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 this problem, we'll start by finding the velocity of the car as a function of time \( t \). The velocity is the derivative of the position function \( x(t) \) with respect to time.

Given:
\[
x(t) = A t^2 - B t^3
\]
where \( A = 1.50 \, \text{m/s}^2 \) and \( B = 0.0500 \, \text{m/s}^3 \).

### Step 1: Find the Velocity Function
The velocity function \( v(t) \) is the derivative of \( x(t) \) with respect to \( t \):

\[
v(t) = \frac{dx(t)}{dt}
\]

Taking the derivative of \( x(t) \):

\[
v(t) = \frac{d}{dt} \left( A t^2 - B t^3 \right)
\]

\[
v(t) = 2A t - 3B t^2
\]

Substituting the values of \( A \) and \( B \):

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

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

### Step 2: Calculate the Velocity for Given Time Intervals

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

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

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

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

1. At \( t = 0 \):
   \[
   v(0) = 0 \, \text{m/s}
   \]

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

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

1. At \( t = 2 \) seconds:
   \[
   v(2) = 5.40 \, \text{m/s}
   \]

2. At \( t = 4 \) seconds:
   \[
   v(4) = 9.60 \, \text{m/s}
   \]

So, the velocities at the specific times are:

- From \( t = 0 \) to \( t = 2 \) seconds: \( v(0) = 0 \, \text{m/s} \) and \( v(2) = 5.40 \, \text{m/s} \).
- From \( t = 0 \) to \( t = 4 \) seconds: \( v(0) = 0 \, \text{m/s} \) and \( v(4) = 9.60 \, \text{m/s} \).
- From \( t = 2 \) to \( t = 4 \) seconds: \( v(2) = 5.40 \, \text{m/s} \) and \( v(4) = 9.60 \, \text{m/s} \).

Would you like more details or have any other questions? Here are some related questions to expand your understanding:

1. How do you calculate the acceleration from the velocity function?
2. What is the physical significance of the coefficients \( A \) and \( B \) in the equation?
3. How would the velocity change if the coefficient \( B \) was larger?
4. Can you find the time at which the velocity of the car is zero again after starting from rest?
5. What is the car's displacement from the stop sign at \( t = 2 \) seconds and \( t = 4 \) seconds?

**Tip:** Always remember that the derivative of the position function gives you the velocity, and the derivative of the velocity function gives you the acceleration.

Learn how to calculate the velocity of a Honda Civic traveling in a straight line using its position function x(t) = At^2 - Bt^3. Detailed solutions provided for time intervals from t=0 to t=2s, t=0 to t=4s, and t=2s to t=4s. Explore the concepts of derivatives and quadratic functions in this math problem.

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