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A body moving ina curved path possessesa velicty 3m//stowards north at any instant of its motion . A fter 10s, the velocity of the body was found to be 4 m//s towards west. Calculate the average acceleration during this interval .

Answer» <html><body><p></p>Solution :To <a href="https://interviewquestions.tuteehub.com/tag/solve-647535" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVE">SOLVE</a> this problem the vector nature of velocity must be <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a> into account. <br/> In the figure, the initial velocity`v_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)`and the final velocity v arfe <a href="https://interviewquestions.tuteehub.com/tag/drawn-2062388" style="font-weight:bold;" target="_blank" title="Click to know more about DRAWN">DRAWN</a> from a common origin . The vecto difference of them is found by the parallelogram method. <br/> The magnitude of difference is <br/> `|v-v_(0)|=OC=sqrt(OA^(2)+AC^(2))`<br/> `=sqrt(4^(2)+3^(2))=5m//s`<br/> The direction is given by <br/> `tan theta =3/4=0.75, theta =37^(0)`<br/> Average acceleration `=|vecv-vecv_(0)|/t=5/(10)=0.5m//s^(2)`at`37^(@)`south of <a href="https://interviewquestions.tuteehub.com/tag/west-1451899" style="font-weight:bold;" target="_blank" title="Click to know more about WEST">WEST</a><br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AI_PHY_XI_V01_A_C04_SLV_010_S01.png" width="80%"/></body></html>


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