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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

In the figure shown acceleration of A is $a_A=15 \hat i+15 \hat j$ then the acceleration of B is

(A remains in contact with B) a)$6 \hat i$ b)$-15 \hat j$ c)$-10 \hat j$ d)$-5 \hat i$  
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2 Answers

–1 vote
in the fig. $a_{AX}$ is actually $a_{AY}$ and vice-versa.
$a_{Ax}\cos 53^{\circ}-a_{ay}\cos 37^{\circ}$
$\quad=a_B \cos 53^{\circ}$
$a_B=-5 m/s$
$a_B=-5 \hat i$
Hence d is the correct answer.
answered Jul 16, 2013 by meena.p
edited May 27, 2014 by lmohan717
–1 vote
<div class="clay6-step-odd"><div id="pr10" class="clay6-image">http://clay6.com/mpaimg/jee%203%2068a.JPG</div><div id="pr11" class="clay6-basic">$a_{Ax}=15$</div><div id="pr12" class="clay6-basic">$a_{Ay}=15$</div><div id="pr13" class="clay6-basic">$a_{Ay}=(a_B)_{\perp}$</div><div id="pr14" class="clay6-basic">$a_{Ax}\cos 53^{\circ}-a_{ay}\cos 37^{\circ}$</div><div id="pr15" class="clay6-basic">$\quad=a_B \cos 53^{\circ}$</div><div id="pr16" class="clay6-basic">$a_B=-5 m/s$</div><div id="pr17" class="clay6-basic">$a_B=-5 \hat i$</div></div>
answered Dec 20, 2013 by nageswari.c
reshown Dec 20, 2013 by nageswari.c

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