A body covers `10 m` in 4 th ` second and ` 15m` in `16 th` second of its uniformly accelerated motion. How far will it travel I the next `3 seconds .

A body covers `10 m` in 4 th ` second and ` 15m` in `16 th` second of

1.	A body covers `10 m` in 4 th ` second and ` 15m` in `16 th` second of its uniformly accelerated motion. How far will it travel I the next `3 seconds .
Answer» Here, `D_(4) =10 m , D_(6) =15 ,` `S_(9) -S _(6)=? ` Let (u) and (a) be the initial velocity and unitorm acceleration of the body. As `D(n) =u +a/2 (2 n-1) :. D_(4) +a/2 (2 xx 4-1) =10 ` …(i) and `D_(6) =u +a/2 (2 xx 6 -1) =15` .. (i) Subtravting (i) from (iii), we get `2a=5 or a=(5//2) ms^-2)` Form (i)., `u=10 -(5//2)/2 (8-1) =10 -(35)/4 =5/4 ms^(-1)` Distance travelled in `t seconds is `S=ut + 1/2 at^(20` When `t=9s`, then S_(9) =5/4 xx 1/2 xx 5/2 xx 9^(2) =112. 5 m` When `t=6 s`, `S_(6)=5/4 xx 6 +1/2 xx 5/2 xx 6^(2) =52.5 m` :. Distance travelled in the next three seconds `S_(9 ) S_(6) =112.5 -52 =60.0 m`.

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A body covers `10 m` in 4 th ` second and ` 15m` in `16 th` second of its uniformly accelerated motion. How far will it travel I the next `3 seconds .

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