The position function of a particle in rectilinear motion is given by s(t) s(t) = t^3 - 9t^2 + 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.

Question

The position function of a particle in rectilinear motion is given by s(t) s(t) = t^3 - 9t^2 + 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.

Answer 1

it reverses direction when its velocity changes sign. That is, when v(t) = 0.

So, find t when

v(t) = s'(t) = 3t^2 - 18t + 24 = 0

then plug that value into s(t) and a(t)

Answer 2

To find the position and acceleration of the particle at the instant it reverses direction, we need to first determine the time when this occurs.

When the particle reverses direction, its velocity changes from positive to negative (or vice versa). In other words, the particle is momentarily stationary at this instant. So, we need to find the time(s) when the velocity of the particle is equal to zero.

The velocity function is the derivative of the position function. To find the velocity function, we differentiate the position function, s(t), with respect to time (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) = 0 to find the times when the velocity is zero:

3t^2 - 18t + 24 = 0

Next, we solve this quadratic equation. We can factor out a common factor of 3:

3(t^2 - 6t + 8) = 0

Now, we factor the quadratic expression inside the parentheses:

3(t - 4)(t - 2) = 0

Setting each factor equal to zero, we get:

t - 4 = 0 --> t = 4
t - 2 = 0 --> t = 2

So, the particle reverses direction at t = 2 and t = 4.

Now that we have the times when the particle reverses direction, we can find the position and acceleration at these instants.

To find the position at t = 2, substitute t = 2 into the position function:

s(2) = 2^3 - 9(2)^2 + 24(2) + 1

Evaluate this expression to find the position at t = 2.

To find the acceleration at t = 2, we differentiate the velocity function v(t) with respect to time:

a(t) = v'(t) = d/dt (3t^2 - 18t + 24)

Differentiating each term separately, we get:

a(t) = 6t - 18

Substitute t = 2 into the acceleration function to find the acceleration at t = 2.

Similarly, repeat the above steps for t = 4 to find the position and acceleration at t = 4.

Remember to include appropriate units in your answers.