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
The correct answer is: A
Explanation
\(u = 27 ms^{-1}; a = -9 ms^{-2}; v = 0; t = ?\)
\(v = u + at; t = \frac{v - u}{a}\)
\(t = \frac{0 - 27}{-9} = \frac{-27}{-9}\)
\(\therefore t = 3 sec\)