","comment":{"@type":"Comment","text":"Resolve the influencing forces into their corresponding components and equate the horizontally contributing ones to arrive at an expression for acceleration. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$0.98\\,ms^{-2}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$19.6\\,ms^{-2}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$9.8\\,ms^{-2}$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4.9\\,ms^{-2}$$","position":3,"answerExplanation":{"@type":"Comment","text":" $$ \\\\ \\text{ From the above diagram, we see that the vertical component of gravity is balanced out by normal reaction.}\\\\ \\text{ The acceleration of the bike however, produces a force that is}\\\\ \\text{ equivalent to the horizontal component of gravity acting down the incline, i.e.,}\\\\ \\Rightarrow ma = mg \\sin\\alpha\\\\ \\Rightarrow a = g \\sin\\alpha\\\\ \\text{Plugging in}\\, \\alpha = 30^{\\circ}:\\\\ a = 9.8 \\times sin (30^{\\circ}) \\\\ \\Rightarrow a= 9.8 \\times 0.5 \\\\ \\therefore a= 4.9\\,ms^{-2} $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Acceleration Quiz 1","text":"Florence Griffith Joyner is known as the fastest woman in history after setting a world record for running a $$100\\,m$$ sprint in $$10.49\\,s$$ in the 1988 U.S Olympics Trials. If she took off from rest upon hearing the starting pistol, and covered the first quarter of the distance in $$3\\,s$$, the second quarter in $$2.5\\,s$$, the third quarter in $$2.2\\,s$$ and the final quarter in $$2.79\\,s$$, what was her acceleration in each quarter? ","comment":{"@type":"Comment","text":"Find the sprinter's final velocity for each quarter, which will be the sprinter's initial velocity for the next quarter, and use that to calculate the acceleration in each case."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2.77\\,ms^{-2}$$, $$0.68\\,ms^{-2}$$, $$0.62\\,ms^{-2}$$, $$0.86\\,ms^{-2}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$2.77\\,ms^{-2}$$, $$4\\,ms^{-2}$$, $$5.68\\,ms^{-2}$$, $$3.21\\,ms^{-2}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$2.77\\,ms^{-2}$$, $$4\\,ms^{-2}$$, $$5.68\\,ms^{-2}$$, $$-3.21\\,ms^{-2}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2.77\\,ms^{-2}$$, $$0.68\\,ms^{-2}$$, $$0.62\\,ms^{-2}$$, $$-0.86\\,ms^{-2}$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$ \\text{ For any quarter, the sprinter’s distance will be d = 25m}\\\\ \\text{ For the first quarter:}\\, u=0\\,ms^{-1}\\,\\text{ and}\\, t = 3\\,s.\\\\ \\text{ Velocity attained:}\\, v= \\dfrac{d}{t} = \\dfrac{25}{3} = 8.3ms^{-1}\\\\ \\text{ Acceleration:}\\, a = \\dfrac{v-u}{t} = \\dfrac{8.3-0}{3} = 2.77\\,ms^{-2}\\\\ \\text{ For the second quarter:}\\, u=8.3ms^{-1}\\,\\text{ and}\\, t = 2.5\\,s.\\\\ \\text{ Velocity attained:}\\,v= \\dfrac{d}{t} = \\dfrac{25}{2.5} = 10\\,ms^{-1}\\\\ \\text{ Acceleration:}\\, a = \\dfrac{v-u}{t} = \\dfrac{10-8.3}{2.5} = 0.68\\,ms^{-2}\\\\ \\text{ For the third quarter:}\\, u=10\\,ms^{-1}\\,\\text{and}\\, t = 2.2\\,s.\\\\ \\text{ Velocity attained:}\\, v= \\dfrac{d}{t} = \\dfrac{25}{2.2} = 11.36\\,ms^{-1}\\\\ \\text{ Acceleration:}\\, a = \\dfrac{v-u}{t} = \\dfrac{11.36-10}{2.2} = 0.62\\,ms^{-2}\\\\ \\text{ For the final quarter:}\\, u=11.36\\,ms^{-1},\\text{ and}\\, t = 2.79s.\\\\ \\text{ Velocity attained:}\\, v= \\dfrac{d}{t} = \\dfrac{25}{2.79} = 8.96\\,ms^{-1}\\\\ \\text{ Acceleration:}\\, a = \\dfrac{v-u}{t} = \\dfrac{8.96-11.36}{2.79} = -0.86\\,ms^{-2}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Acceleration Quiz 1","text":" Fill in the blanks: You decide to attend a Formula 1 practice session for which you have an all-access pass, taking place this weekend at the Buddh Circuit in Noida. This is where the teams test out the performance of their cars and tweak the car’s wings and suspension settings to ensure they’re all set for the actual race. One of the teams decides to test out a new turbocharger that they claim can harness more power from the same engine and provide more acceleration. From the test data, you see that without the turbocharger, the car goes from $$0-100kmh^{-1}$$ in $$3s$$ but does the same with the turbocharger in $$2.6\\,s$$. The acceleration of the car with the turbocharger was found to be nearly __ times the acceleration of the car without it, and the power increased by __ percent. ","comment":{"@type":"Comment","text":" Find the acceleration in each case and compare the values. Remember that power is proportional to the force acting on a body causing it to move at a certain velocity. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1.5$$, $$15.3\\%$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$1.15$$, $$50\\%$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$1.25$$, $$5\\%$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1.15$$, $$15.3\\%$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$ \\text{ In any case, the initial and final velocities of the car are,}\\\\ u = 0\\,ms^{-1}\\,\\text{and}\\, v = 100kmh^{-1} = 100 \\times \\dfrac{5}{18} = 27.8ms^{-1}\\\\ \\text{ Without the turbocharger: t = 3s.}\\\\ \\text{ The acceleration of the car will be:}\\\\ a = \\dfrac{v-u}{t} \\\\ \\Rightarrow a= \\dfrac{27.8-0}{3} \\\\ \\Rightarrow a= 9.27ms^{-2}\\\\ \\text{ Power of the car will be:}\\\\ P = Fv \\\\ \\Rightarrow P= mav \\\\ \\Rightarrow P= 9.27mv\\\\ \\text{ With the turbocharger: t = 2.6s.}\\\\ \\text{ The acceleration of the car will be:}\\\\ a^{\\prime} = \\dfrac{v-u}{t} \\\\ \\Rightarrow a^{\\prime} = \\dfrac{27.8-0}{2.6}\\\\ \\Rightarrow a^{\\prime} = 10.69ms^{-2}\\\\ \\text{ Power of the car will be:}\\\\ P^{\\prime} = Fv \\\\ \\Rightarrow P^{\\prime} = mav \\\\ \\Rightarrow P^{\\prime} = 10.69mv\\\\ \\text{ Now,}\\\\ \\dfrac{a^{\\prime}}{a} = \\dfrac{10.69}{9.27} \\\\ \\Rightarrow a^{\\prime} = 1.15 \\times a \\\\ \\text{ The percentage increase in power will be:}\\\\ \\dfrac{P^{\\prime} - P}{P} \\times 100\\% = \\dfrac{(10.69-9.27)mv}{9.27mv} \\times 100\\% \\\\ \\therefore\\dfrac{P^{\\prime} - P}{P} \\times 100\\% = 15.3\\%\\\\ \\text{In the context of the question:}\\\\ \\text{“…The acceleration of the car with the turbocharger was found to be nearly 1.15 times}\\\\ \\text{ the acceleration of the car without it, and the power increased by 15.3% percent.”}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}