A particle moves along the x-axis so that at time t, is greater than or equal to, 0t≥0 its velocity is given by v, left bracket, t, right bracket, equals, minus, 4, t, cubed, plus, 12, t, .v(t)=−4t 3 +12t. Determine all intervals when the acceleration of the particle is positive.
Question
A particle moves along the x-axis so that at time t, is greater than or equal to, 0t≥0 its velocity is given by v, left bracket, t, right bracket, equals, minus, 4, t, cubed, plus, 12, t, .v(t)=−4t 3 +12t. Determine all intervals when the acceleration of the particle is positive.
Solution
The acceleration of a particle is the derivative of its velocity. Given the velocity function v(t) = -4t^3 + 12t, we can find the acceleration function a(t) by taking the derivative of v(t).
The derivative of -4t^3 is -12t^2 and the derivative of 12t is 12. So, the acceleration function is a(t) = -12t^2 + 12.
To find when the acceleration is positive, we need to solve the inequality -12t^2 + 12 > 0.
First, we can simplify the inequality by dividing every term by -12 (and remember to flip the inequality sign because we're dividing by a negative number):
t^2 - 1 < 0
This inequality is equivalent to:
(t - 1)(t + 1) < 0
The solutions to the equation (t - 1)(t + 1) = 0 are t = 1 and t = -1. These are the points where the expression changes sign.
We can test the sign of the expression in each of the intervals (-∞, -1), (-1, 1), and (1, ∞) by choosing a test point in each interval and substituting it into the expression.
For the interval (-∞, -1), we can choose t = -2. Substituting t = -2 into the expression gives (-2 - 1)(-2 + 1) = 3 > 0.
For the interval (-1, 1), we can choose t = 0. Substituting t = 0 into the expression gives (0 - 1)(0 + 1) = -1 < 0.
For the interval (1, ∞), we can choose t = 2. Substituting t = 2 into the expression gives (2 - 1)(2 + 1) = 3 > 0.
Therefore, the acceleration is positive for t in the intervals (-∞, -1) and (1, ∞).
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