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a particle moving with acceleration and then retardation covers a distance of 4 km prove that 1/a1+1/a2=4min^2/km |
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Answer» 1) Direct Proportionality between Time and Distance (When Speed is CONSTANT) When speed is constant distance will vary as time Car A moves for 2 hours @ 25kmph ---------> d = s X t -------> 50 = 25 X 2 Car B moves for 3 hours @ 25kmph ---------> d = s X t --------> 75 = 25 X 3 Here (tA / tB) = (dA / dB) -----> time ratio = distance ratio = 2 : 3 2) Direct Proportionality between Speed and Distance (When Time is constant) If time of two bodies or motion is constant, then distance will vary as speed Here (S1 / S2) = (d1 / d2) If two cars start simultaneously from opposite ends towards each other. They meet at point ‘C’ In this case distance covered by each car will vary as their speeds since they travelled for equal period. Suppose the distance AB = 900 km Speed A = 25kmph Speed B = 20kmph In this case their meeting point (C) will be (S1 / S2) = (d1 / d2) ------> 25/20 ----> 5/4 = 500 / 400 i.e. 500 kms from A 3) Inverse Proportionality between Speed and Time (when Distance is constant) If the distance to be covered is constant, then time will vary inversely as speed i.e. as speed increases, time decreases and vice versa. (S1 / S2) = (t2 / t1) Example 1 :- A Train meets with an ACCIDENT and moves at ¾ of its original speed. Because of this it delayed by 20 minutes. What is the original time for the journey beyond the point of accident. -----> Speed becomes ¾ hence time becomes 4/3 -------------------- remember (S1 / S2) = (t2 / t1)) -----> increased time is 1/3 = 20 minutes. So original time = 60 minutes. (1 belongs to 20mins, then 3 belongs to 60mins) In other WORDS ------> speed dropped to 75% and time increased to 133.33% ----> increased time = 33.33% = 20 minutes Now 33.33 % belongs to 20 minutes, So 100% belongs to 60 minutes. Example 2 :- A Man walked from his HOUSE to office at 5kmph and got 20 minutes late. if he had travelled at 7.5kmph, he would have reached 12 minutes early. The distance from his house to office is? Here s1 = 5kmph t1 = t2 + 20 + 12 s2 = 7.5kmph t2 = t2 (S1 / S2) = (t2 / t1) ------> (5/7.5) = (t2 / t2 + 32) ---------> 5t2 + 160 = 7.5t2 -------> t2 = 64 minutes So distance = 7.5 X 1.7 (64 mins) = 8km Relative Speed Relative Speed can be viewed as a movement of one body relative to another moving body. 1. Relative Speed of bodies moving in same direction. In the case of the bodies moving to and fro between two points A and B, The faster body will reach the end first and will meet the second body on its way back. The relative speed S1 – S2 will apply till the point of reversal of the faster body and after that the two bodies will start to move in the opposite directions at a relative speed of S1 + S2. The relative speed governing the movement of the two bodies will alternate between S1 – S2 and S1 + S2 everytime anyone of the bodies reverses the direction. However, if both the bodies reverse their direction at the same instant, there will be no change in the relative speed equation. In this case, the description of the motion of the two bodies between two consecutive meetings will also be governed by the proportionality between speed and distance – since the time of movement between any two meetings will be constant. In this case, for every meeting, the total distance covered by the two bodies will be 2d (d = distance between the extreme points). The respective coverage of the distance will in the ratio of the individual speeds. THUS for the 9th meeting the total distance covered will be 9 X 2d = 18d 2. Relative Speed of bodies moving in opposite direction. In the case of the bodies moving to and fro between two points A and B starting from opposite ends of the path, The two bodies will meet at a point in between A and B, then move apart away from each other. The faster body will reach its extreme point first followed by the slower body reaching its extreme point next. Relative speed will change every time ; one of the bodies reverses direction. In this case, the position of the meeting point will be determined by the ratio of speeds of the bodies – since the time travelled of both the bodies is same. (Remember (S1 / S2) = (d1 / d2)) |
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