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In the diagram above, a mass $m$ starting at point $A$ is projected with the same initial horizontal velocity $v_{0}$ along each of the three tracks shown here (with negligible friction) sufficient in each case to allow the mass to reach the end of the track at point $B .$ (Path 1 is directed up, path 2 is directed horizontal, and path 3 is directed down.) The masses remain in contact with the tracks throughout their motions. The displacement $A$ to $B$ is the same in each case, and the total path length of path 1 and 3 are equal. If $t_{1}, t_{2},$ and $t_{3}$ are the total travel times between $A$ and $B$ for paths $1,2,$ and $3,$ respectively, what is the relation among these times?(A) $t_{3} < t_{2} < t_{1}$(B) $t_{2} < t_{3} < t_{1}$(C) $t_{2} < t_{1}=t_{3}$(D) $t_{2}=t_{3} < t_{1}$

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Chapter 12

Practice Test 2

Section 1

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Hi in the given problem there are three possible parts between two point E and be the first part curved upward, vertically, straight, horizontal. And the 3rd part garb like this girl vertically downward. An object is projected from point here and it has to reach up to POINT B. So we have to compare the time taken by this object to cover All these three parts No and long but one august to climb up against gravity. And then after reaching up to the maximum height then it will come down again to recheck point B. But its overall average velocity will reduce and long but two being horizontal. The velocity average velocity well, really insane while at last a long but three the object goes down first with gravity so its average speed average velocity will increase. Hence we conclude that we win means the average velocity along Part three. V three will be greatest. It will be more than that a long path to and it will be more than that a long path, one, So the time will be reversed. Hence, time taken by the particle along. Part three will be shortest, then intermediate along path to, and then maximum time will be taken along. But one so you can say here. Our option E. Is correct. Thank you.

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