A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).
Calculate the distance travelled in the third second.
Calculate the distance travelled in the third second.
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Correct Answer: Option A
Explanation:
From S = 12t + \(\frac{5}{2t^2}\) - t\(^3\);
Distance traveled in 2 seconds
S = 12[2] + \(\frac{5}{2[2]^2}\) - 2\(^3\).
= 24 + 10 - 8 = 26m
Distance traveled in 3 seconds
S = 12[3] + \(\frac{5}{2[3]^2}\) - 3\(^3\)
= 36 + 22.5 - 27 = 31.5m
Distance traveled in the 3rd second
= 31.5m - 26m
: S = 5.5m
From S = 12t + \(\frac{5}{2t^2}\) - t\(^3\);
Distance traveled in 2 seconds
S = 12[2] + \(\frac{5}{2[2]^2}\) - 2\(^3\).
= 24 + 10 - 8 = 26m
Distance traveled in 3 seconds
S = 12[3] + \(\frac{5}{2[3]^2}\) - 3\(^3\)
= 36 + 22.5 - 27 = 31.5m
Distance traveled in the 3rd second
= 31.5m - 26m
: S = 5.5m