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The problem provided involves a particle moving according to the law of motion \( s = f(t) \), where \( s \) is the displacement in feet, and \( t \) is time in seconds. Let's go through the questions step-by-step and solve them. The equation of motion is given as:

\[
f(t) = t^3 - 9t^2 + 24t
\]

### (a) Find the velocity \( v(t) \) at any time \( t \):

The velocity is the derivative of the position function with respect to time:

\[
v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 9t^2 + 24t) = 3t^2 - 18t + 24
\]

### (b) What is the velocity \( v(1) \) after 1 second?

Substitute \( t = 1 \) into the velocity equation:

\[
v(1) = 3(1)^2 - 18(1) + 24 = 3 - 18 + 24 = 9 \, \text{ft/s}
\]

### (c) When is the particle at rest?

The particle is at rest when the velocity is zero:

\[
v(t) = 3t^2 - 18t + 24 = 0
\]

To solve this quadratic equation:

\[
t^2 - 6t + 8 = 0
\]

Factoring:

\[
(t - 4)(t - 2) = 0
\]

So, the particle is at rest at \( t = 2 \) seconds and \( t = 4 \) seconds.

### (d) When is the particle moving in the positive direction?

The particle moves in the positive direction when \( v(t) > 0 \). We can analyze the sign of \( v(t) = 3t^2 - 18t + 24 \) by factoring it:

\[
v(t) = 3(t - 4)(t - 2)
\]

The critical points are \( t = 2 \) and \( t = 4 \). We check the intervals:

- For \( t < 2 \), say \( t = 1 \), \( v(1) = 9 \), so the velocity is positive.
- For \( 2 < t < 4 \), say \( t = 3 \), \( v(3) = -3 \), so the velocity is negative.
- For \( t > 4 \), say \( t = 5 \), \( v(5) = 9 \), so the velocity is positive.

Thus, the particle is moving in the positive direction in the intervals \( (0, 2) \cup (4, \infty) \).

### (e) Draw a diagram to illustrate the motion of the particle and find the total distance traveled in the first 6 seconds.

We already know when the particle is moving positively and negatively. To find the total distance traveled, we need to compute the position at \( t = 0 \), \( t = 2 \), \( t = 4 \), and \( t = 6 \).

The position function is:

\[
f(t) = t^3 - 9t^2 + 24t
\]

- \( f(0) = 0^3 - 9(0)^2 + 24(0) = 0 \)
- \( f(2) = 2^3 - 9(2)^2 + 24(2) = 8 - 36 + 48 = 20 \, \text{ft} \)
- \( f(4) = 4^3 - 9(4)^2 + 24(4) = 64 - 144 + 96 = 16 \, \text{ft} \)
- \( f(6) = 6^3 - 9(6)^2 + 24(6) = 216 - 324 + 144 = 36 \, \text{ft} \)

To find the total distance, we add the absolute values of changes in position:

- From \( t = 0 \) to \( t = 2 \): \( |f(2) - f(0)| = |20 - 0| = 20 \, \text{ft} \)
- From \( t = 2 \) to \( t = 4 \): \( |f(4) - f(2)| = |16 - 20| = 4 \, \text{ft} \)
- From \( t = 4 \) to \( t = 6 \): \( |f(6) - f(4)| = |36 - 16| = 20 \, \text{ft} \)

Total distance traveled = \( 20 + 4 + 20 = 44 \, \text{ft} \).

### (f) Find the acceleration \( a(t) \) at any time \( t \):

The acceleration is the derivative of the velocity:

\[
a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 18t + 24) = 6t - 18
\]

### Find the acceleration \( a(1) \) after 1 second:

Substitute \( t = 1 \) into the acceleration equation:

\[
a(1) = 6(1) - 18 = -12 \, \text{ft/s}^2
\]

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Would you like further details on any part of this solution? Here are some related questions:

1. How would the solution change if the equation of motion were different?
2. What happens if we change the units of time or displacement?
3. Could the particle ever reach a maximum distance? If so, how can we find it?
4. How do we calculate the total distance when the velocity changes direction multiple times?
5. Can we solve these problems using numerical methods instead of analytical solutions?

**Tip**: When solving motion problems, always identify where the velocity is zero to determine when the particle is at rest and where the motion changes direction.

A particle moves according to the law of motion s = f(t) = t^3 - 9t^2 + 24t, where t is measured in seconds and s in feet. Find velocity, acceleration, and distance traveled.

Solve a particle motion problem using calculus. Find the velocity, acceleration, and total distance traveled over time for the function s = t^3 - 9t^2 + 24t. Key concepts include differentiation and quadratic equations. Suitable for high school calculus students.

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