A particle moves along a straight line such that its displacement at any time t is given by S = t3 – 6t2 + 3t + 4 metres. The velocity when the acceleration is zero is :4 ms–1– 12 ms–1 42 ms–1– 9 ms–1
Question
A particle moves along a straight line such that its displacement at any time t is given by S = t3 – 6t2 + 3t + 4 metres. The velocity when the acceleration is zero is :4 ms–1– 12 ms–1 42 ms–1– 9 ms–1
Solution
The displacement of the particle is given by S = t³ - 6t² + 3t + 4.
The velocity of the particle is the first derivative of the displacement with respect to time, which is v = ds/dt = 3t² - 12t + 3.
The acceleration of the particle is the first derivative of the velocity with respect to time, which is a = dv/dt = 6t - 12.
Setting the acceleration equal to zero gives 6t - 12 = 0, which solves to t = 2.
Substituting t = 2 into the velocity equation gives v = 3(2)² - 12(2) + 3 = 12 - 24 + 3 = -9 ms⁻¹.
So, the velocity when the acceleration is zero is -9 ms⁻¹.
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