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Given that Avector +Bvector=R and A^2+B^2=R^2, find the angle between them

Given that Avector +Bvector=R and A^2+B^2=R^2, find the angle between them

Grade:11

2 Answers

Sitanshu kumar Singh
53 Points
5 years ago
A vector +B vector =R,then magnitude of A vector +B vector is ✓A^2+B^2+2ABcostheta=R. Squaring both sides we get,. A^2+B^2+2ABcos theta=R. As in the question for A^2+B^2=R 2ABcos theta should be 0 so cos theta =0 this theta =90° .Thus the angle between A vector and B vector is 90°. Please upvote
Gitanjali Rout
184 Points
5 years ago
First let`s find the velocity. First let`s write all the informational given in the question . Initial velocity =0,acceleration=2icap(as it moves towards positive x axis).and time =2s. then we can find the velocity using v=u+at and we get v=0+2icap×2=4icap.This is the velocity at the end of 2s.Now an additional acceleration starts acting so now a becomes 2icap-4jcap. So to find velocity after next 2 s we use v=u+at and it gives v=4icap(as calculated above)+(2icap-4jcap)×2= (8icap-8jcap). So now move further and calculate co-ordinate of the particle x=xnot +it+1/2at^2 and xnot is given as (2icap+4jcap) and u=0 and a =2icap and t=2s.so putting all these we get x=6icap+4jcap .this is the co-ordinate of particle after 2 s. Now to find the coordinates after next 2 s. we use the same equation and putting all values we get 18 icap-4jcap it means 18m,-4m Hope it helps u ask any queries relating to this if you didn`t understand please approved

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