Answer:
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.