Question

T
5. A body covers a distance of 20m in 7th second and 24m in 9th second how much
shall it cover in 15th s?

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Answer 1

Answer:-

36m

Explanation:-

Let the initial velocity of the body be u and acceleration of the body be a.

Distance travelled by a body in the nth second is given by :-

=> Sₙₜₕ = u + a/2(2n-1)

When n = 7th sec :-

=> 20 = u + a/2(2×7-1)

=> 20 = u + 13a/2

=> 20 = u + 6.5a ----(1)

When n = 5th sec :-

=> 24 = u + a/2(2×9-1)

=> 24 = u + 17a/2

=> 24 = u + 8.5a ----(2)

On subtracting eq.1 from eq.2, we get :-

=> 24-20 = u+8.5a-(u+6.5a)

=> 4 = u+8.5a-u-6.5a

=> 4 = 2a

=> a = 4/2

=> a = 2m/s²

Now, let's put the value of 'a' in eq.1 to find the initial velocity of the body.

=> 20 = u + 2/2(2×7-1)

=> 20 = u + 13

=> u = 20-13

=> u = 7m/s

Hence, distance covered in 15th sec :-

=> S₁₅ₜₕ = 7 + 2/2(2×15-1)

=> S₁₅ₜₕ = 7 + 29

=> S₁₅ₜₕ = 36m

Thus, distance covered by the body in the 15th second is 36m.