A body moving in a straight line cones a distance of 14m in 5th sec and 20m in 8th sec . how much distance will it cover in 15th sec
Asked by drdilipmane1968 | 29th Jul, 2017, 10:05: PM
This query becomes simple if you consider the distance formula and find u and a from it and solve for the asked time interval.
The following steps will aid you in the query:
1) Consider distance formula, s = ut + at2/2. This gives the total distance travelled in t seconds. But, the given data is of tth second and not total t seconds.
2) So, consider for 5th second first. S5 = distance travelled in 5 seconds - distance travelled in 4 seconds. S5 = [(5u + 25a/2) - (4u + 16a/2)]. You will get S5 = 14 = u + 9a/2.
3) Similarly, find S8 = 20 = u + 15a/2.
4) So, now you have 2 equations. Solve them to get the values of 'u' and 'a'. Once you subtract them, you will get 'a' = 2 m/s2.
5) Then you multiply one equation by 10 and the other by 6. Then subtract the two equations and solve to get 'u'. You will get u = 5 m/s.
6) Finally use the values of u and a to get S15 = [(15u + 225a/2) - (14u + 196a/2)] = u + 29a/2 = 34 m.
Answered by Romal Bhansali | 30th Jul, 2017, 12:32: PM
- <div>An object starting from rest Travels 20 m in first 2 seconds and 160 m in next 4 seconds. what will be the velocity after 7 seconds?</div>
- <div> <div>Two balls are projected at different angles from the same place and with the same initial speed</div> <div>of 50 m/s. Both balls have the same range of 216 m. The difference in their times of flight is</div> <div>close to</div> <div>(A) 14.4 s (B) 7.8 s (C) 3.6 s (D) 0 s</div> </div> <div> </div> <div>Please give detailed solution.</div> <div> </div>
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