","comment":{"@type":"Comment","text":" The change in velocity with respect to time is defined as acceleration. If there is no acceleration the velocity will be constant "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$16\\text{ m}{{\\text{s}}^{-1}}\\text{ , accelerating}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$16\\text{ m}{{\\text{s}}^{-1}}\\text{ , decelerating}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$15\\text{ m}{{\\text{s}}^{-1}}\\text{ , decelerating}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$15\\text{ m}{{\\text{s}}^{-1}}\\text{ , accelerating}$$","position":2,"answerExplanation":{"@type":"Comment","text":" $$\\text{Acceleration is given by;} \\\\ \\dfrac{dv}{dt}=a \\\\ \\Rightarrow dv=adt \\\\ \\Rightarrow \\int{dv}=\\int{adt} \\\\ \\Rightarrow v\\left( t \\right)=2\\int{dt}=2t+C \\\\ \\text{For }t=0\\text{ velocity is given as;} \\\\ v\\left( 0 \\right)=0 \\\\ \\Rightarrow C=0 \\\\ \\Rightarrow v\\left( t \\right)=2t \\\\ \\therefore v\\left( 8 \\right)=16\\text{ m}{{\\text{s}}^{-1}} \\\\ \\text{Acceleration is positive.} \\\\ \\therefore \\text{The object is accelerating}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Acceleration-time graphs Quiz 1","text":" Referring to the $$a-t$$ diagram given below, determine the change in velocity.","comment":{"@type":"Comment","text":"The differentiation of velocity of an object with respect to time is termed as acceleration of the object. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$36\\text{ m}{{\\text{s}}^{-1}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$34\\text{ m}{{\\text{s}}^{-1}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" None of the above.","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$35\\text{ m}{{\\text{s}}^{-1}}$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$\\text{Change in velocity is given by;} \\\\ \\Delta v=\\text{ Area under the curve} \\\\ \\Rightarrow \\Delta v=\\dfrac{1}{2}\\times 7\\times 4+7\\times 3 \\\\ \\Rightarrow \\Delta v=14+21 \\\\ \\therefore \\Delta v=35\\text{ m}{{\\text{s}}^{-1}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Acceleration-time graphs Quiz 1","text":"A ball is moving in the positive $$X$$ direction such that $$x\\left( t \\right)=\\left( 4{{t}^{3}}-2t+2 \\right)\\text{ m}$$ . What is the acceleration time relationship, also find the change in acceleration for $$t=3\\text{ sec}$$ to $$t=6\\text{ sec}$$. ","comment":{"@type":"Comment","text":"The change in displacement with respect to time is velocity. The change in velocity with respect to time is acceleration. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$a=62t\\text{ , 23 m}{{\\text{s}}^{-2}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$a=72t\\text{ , 24 m}{{\\text{s}}^{-2}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$a=23t\\text{ , 6}2\\text{ m}{{\\text{s}}^{-2}}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$a=24t\\text{ , }72\\text{ m}{{\\text{s}}^{-2}}$$","position":2,"answerExplanation":{"@type":"Comment","text":" $$\\text{Given that;} \\\\ x\\left( t \\right)=4{{t}^{3}}-2t+2 \\\\ \\text{velocity of an object is given by;} \\\\ v=\\dfrac{dx}{dt}=12{{t}^{2}}-2 \\\\ \\text{Acceleration of a body is given as;} \\\\ a=\\dfrac{dv}{dt}=\\dfrac{{{d}^{2}}x}{d{{t}^{2}}}=24t \\\\ \\Rightarrow a\\left( 3 \\right)=3\\times 24=72\\text{ m}{{\\text{s}}^{-2}} \\\\ \\Rightarrow a\\left( 6 \\right)=24\\times 6=144\\text{ m}{{\\text{s}}^{-2}} \\\\ \\Rightarrow \\Delta a=a\\left( 6 \\right)-a\\left( 3 \\right)=144-72 \\\\ \\therefore \\Delta a=72\\text{ m}{{\\text{s}}^{-2}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Acceleration-time graphs Quiz 1","text":"A body of mass $$m$$ initially at rest is accelerating with respect to time as given: $$a\\left( t \\right)=\\left( 4t-8 \\right)\\text{ m}{{\\text{s}}^{-2}}$$. After travelling for $$3\\text{ sec}$$ it collides with a ball, finding the impulse delivered to the ball by the wall. ","comment":{"@type":"Comment","text":"The impulse delivered to a body is the area of the force-time curve. This simply means the product of mass and area of acceleration-time curve. