A body covers 24 m in the 6th second and 32 m in the 8th second of its motion. If the body is uniformly accelerated, the distance traveled in the 10th second will be:

The distance covered in the 10th second is 40 m

Explanation:

The parameters given are;

Distance traveled in the 6th second = 24 m

Distance traveled in the 8th second = 32 m

Therefore, we have;

v² = u² + 2·a·s

s = u·t + 1/2·a·t²

Where:

s = Distance traveled

t = Time taken = 1 second each

a = Acceleration

Plugging in the values, we obtain the following equations;

24 = u₆ × 1 + 1/2 × a × 1² = u₆ + a/2.....(1)

Also, 32 = u₈ + a/2

However, in the 8th second, the initial speed is given as follows;

u₈ = u₆ + a×t = u + a × (t₂ - t₁) = u₆ + a × (8 - 6) = u₆ + 2·a

Therefore, we have;

32 = u₆ + 2·a + a/2  

....................(2)

Subtracting equation (1) from equation (2), we have;

∴ 2·a = 8

a = 8/2 = 4 m/s²

The velocity in the 8th second is given as follows;

32 = u₈ + a/2

32 = u₈ + 4/2

u₈ = 32 - 2 = 30 m/s

The velocity in the 10th second is therefore, given as follows;

u₁₀ = u₈ + a×t = u₈ + a × (t₂ - t₁) = u₈ + a × (8 - 6) = u₈ + 2·a = 30 + 2×4 = 38 m/s

The distance covered in the 10th second is therefore given as follows

s₁₀ = u₁₀ + a/2 = 38 + 4/2 = 40 m.