A particle starts from rest and moves in a straight line such that its velocity,...
FURTHER MATHEMATICS
WAEC 2019
A particle starts from rest and moves in a straight line such that its velocity, V ms\(^{-1}\), at time t second is given by V = 3t\(^2\) - 6t. Calculate the acceleration in the 3rd second.
- A. 12m
- B. 16m
- C. 64m
- D. 96m
Correct Answer: B. 16m
Explanation
V = 3t\(^2\) - 6t
\(\frac{ds}{dt} = 3t^2 - 6t\)
s = \(\int 3t^2 - 6t\)
s = \(\frac{3t^3}{3} - \frac{6t^2}{2} + k\)
s = t\(^3\) - 3t\(^2\) + k
s = 0, t = 0
s = t\(^3\) - 3t\(^2\)
s = 4\(^3\) - 3t\(^2\)
s = 4\(^3\) - 3(4)\(^2\)
= 64 - 48 = 16m
Post an Explanation Or Report an Error
If you see any wrong question or answer, please leave a comment below and we'll take a look. If you doubt why the selected answer is correct or need additional more details? Please drop a comment or Contact us directly. Your email address will not be published. Required fields are marked *