10 = 9 t + . 5 ( 5 2 3 ) t 2 10 = 9t + .5*(5-2√3)*t^2 Therefore t comes out to be nearly 1, not exactly 1

Aditya Kumar - 6 years ago

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@Aditya Kumar I gave the coefficient as (√3)/5

Ashutosh Mishra - 6 years ago

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@Ashutosh Mishra My mistake :p

Aditya Kumar - 6 years ago

Okay, will use PC from now on....but is my solution wrong?

Ashutosh Mishra - 6 years ago

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