QuestionWatch VideoShow ExamplesA 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, 8, t, cubed, minus, 48, t, squared, .v(t)=8t 3 −48t 2 . Determine all intervals when the acceleration of the particle is positive.
Question
QuestionWatch VideoShow ExamplesA 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, 8, t, cubed, minus, 48, t, squared, .v(t)=8t 3 −48t 2 . 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) = 8t^3 - 48t^2, we can find the acceleration function a(t) by taking the derivative of v(t).
The derivative of 8t^3 is 24t^2 and the derivative of -48t^2 is -96t. So, the acceleration function is a(t) = 24t^2 - 96t.
To find when the acceleration is positive, we set a(t) > 0 and solve for t:
24t^2 - 96t > 0 t(24t - 96) > 0 t(24(t - 4)) > 0
This gives us two critical points: t = 0 and t = 4. We can test these intervals on the number line:
- For t < 0, if we plug in t = -1 into a(t), we get a negative number, so the acceleration is negative.
- For 0 < t < 4, if we plug in t = 2 into a(t), we get a negative number, so the acceleration is negative.
- For t > 4, if we plug in t = 5 into a(t), we get a positive number, so the acceleration is positive.
Therefore, the acceleration of the particle is positive when t > 4.
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