CBSE Class 11-science Answered
On a two lane road,car A is travelling with a speed of 36km/h.Two cars B and C approach car A in opposite directions with a speed of 54km/h each. At a certain instant when the distance AB=AC,both being 1 km,B decides to overtake A before C does.What minimum acceleration of car B ia required to avoid an accident?
Asked by suchananag2002 | 18 Sep, 2019, 12:09: PM
Expert Answer
Speed of car-A = 36 km/hr = 36×(5/18) = 10 m/s
Speed of car-C = 54 km/hr = 54×(5/18) = 15 m/s
Relative speed of car-C with respect to car-A = 25 m/s
when the distance btween car-A and Car-C is 1000 m, time taken to meet for both the car = 1000/25 = 40 s.
Hence car-B should travel to meeting point before 40 s in order to overtake car-A .
Let us use the formula, " S = u t - [ (1/2) a t2 ] " to get acceleration a of car-B
In the above formula speed u is relative speed of car-B with respect to car-A, i.e. u = 15 - 10 = 5 m/s
Hence, we have, 1000 = 5×40 - [ (1/2)×a×1600] or a = 1 m/s2
Answered by Thiyagarajan K | 18 Sep, 2019, 03:15: PM
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