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 

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 t^th second and not total t seconds.

 

2) So, consider for 5^th 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.