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$-4m\\text{ Ns}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$6m\\text{ Ns}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$4m\\text{ Ns}$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$-6m\\text{ Ns}$$","position":3,"answerExplanation":{"@type":"Comment","text":" $$\\text{Impulse is given as;} \\\\ J=\\int\\limits_{{{t}_{1}}}^{{{t}_{2}}}{Fdt} \\\\ F=ma=m\\left( 4t-8 \\right) \\\\ \\Rightarrow J=m\\int\\limits_{0}^{3}{\\left( 4t-8 \\right)dt} \\\\ \\Rightarrow J=m\\left( \\left. \\dfrac{4{{t}^{2}}}{2} \\right|_{0}^{3}-\\left. 8t \\right|_{0}^{3} \\right) \\\\ \\Rightarrow J=m\\left( 18-24 \\right) \\\\ \\therefore J=-6m\\text{ Ns}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Acceleration-time graphs Quiz 1","text":"If a small mass $$m$$ is accelerated by an acceleration of $$a\\left( t \\right)=\\left( 2{{t}^{3}}-4{{t}^{2}}-t +1 \\right)\\text{ m}{{\\text{s}}^{-2}}$$. Calculate the change in momentum for $$8\\text{ sec}$$ and the nature of the motion. ","comment":{"@type":"Comment","text":" The slope of the acceleration time graph will be positive after a certain time. The value of acceleration will always be positive."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1000m\\text{ , accelerated motion with decreasing acceleration }t=\\dfrac{8+\\sqrt{40}}{12}\\text{.}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$2000m\\text{ , always decelerated motion with increasing deceleration}\\text{.}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" Data is insufficient.","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1000m\\text{ , accelerated motion with increasing acceleration after }t=\\dfrac{8+\\sqrt{40}}{12}\\text{.}$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$\\text{Change in velocity is given by;} \\\\ \\Delta v=\\int\\limits_{{{t}_{1}}}^{{{t}_{2}}}{adt} \\\\ \\Rightarrow \\Delta v=\\int\\limits_{0}^{8}{\\left( 2{{t}^{3}}-4{{t}^{2}}-t+1 \\right)dt} \\\\ \\Rightarrow \\Delta v=\\left. \\left( \\dfrac{2{{t}^{4}}}{4}-\\dfrac{4{{t}^{3}}}{3}-\\dfrac{{{t}^{2}}}{2}+t \\right) \\right|_{0}^{8} \\\\ \\Rightarrow \\Delta v=2048-1024-32+8 \\\\ \\Rightarrow \\Delta v=1000\\text{ m}{{\\text{s}}^{-1}} \\\\ \\text{Change in momentum is given by;} \\\\ \\Delta p=m\\Delta v \\\\ \\Rightarrow \\Delta p=1000m\\text{ kgm}{{\\text{s}}^{-1}} \\\\ \\text{For the acceleration versus time relation;} \\\\ \\text{slope}=\\dfrac{da}{dt}=6{{t}^{2}}-8t-1 \\\\ \\text{The roots of the above quadratic equation are;} \\\\ t=\\dfrac{8+\\sqrt{40}}{12}\\text{ and }t=\\dfrac{8-\\sqrt{40}}{12} \\\\ \\Rightarrow t=\\dfrac{8-\\sqrt{40}}{12}<0 \\\\ \\text{Time cannot be negative, therefore discarding }t=\\dfrac{8-\\sqrt{40}}{12} \\\\ t=\\dfrac{8+\\sqrt{40}}{12} \\\\ \\Rightarrow \\text{For }t=\\dfrac{8+\\sqrt{40}}{12}\\text{ there is a constant deceleration of 1}\\text{.3 m}{{\\text{s}}^{-2}}\\text{ and} \\\\ \\text{starts accelerating after }t=\\dfrac{8+\\sqrt{40}}{12} \\\\ \\text{For }t>\\dfrac{8+\\sqrt{40}}{12} \\\\ \\dfrac{da}{dt}>0 \\\\ \\text{Therefore, the motion is accelerated motion with increasing velocity after }t=\\dfrac{8+\\sqrt{40}}{12}\\text{ seconds}\\text{.}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}