the position of a particle is given by x=t^3-6t^2+9t. find its position at which velocity becomes zero
Answers
Answered by
78
Answer:
4 and 0
Explanation:
Velocity = dx/dt
Velocity = d(t³ - 6t² + 9t) /dt
Velocity = 3t² - 12t + 9
Let the velocity be 0 at t = T.
=> 3T² - 12T + 9 = 0
=> 3(T² - 4T + 3) = 0
=> T² - 3T - T + 3 = 0
=> T(T - 3) - 1(T - 3) = 0
=> (T - 3)(T - 1) = 0
=> T = 3 and 1 second.
Hence, at t = 3 and 1, its position is:
=> x = (1)³ - 6(1)² + 9(1) or (3)³ - 6(3)² + 9(3)
=> x = 4 and 0
Answered by
7
Answer:
Hope it is your helping..
Attachments:
Similar questions