An object moves along a straight path at a speed Vt = 3t + 4t^2 in m/s, where Vt = velocity in m/s and t = time in seconds. Find the distance traveled after 2.73seconds.

To find the distance traveled by the object after 2.73 seconds, given the velocity function \( V(t) = 3t + 4t^2 \) in meters per second, we need to integrate the velocity function with respect to time. The distance traveled, \( S(t) \), can be found by computing the definite integral of the velocity function over the interval from \( t = 0 \) to \( t = 2.73 \) seconds.

### 1. Velocity Function
The given velocity function is:

\[
V(t) = 3t + 4t^2
\]

### 2. Distance Function
The distance traveled is the integral of the velocity function:

\[
S(t) = \int V(t) \, dt = \int (3t + 4t^2) \, dt
\]

### 3. Perform the Integration
Let's integrate the function:

\[
S(t) = \int 3t \, dt + \int 4t^2 \, dt
\]

\[
S(t) = \left(\frac{3t^2}{2}\right) + \left(\frac{4t^3}{3}\right) + C
\]

Since we're calculating the definite integral from \( t = 0 \) to \( t = 2.73 \) seconds, we do not need the constant \( C \). The distance traveled after \( t = 2.73 \) seconds is:

\[
S(2.73) = \left[\frac{3t^2}{2} + \frac{4t^3}{3}\right]_{0}^{2.73}
\]

### 4. Evaluate the Distance
Substituting \( t = 2.73 \) into the expression:

\[
S(2.73) = \frac{3(2.73)^2}{2} + \frac{4(2.73)^3}{3}
\]

Let's calculate this step-by-step:

\[
\text{First term: } \frac{3(2.73)^2}{2} = \frac{3 \times 7.4529}{2} = \frac{22.3587}{2} = 11.17935 \, \text{m}
\]

\[
\text{Second term: } \frac{4(2.73)^3}{3} = \frac{4 \times 20.3571}{3} = \frac{81.4284}{3} = 27.1428 \, \text{m}
\]

### 5. Add the Two Results
Finally, adding the two results:

\[
S(2.73) = 11.17935 + 27.1428 = 38.32215 \, \text{m}
\]

### Conclusion
The distance traveled by the object after 2.73 seconds is approximately **38.32 meters**.

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Would you like any further details, or do you have any questions?

Here are 8 follow-up questions:
1. How would the result change if the time were 5 seconds instead of 2.73 seconds?
2. What is the significance of integrating the velocity function?
3. Can you determine the average speed over the interval from \( t = 0 \) to \( t = 2.73 \)?
4. What would happen if the velocity function was more complex?
5. How would you find the displacement if the velocity function included negative values?
6. How does calculus help in understanding motion in physics?
7. Can this method be applied to non-linear velocity functions?
8. What if the initial velocity was not zero? How would that affect the distance calculation?

**Tip:** When integrating a velocity function, remember that the result gives the total distance traveled, regardless of the path's direction.

Learn how to calculate the distance traveled by an object moving with velocity function V(t) = 3t + 4t^2 after 2.73 seconds. This problem involves integrating the velocity function and applying definite integration to find the distance. Suitable for college-level students and physics enthusiasts.

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