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a particle moving in a circle with a constant speed covers an angle of 90degrees .the change in its velocity is ??????in terms of v
Dear Aditya In 90 degree it will cover one fourth of the perimeter suppose speed is v then time taken t=2∏r/4v=∏r/2v velocity=displacement/time displacement=√r2+r2 =r√2 velocity=r√2/t = 2√2 v/∏ All the best. AKASH GOYAL AskiitiansExpert-IITD Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
Dear Aditya
In 90 degree it will cover one fourth of the perimeter
suppose speed is v then time taken t=2∏r/4v=∏r/2v
velocity=displacement/time
displacement=√r2+r2 =r√2
velocity=r√2/t = 2√2 v/∏
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
initial velocity is vi and final velocity is vj,,,,,so the net velocity is vj-vi,,,,whose magnitude is(v2+v2).5
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