Math Problem Statement

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

.

Solution

To find the total displacement from t1=1.5st_1 = 1.5 \, \text{s} to t2=3.1st_2 = 3.1 \, \text{s} given the velocity function v(t)=29+20tv(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+20tv(t) = 29 + 20t

  2. Displacement Formula Using Calculus:

    The displacement ss over a time interval [t1,t2][t_1, t_2] can be calculated using the definite integral of the velocity function:

    s=t1t2v(t)dts = \int_{t_1}^{t_2} v(t) \, dt

  3. Set Up the Integral:

    Substitute the given values for t1t_1 and t2t_2:

    s=1.53.1(29+20t)dts = \int_{1.5}^{3.1} (29 + 20t) \, dt

  4. Integrate the Velocity Function:

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

    (29+20t)dt=29dt+20tdt\int (29 + 20t) \, dt = \int 29 \, dt + \int 20t \, dt

    The integrals of the individual terms are:

    29dt=29t\int 29 \, dt = 29t 20tdt=10t2\int 20t \, dt = 10t^2

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

    F(t)=29t+10t2F(t) = 29t + 10t^2

  5. Evaluate the Integral from t1=1.5t_1 = 1.5 to t2=3.1t_2 = 3.1:

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

    s=F(3.1)F(1.5)s = F(3.1) - F(1.5)

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

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

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

  6. Calculate the Displacement:

    s=18666=120s = 186 - 66 = 120

So, the total displacement from t1=1.5st_1 = 1.5 \, \text{s} to t2=3.1st_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+5t2v(t) = 10 + 5t^2?
  2. How does the acceleration relate to the velocity function v(t)=29+20tv(t) = 29 + 20t?
  3. What is the instantaneous velocity at t=2st = 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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration

Formulas

Definite integral

Theorems

-

Suitable Grade Level

College