Given 

v(t)

v

(

t

)

 = 29 + 20 

t

t

, where 

v

v

 is in 

m/s

m

/

s

 and 

t

t

 is in 

s

s

, use calculus to determine the total displacement from 

t

1

t

1

 = 1.5 

s

s

 to 

t

2

t

2

 = 3.1 

s

s

.

To find the total displacement from \( t_1 = 1.5 \, \text{s} \) to \( t_2 = 3.1 \, \text{s} \) given the velocity function \( v(t) = 29 + 20t \), we need to integrate the velocity function over this time interval. 

### Step-by-Step Solution

1. **Given Velocity Function:**

   \[
   v(t) = 29 + 20t
   \]

2. **Displacement Formula Using Calculus:**

   The displacement \( s \) over a time interval \([t_1, t_2]\) can be calculated using the definite integral of the velocity function:

   \[
   s = \int_{t_1}^{t_2} v(t) \, dt
   \]

3. **Set Up the Integral:**

   Substitute the given values for \( t_1 \) and \( t_2 \):

   \[
   s = \int_{1.5}^{3.1} (29 + 20t) \, dt
   \]

4. **Integrate the Velocity Function:**

   To integrate, apply the rules of integration to each term separately:

   \[
   \int (29 + 20t) \, dt = \int 29 \, dt + \int 20t \, dt
   \]

   The integrals of the individual terms are:

   \[
   \int 29 \, dt = 29t
   \]
   \[
   \int 20t \, dt = 10t^2
   \]

   Therefore, the antiderivative \( F(t) \) of \( v(t) = 29 + 20t \) is:

   \[
   F(t) = 29t + 10t^2
   \]

5. **Evaluate the Integral from \( t_1 = 1.5 \) to \( t_2 = 3.1 \):**

   Calculate the definite integral by evaluating the antiderivative at the upper and lower limits:

   \[
   s = F(3.1) - F(1.5)
   \]

   Compute \( F(3.1) \) and \( F(1.5) \):

   \[
   F(3.1) = 29(3.1) + 10(3.1)^2
   \]
   \[
   F(3.1) = 89.9 + 96.1 = 186
   \]

   \[
   F(1.5) = 29(1.5) + 10(1.5)^2
   \]
   \[
   F(1.5) = 43.5 + 22.5 = 66
   \]

6. **Calculate the Displacement:**

   \[
   s = 186 - 66 = 120
   \]

So, the total displacement from \( t_1 = 1.5 \, \text{s} \) to \( t_2 = 3.1 \, \text{s} \) is **120 meters**.

Would you like further details on this process or have any additional questions?

Here are some related questions to further explore:

1. What would the displacement be if the velocity function were \( v(t) = 10 + 5t^2 \)?
2. How does the acceleration relate to the velocity function \( v(t) = 29 + 20t \)?
3. What is the instantaneous velocity at \( t = 2 \, \text{s} \)?
4. How do you find the total distance traveled if the velocity function changes direction?
5. What are the units of displacement when integrating a velocity function?

**Tip:** When integrating velocity to find displacement, ensure you understand the difference between displacement (net change in position) and distance traveled (total path length), especially if the velocity changes direction.

Learn how to calculate the total displacement from t1 = 1.5 s to t2 = 3.1 s using the velocity function v(t) = 29 + 20t. This problem involves integrating the velocity function over the specified time interval and computing the resultant displacement. Suitable for college-level students studying calculus.

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