Dare u not to go back to basics

a car is revolving around a circular orbit with a velocity is given by ω = e t \omega=e^t
what is the acceleration at t=0?.

Derails and assumptions:
Neglect friction.
Radius of the orbit is 200 m.
Report your answer with two decimal places.
Good luck!


The answer is 282.842.

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2 solutions

a ( t ) = ( a n 2 ( t ) + a t 2 ( t ) ) 1 2 a(t)=(a_n^2(t)+a_t^2(t))^{\frac{1}{2}}

a n ( t = 0 ) = ω 2 ( t = 0 ) r = r a_n(t=0)=\omega^2(t=0) r=r

a t ( t = 0 ) = r ( d ω d t ) t = 0 = r a_t(t=0)=r(\frac{d\omega}{dt})_{t=0}=r

So a ( t = 0 ) = r 2 1 2 a(t=0)=r 2^{\frac{1}{2}}

Jafar Badour
Sep 14, 2015

x = R c o s ( e t ) x=Rcos(e^t) and y = R s i n ( e t ) y=Rsin(e^t) so the acceleration on the x axis is R ( e t s i n ( e t ) e 2 t c o s ( e t ) R( -e^t sin(e^t) -e^2t cos(e^t) and the acceleration on y axis is R ( e t c o s ( e t ) e 2 t s i n ( e t ) R(e^t cos(e^t) -e^2t sin(e^t) so the acceleration on the x axis is 276.354 and on the y axis 60.2337 so the acceleration is the radical of 276.35 4 2 + 60.233 7 2 276.354^2 +60.2337^2 equal 282.842

hmm... they re just asking magnitude... it would ve been easier if u considered tangential and normal accelerations...

Sriram Vudayagiri - 5 years, 8 months ago

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yeah but I wanted to get back to the basics!

jafar badour - 5 years, 8 months ago

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Overrated ques

Gauri shankar Mishra - 5 years, 4 months ago

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@Gauri shankar Mishra why? I guess the rating system in brilliant is good

jafar badour - 5 years, 4 months ago

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