","comment":{"@type":"Comment","text":" Displacement in circular motion can be calculated using formula $S=2R\\sin \\left( \\dfrac{\\theta }{2} \\right)$ where $''\\theta ''$ is angular displacement. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{3 m}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{2}{\\sqrt{3}}\\,m$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{\\sqrt{3}}{2}\\,m$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1\\,m$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$ \\text{Given: R=1m} \\\\ \\Rightarrow\\theta =60 \\\\ \\Rightarrow\\dfrac{\\theta }{2}=30 \\\\ \\Rightarrow\\sin \\dfrac{\\theta }{2}=1/2 \\\\ \\text{Displacement in circle given by,} \\\\S=2R\\sin \\dfrac{\\theta }{2}\\\\ \\Rightarrow S =2\\times 1\\times \\dfrac{1}{2}\\\\ \\therefore S =1m$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Displacement and distance Quiz 1","text":"From $$V-t$$ graph,calculate the distance travelled by particle from $$0-6$$ sec. ","comment":{"@type":"Comment","text":"The modulus of area under $v-t$ graph gives value of distance travelled. Negative sign is not considered because we are interested in the path followed not in initial and final position."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4m$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$8m$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" None of these","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$12m$$","position":2,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ $$ The modulus of area under the velocity time graph gives distance. Total distance covered from $$0\\text{ to }6\\text{ sec} =ar\\left( ABCO \\right)+ar\\left( CDEF \\right) +ar\\left( FGHI \\right)$$ $$ \\\\ $$ Total distance covered from $$0\\text{ to }6\\text{ sec}=\\left( 2\\times 2 \\right)+\\left( 2\\times 2 \\right)+\\left( 2\\times 2 \\right)$$ $$ \\\\ $$ Total distance covered from $$0\\text{ to }6\\text{ sec}=4+4+4$$ $$ \\\\ $$ $$\\therefore$$ Total distance covered from $$0\\text{ to }6\\text{ sec}=12\\text{ metres}.$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Displacement and distance Quiz 1","text":" A man on a day follows a sinusoidal direction to perform a magic trick as $$Y=\\sin \\left( \\dfrac{2}{\\pi }t \\right)$$ where $$t$$ is in sec, and the displacement of the man from $$t=0$$ to $$\\dfrac{{{\\pi }^{2}}}{6}$$ sec is: ","comment":{"@type":"Comment","text":"Displacement for two time instances is ${{S}_{{{t}_{2}}}}-{{S}_{t1}}=\\Delta S$,where ${{S}_{{{t}_{2}}}}$ is displacement at time ${t_2}$ and ${{S}_{t_1}}$ is displacement at time ${t_1}$."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{\\sqrt{3}}{2}-1$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{1}{2}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$-\\dfrac{1}{2}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{\\sqrt{3}}{2}$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$\\text{As }y\\left( t \\right)=\\sin \\left( \\dfrac{2}{\\pi }t \\right) \\\\ \\text{for }t=0 \\\\y\\left( 0 \\right)=\\sin \\left( \\dfrac{2}{\\pi }\\times 0 \\right)=0\\text{ units} \\\\ \\text{For }t=\\dfrac{{{\\pi }^{2}}}{6} \\\\ \\Rightarrow y\\left( \\dfrac{{{\\pi }^{2}}}{6} \\right)=\\sin \\left( \\dfrac{2}{\\pi }\\times \\dfrac{{{\\pi }^{2}}}{6} \\right)=\\sin \\left( \\dfrac{\\pi }{3} \\right) \\\\ \\Rightarrow y\\left( \\dfrac{{{\\pi }^{2}}}{6} \\right)=\\dfrac{\\sqrt{3}}{2}\\text{ units} \\\\ \\Rightarrow\\text{Displacement}=y\\left( \\dfrac{{{\\pi }^{2}}}{6} \\right)-y\\left( 0 \\right) \\\\ \\therefore\\text{Displacement}=\\dfrac{\\sqrt{3}}{2}\\text{ units}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Displacement and distance Quiz 1","text":" On a random day, a boy starts walking west and after $$5$$ metre, he turns right and walks straight for $$7$$ metre. He takes a right again and travels $$2$$ meters straight. Finally, he turns right and travels $$3$$ metres straight stops. How far is he from the starting point?","comment":{"@type":"Comment","text":" Use direction sense and calculate shortest distance between initial and final point. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$7\\,m$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$12\\,m$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" zero","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$5\\,m$$","position":2,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ $$ By using direction sense, the figure for the path is as above. The starting and ending points are $$O$$ and $$P$$ respectively. $$ \\\\ \\Rightarrow OX=3m,PX=4m \\\\ $$ By pythagoras theorem $$ \\\\ \\Rightarrow O{{P}^{2}}=O{{X}^{2}}+P{{X}^{2}} \\\\ \\Rightarrow OP=\\sqrt{{{4}^{2}}+{{3}^{2}}} \\\\ \\therefore OP=5\\,m$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Displacement and distance Quiz 1","text":"From $$V-t$$ graph the value of displacement during $$0-5$$ sec is $$ \\\\ $$
","comment":{"@type":"Comment","text":"Area under $\\left( v-t \\right)$ graph gives the value of displacement considering the $-ve$ sign if persisting. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" zero","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$-2\\,m$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$+4\\,m$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$+2\\,m$$","position":2,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ \\text{Area under }v-t\\text{ graph gives the displacement,} \\\\ \\text{Thus total displacement from }t=0\\text{ to }t=5\\text{ seconds is} \\\\ =area\\left( ABCO \\right)+area\\left( \\Delta CED \\right) \\\\ \\Rightarrow \\left( AB\\times BC \\right)+\\dfrac{1}{2}\\times CE\\times CD \\\\ \\Rightarrow \\left( 2\\times 4 \\right)+\\dfrac{1}{2}\\times 3\\times \\left( -4 \\right) \\\\ \\Rightarrow 8-6 = 2\\text{ units}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}