Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
+1 vote

An aeroplane is moving on a circular path with a speed of $250 \;kmhr^{-1}$ The change in its velocity in half revolution is

$\begin{array}{1 1}(A)\;0 \\(B)\;250\;km\;hr^{-1} \\(C)\;500\;km\;hr^{-1}\\(D)\;250 \sqrt 2 \;km\;hr^{-1} \end{array} $

Can you answer this question?

1 Answer

0 votes
If $\overrightarrow{v_1} =(250 km hr^{-1} ) \hat i$
then $\overrightarrow{v_2} =-(2250\;hm\;hr^{-1} ) \hat i$
Hence $ \Delta \overrightarrow{v}= (\overrightarrow{v_2}-\overrightarrow{v_1})$
$\qquad= (-500 km hr^{-1} ) \hat i $
So, $ | \Delta \overrightarrow{v}|=500 \;km\;hr^{-1} $
Hence C is the correct answer.
answered Jul 16, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App