A Honda Civic travels in a straight line a long a road. The car's distance x from a stop sign is given as a function of the tby the equation x(t) =At^2-Bt^3, where A=1.50 m/s^2 and B=0.0500 m/s^3 calculate the velocity of car for a) t=0 to t=2s. b) t=0 to t=4s .  c) t=2s to t=4s

To solve the problem, we'll first determine the velocity function for the Honda Civic by differentiating the given distance function \( x(t) \) with respect to time \( t \). Then, we'll use the velocity function to calculate the velocity of the car over the specified time intervals.

### Step 1: Given Information

The distance function is given as:

\[
x(t) = At^2 - Bt^3
\]

Where:
- \( A = 1.50 \, \text{m/s}^2 \)
- \( B = 0.0500 \, \text{m/s}^3 \)

### Step 2: Find the Velocity Function

The velocity \( v(t) \) is the first derivative of the position function \( x(t) \) with respect to time \( t \):

\[
v(t) = \frac{dx(t)}{dt} = \frac{d}{dt} \left( 1.50t^2 - 0.0500t^3 \right)
\]

Differentiate each term:

\[
v(t) = 2(1.50)t - 3(0.0500)t^2
\]

Simplify the expression:

\[
v(t) = 3.00t - 0.150t^2
\]

This is the velocity function for the car.

### Step 3: Calculate the Velocity Over Specific Time Intervals

#### a) From \( t = 0 \) to \( t = 2 \) seconds

To find the velocity at specific times, we can evaluate the velocity function at \( t = 0 \) and \( t = 2 \):

\[
v(0) = 3.00(0) - 0.150(0)^2 = 0 \, \text{m/s}
\]

\[
v(2) = 3.00(2) - 0.150(2)^2 = 6.00 - 0.600 = 5.40 \, \text{m/s}
\]

So, the velocity at \( t = 0 \) is \( 0 \, \text{m/s} \) and at \( t = 2 \) seconds is \( 5.40 \, \text{m/s} \).

#### b) From \( t = 0 \) to \( t = 4 \) seconds

Similarly, evaluate the velocity at \( t = 4 \):

\[
v(4) = 3.00(4) - 0.150(4)^2 = 12.00 - 2.40 = 9.60 \, \text{m/s}
\]

So, the velocity at \( t = 4 \) seconds is \( 9.60 \, \text{m/s} \).

#### c) From \( t = 2 \) to \( t = 4 \) seconds

For this interval, we've already calculated the velocities at \( t = 2 \) and \( t = 4 \):

- \( v(2) = 5.40 \, \text{m/s} \)
- \( v(4) = 9.60 \, \text{m/s} \)

### Summary of Results

- **a) From \( t = 0 \) to \( t = 2 \) seconds:**
  - Velocity at \( t = 0 \) is \( 0 \, \text{m/s} \).
  - Velocity at \( t = 2 \) is \( 5.40 \, \text{m/s} \).

- **b) From \( t = 0 \) to \( t = 4 \) seconds:**
  - Velocity at \( t = 4 \) is \( 9.60 \, \text{m/s} \).

- **c) From \( t = 2 \) to \( t = 4 \) seconds:**
  - Velocity changes from \( 5.40 \, \text{m/s} \) to \( 9.60 \, \text{m/s} \).

Would you like further explanation on any of these points or have any questions?

Here are some related questions that might interest you:
1. How can we calculate the average velocity of the car over the time intervals?
2. What would be the car's acceleration at different times?
3. How does the negative term \( -Bt^3 \) affect the car's motion over time?
4. What would happen to the velocity if \( B \) were increased or decreased?
5. Can we determine the time at which the car's velocity is zero?

**Tip:** When calculating velocities over intervals, remember to differentiate the position function first to obtain the velocity function, which gives you the instantaneous velocity at any specific time.

Learn how to calculate the velocity of a Honda Civic using the position function x(t) = 1.50t^2 - 0.0500t^3. Solve for velocity at specific time intervals: 0 to 2 seconds, 0 to 4 seconds, and 2 to 4 seconds. Understand the differentiation process and its application in kinematics. Suitable for high school students studying physics or calculus.

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