Question

A body moving with a constant retardation in straight line travels 5.7m and 3.9m in the 6th and 9th second, respectively. When will the body come momentarily to rest?
(a) 10 s
(b) 15 s
(c) 30 s
[please tell the way also]

Answer 1

6th second: 5.7m9th second: 3.9mfor a difference of (9 - 6) = 3 sec,distance travelled = (5.7 - 3.9) = 1.8 mtrat (9 + 3) = 12 sec,distance travelled = (3.9 - 1.8) = 2.1 mtrat (12 + 3) = 15 sec,distance travelled = (2.1 - 1.8) = 0.3 mtrfor a distance of the last 0.3 mtr to be travelled, requires x sec to be added to the 15th sec.0.3 mtr.............x sec______... =...._____1.8 mtr.............3 secx = (0.3 mtr * 3 sec) / 1.8 mtr = 0.5 sec.Body will momentarily come to rest at the 15½th second.

Answer 2

Answer:

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