","comment":{"@type":"Comment","text":" The first equation of motion is best suited for time varying uniform motion in a straight line with some initial velocity and acceleration. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$u=2 m{{s}^{-1}},a=2 m{{s}^{-2}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$u=5 m{{s}^{-1}},a=10 m{{s}^{-2}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$u=10 m{{s}^{-1}},a=5 m{{s}^{-2}}$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$u=5 m{{s}^{-1}},a=5 m{{s}^{-2}}$$","position":3,"answerExplanation":{"@type":"Comment","text":" First equation of motion is given by, $$ \\\\v=u+at\\\\ $$ From the above problem by putting the values in two scenarios we have, $$ \\\\ 15=u+a\\times 2.......(1) \\\\ \\Rightarrow 25=u+a\\times 4.......(2) \\\\ $$ Subtracting equation $$(1)$$ by $$(2)$$ we have, $$ \\\\ u=5 m{{s}^{-1}}\\\\ $$ And putting this value in any one of the equation we get, $$ \\\\ \\therefore a=5 m{{s}^{-2}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Kinematic equations Quiz 1","text":"Tesla launches a new automatic car in India whose displacement on empty road is given by $$s=7{{t}^{3}}+4{{t}^{2}}+3t+4$$m, due to this variation in displacement only physics professors can drive this car. So Mehta sir went to the showroom and did a test drive of the car and wants to calculate instantaneous velocity of car at $$t=2 s$$. the calculated velocity will be- ","comment":{"@type":"Comment","text":"We are calculating instantaneous velocity by taking the first derivative of the displacement function and the second derivative will be used to calculate acceleration. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$83 m{{s}^{-1}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$90 m{{s}^{-1}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$72 m{{s}^{-1}}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$103 m{{s}^{-1}}$$","position":0,"answerExplanation":{"@type":"Comment","text":" Velocity is given by change in displacement with respect to time .i.e.$$ \\\\v=\\dfrac{ds}{dt}\\\\\\Rightarrow v=\\dfrac{d}{dt}\\left( 7{{t}^{3}}+4{{t}^{2}}+3t+4 \\right)=21{{t}^{2}}+8t+3\\\\ $$ Now instantaneous velocity at $$t=2$$ sec will be, $$ \\\\v=21\\times {{2}^{2}}+8\\times 2+3\\\\\\therefore v=103 m{{s}^{-1}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Kinematic equations Quiz 1","text":"A high speed vehicle id travelling along a horizontal direction. The acceleration-time graph of its motion is shown in figure. Now he is measuring velocity at $$t=10 s$$ and $$t=20 s$$ so he can plot a velocity-time graph of its motion. The $$V-t$$ graph will be given by-","comment":{"@type":"Comment","text":" In first $10 s$ acceleration is varying with time so that velocity – time graph can’t be a straight line in that time. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"
","position":0},{"@type":"Answer","encodingFormat":"text/html","text":"
","position":1},{"@type":"Answer","encodingFormat":"text/html","text":"
","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"
","position":2,"answerExplanation":{"@type":"Comment","text":" From the given graph in $$ t=0s$$ to $$t=10s$$ the acceleration as a function of time will be,$$a=2t$$ $$ \\\\ $$ Velocity at $$t=20s$$ will be, $$ \\\\ \\Rightarrow a=\\dfrac{dv}{dt}\\\\\\Rightarrow dv=\\int_{0}^{10}{2t.dt}\\\\\\Rightarrow v=\\int_{0}^{10}{2t.dt}=\\left[ {{t}^{2}} \\right]_{0}^{10}=100 m{{s}^{-1}}\\\\ $$ Now after $$10s$$ acceleration is constant so to calculate velocity at $$t=20s$$ we can apply first equation of motion so, $$ \\\\ \\Rightarrow v=u+at \\\\ \\therefore v=100+20\\times 10=300m{{s}^{-1}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Kinematic equations Quiz 1","text":"A man $$A$$ is standing on a hill of height of $$15 m$$ from the ground and his friend $$B$$ is standing on the ground just nearby the hill as shown in figure. Now, $$B$$ wants to catch a ball thrown by his friend $$A$$ but with one condition that he can catch a ball with a maximum of $$20 m{{s}^{-1}}$$ of speed. Now man $$A$$ throws a ball in an upward direction which eventually comes down to $$A$$ after some time. The maximum speed by which $$A$$ throws a ball so $$B$$ can catch it easily will be $$ \\\\ $$
","comment":{"@type":"Comment","text":"Third equation of motion is used in the above solution. Total energy conservation can also be used to calculate the required answer."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$20 m{{s}^{-1}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$15 m{{s}^{-1}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$25 m{{s}^{-1}}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$10 m{{s}^{-1}}$$","position":2,"answerExplanation":{"@type":"Comment","text":" $$B$$ can catch a ball of maximum velocity $$ 20 m{{s}^{-1}}$$ $$ \\\\ $$ So, let us suppose A throws a ball in upward direction then it reaches some height H and starts to fall down eventually.$$ \\\\ $$ Calculating H $$ {{v}^{2}}={{u}^{2}}+2as\\\\\\Rightarrow {{20}^{2}}={{0}^{2}}+2\\times 10\\times (H+15)\\\\\\Rightarrow 400=20H+300\\\\\\Rightarrow H=5m\\\\ $$ Now the speed by which A can through a ball to reach a height of 5m is $$ \\\\ {{0}^{2}}={{u}^{2}}-2\\times 10\\times 5\\\\\\Rightarrow {{u}^{2}}=2\\times 10\\times 5\\\\\\therefore u=10 m{{s}^{-1}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}