The position function of a particle in rectilinear motion is given by s(t) s(t) = t3 – 9t2 + 24t + 1 for t ≥ 0. Find the position and acceleration of the particle at the instant the when the particle reverses direction. Include units in your answer.
it reverses direction when v changes from + to -
v(t) = 3t^2-18t+24 = 3(t^2-6t+8) = 3(t-2)(t-4)
So, where does v(t) cross the axis and become negative?
Now use that value to find s(t) and a(t)
To find the position and acceleration of the particle at the instant it reverses direction, we need to find the time when the velocity changes sign. In other words, we need to find the value(s) of t for which the velocity is equal to zero.
The velocity function can be obtained by taking the derivative of the position function with respect to time, s(t):
v(t) = s'(t) = d/dt (t^3 - 9t^2 + 24t + 1).
Differentiating each term separately, we get:
v(t) = 3t^2 - 18t + 24.
Now, we set v(t) equal to zero and solve for t:
0 = 3t^2 - 18t + 24.
Dividing both sides by 3, we get:
0 = t^2 - 6t + 8.
This quadratic equation can be factored as:
0 = (t - 2)(t - 4).
Setting each factor equal to zero, we find two solutions:
t - 2 = 0, t - 4 = 0.
Solving for t, we get t = 2 and t = 4. These are the times when the particle changes direction.
To find the position and acceleration at these times, we substitute these values of t back into the position and acceleration functions.
First, let's find the position at t = 2. Substituting t = 2 into the position function s(t), we get:
s(2) = (2)^3 - 9(2)^2 + 24(2) + 1.
Evaluating this expression, we find:
s(2) = 8 - 36 + 48 + 1 = 21.
Therefore, the position of the particle at the instant it reverses direction (t = 2) is 21 units.
Next, let's find the acceleration at t = 2. To do so, we take the derivative of the velocity function with respect to time:
a(t) = v'(t) = d/dt (3t^2 - 18t + 24).
Differentiating each term separately, we get:
a(t) = 6t - 18.
Substituting t = 2 into the acceleration function, we find:
a(2) = 6(2) - 18 = 12 - 18 = -6.
Therefore, the acceleration of the particle at the instant it reverses direction (t = 2) is -6 units/s^2.
Similarly, we can find the position and acceleration at t = 4:
s(4) = (4)^3 - 9(4)^2 + 24(4) + 1,
and
a(4) = 6(4) - 18.
Evaluating these expressions, we can find the position and acceleration at t = 4.
Therefore, at the instant the particle reverses direction, the position is 21 units and the acceleration is -6 units/s^2.