Stop! Someone met with an accident....

A driver applies breaks after seeing an accident site 400m ahead.At the instant he applies the breaks ,the vehicle was moving with a speed of 15m/s and after it started to retard with 0.3m/s^2 .Find the distance of vehicle after 1 minute ,from the accident site(Answer in metre)?

25 60 375 40

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1 solution

Raven Herd
May 20, 2014

u=15,r=0.3 v^2 - u^2=2as since final velocity v =0, thus u^2=2as s=u^2/2r = 375 v=u+at u=at t= 15/0.3 =50 s after 50 s it comes to a halt so distance from site=400-375=25

Demonstrating once again, especially on multiple-choice questions, the importance of reading exactly what the question asks: here, 375 m. was one of the choices.

Gregory Ruffa - 7 years ago

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READ AGAIN ; It says find the distance FROM accident site and not it's actual POSITION ON THE SCENE.

Raven Herd - 7 years ago

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That's just what I'm saying: it's a typical example of where you shouldn't jump on the choice that matches the number you got from the "hardest" part of the analysis. There's one more step to finish before making your choice. (Test writers do this to students all the time...)

Gregory Ruffa - 7 years ago

With acc. = .3, V after 60 sec. is 15 - .3*60 = _ 3m/s. <..> means it stops before 60 sec. <..> So Vf = 0.

Vf^2 = Vo^2 - 2 * acc * dis.
0^2 = 15^2 - 2 * (.3) * dis. <..> dis. 375m
ans. = 400 - 375 = 25m.

Just to put thing clearly.

Niranjan Khanderia - 7 years ago

Why doesn't this work?

s=ut + (at^2)/2 = 15x60 + ((-.3)(60)(60))/2) = 900 - 540 = 360

400 - 360 = 40m before accident site

Adarsh Narayanan - 6 years, 10 months ago

Yup . The third equation of motion will give the answer . An easy one!

Abhiram Rao - 5 years, 1 month ago

Here in the question ....distance after 60 seconds is askd....if u use d v^2-u^2=2as and substitute v=0....u wil only get the distance when car comes to rest...and not d distance after 1 minute. ..am I rit? I think s=ut+1/2at^2 is needed...

Vishnu P Nair - 7 years ago

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No, that is the trick involved. The car will stop before one minute itself and since after stopping it will not move further ahead therefore if the distance has to be calculated using s = u t + 1 2 a t 2 \displaystyle s= ut + \frac{1}{2} at^2 then the time used must be the time for which the body is under the influence of the constant acceleration which here is 50 50 seconds and not one minute.
Hope that clears your doubt.

Sudeep Salgia - 7 years ago

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Thanks for the explanation...

Tanya Gupta - 7 years ago

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