WAEC - Further Mathematics (2023 - No. 10)
A particle began to move at \(27 ms^{-1}\) along a straight line with constant retardation of \(9 ms^{-2}\). Calculate the time it took the particle to come to a stop.
3 sec
2 sec
4 sec
1 sec
คำอธิบาย
\(u = 27 ms^{-1}; a = -9 ms^{-2}; v = 0; t = ?\)
\(v = u + at; t = \frac{v - u}{a}\)
\(t = \frac{0 - 27}{-9} = \frac{-27}{-9}\)
\(\therefore t = 3 sec\)