The position of the front bumper of the test car under microprocessor control is given by x(t)=2.17m+(4.80 m/s^2)t^2-(0.100m/s^6)t^6. Find the position, velocity, and acceleration when t=2.0s

To solve this problem, we need to find the position, velocity, and acceleration of the car at \( t = 2.0 \, \text{s} \) using the given position function:

\[
x(t) = 2.17 \, \text{m} + (4.80 \, \text{m/s}^2) t^2 - (0.100 \, \text{m/s}^6) t^6
\]

### 1. Position at \( t = 2.0 \, \text{s} \)

The position of the car at \( t = 2.0 \, \text{s} \) can be found by substituting \( t = 2.0 \) into the position function \( x(t) \).

\[
x(2) = 2.17 + (4.80)(2.0)^2 - (0.100)(2.0)^6
\]

### 2. Velocity at \( t = 2.0 \, \text{s} \)

The velocity is the first derivative of the position function \( x(t) \) with respect to time \( t \). Let's calculate:

\[
v(t) = \frac{dx(t)}{dt} = \frac{d}{dt} \left[ 2.17 + 4.80 t^2 - 0.100 t^6 \right]
\]

\[
v(t) = 0 + (4.80 \times 2)t - (0.100 \times 6)t^5
\]

\[
v(t) = 9.60t - 0.60t^5
\]

Substitute \( t = 2.0 \):

\[
v(2) = 9.60(2.0) - 0.60(2.0)^5
\]

### 3. Acceleration at \( t = 2.0 \, \text{s} \)

The acceleration is the second derivative of the position function, or the first derivative of the velocity function:

\[
a(t) = \frac{d^2 x(t)}{dt^2} = \frac{d}{dt} \left( 9.60t - 0.60t^5 \right)
\]

\[
a(t) = 9.60 - (0.60 \times 5)t^4
\]

\[
a(t) = 9.60 - 3.00t^4
\]

Substitute \( t = 2.0 \):

\[
a(2) = 9.60 - 3.00(2.0)^4
\]

Now, let's compute these values.Here are the calculated values for the position, velocity, and acceleration of the car at \( t = 2.0 \, \text{s} \):

1. **Position**: \( x(2) = 14.97 \, \text{m} \)
2. **Velocity**: \( v(2) = 0.0 \, \text{m/s} \)
3. **Acceleration**: \( a(2) = -38.4 \, \text{m/s}^2 \)

Would you like further details on these calculations or have any other questions?

### Related Questions:
1. How does the position change over a range of times, say from \( t = 0 \) to \( t = 5 \) seconds?
2. What is the significance of a negative acceleration in this context?
3. How would the velocity change if a different coefficient were used for the \( t^5 \) term in the position function?
4. What is the impact of the \( t^6 \) term on the overall behavior of the car's position?
5. How can we graph the position, velocity, and acceleration functions to better visualize their changes over time?

**Tip:** Understanding derivatives can help you interpret how quantities like velocity and acceleration change with respect to time!

Calculate the position, velocity, and acceleration of a car at t = 2.0 seconds using the given position function x(t). Understand how derivatives are used to find velocity and acceleration in a polynomial function context. Suitable for high school students studying kinematics and polynomial functions.

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