Please explain how to solve, thank you! **Calculating Position from Velocity using Integration**Suppose an object moves along a linear path according to the velocity function \( v(t) = 3t^2 - 2t \), where \( v(t) \) is in meters per second.Given:If the object’s position at \( t = 1 \) is \( s = 5 \), what is the function in terms of \( t \) that represents the position at any time \( t \)?**Options:**1. \( \int_{1}^{5} \left( 3u^2 - 2u \right) du \)2. \( 1 + \int_{1}^{5} \left( 3u^2 - 2u \right) du \)3. \( \boxed{5 - \int_{1}^{t} \left( 3u^2 - 2u \right) du \text{ (Highlighted as the correct option) }} \)4. \( 5 + \int_{1}^{t} \left( 3u^2 - 2u \right) du \)### Explanation:To find the object's position \( s(t) \) at any time \( t \), we need to integrate the velocity function \( v(u) = 3u^2 - 2u \) with respect to \( u \).Since the position at \( t = 1 \) is given as \( s(1) = 5 \), we can set up the integral to find the position \( s \) at any time \( t \) as follows:\[ s(t) = s(1) + \int_{1}^{t} v(u) \, du \]Substituting the given values and function into the equation:\[ s(t) = 5 + \int_{1}^{t} \left( 3u^2 - 2u \right) du \]Thus, the correct expression representing the function in terms of \( t \) for the position is:\[ \boxed{5 - \int_{1}^{t} \left( 3u^2 - 2u \right) du} \]This is the function that represents the position \( s(t) \) of the object at any time \( t \).

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Suppose an object moves along a linear path according to the velocity function v(t) = 3x² - 2x, where v(t) is in meters per second. If the object's position at t = 1 is s = 5, what is the function in terms of t that represents the position at any time t? O 1,³ (3u² – 2u) du O 1+5 (3u² - 2u) du • Si 5- (3u²-2u) du 0 5+ (3u² - 2u) du

Suppose an object moves along a linear path according to the velocity function v(t) = 3x² - 2x, where v(t) is in meters per second. If the object's position at t = 1 is s = 5, what is the function in terms of t that represents the position at any time t? O 1,³ (3u² – 2u) du O 1+5 (3u² - 2u) du • Si 5- (3u²-2u) du 0 5+ (3u² - 2u) du

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Please explain how to solve, thank you!

**Calculating Position from Velocity using Integration**

Suppose an object moves along a linear path according to the velocity function \( v(t) = 3t^2 - 2t \), where \( v(t) \) is in meters per second.

Given:
If the object’s position at \( t = 1 \) is \( s = 5 \), what is the function in terms of \( t \) that represents the position at any time \( t \)?

**Options:**

1. \( \int_{1}^{5} \left( 3u^2 - 2u \right) du \)
2. \( 1 + \int_{1}^{5} \left( 3u^2 - 2u \right) du \)
3. \( \boxed{5 - \int_{1}^{t} \left( 3u^2 - 2u \right) du \text{ (Highlighted as the correct option) }} \)
4. \( 5 + \int_{1}^{t} \left( 3u^2 - 2u \right) du \)

### Explanation:

To find the object's position \( s(t) \) at any time \( t \), we need to integrate the velocity function \( v(u) = 3u^2 - 2u \) with respect to \( u \).

Since the position at \( t = 1 \) is given as \( s(1) = 5 \), we can set up the integral to find the position \( s \) at any time \( t \) as follows:

\[ s(t) = s(1) + \int_{1}^{t} v(u) \, du \]

Substituting the given values and function into the equation:

\[ s(t) = 5 + \int_{1}^{t} \left( 3u^2 - 2u \right) du \]

Thus, the correct expression representing the function in terms of \( t \) for the position is:

\[ \boxed{5 - \int_{1}^{t} \left( 3u^2 - 2u \right) du} \]

This is the function that represents the position \( s(t) \) of the object at any time \( t \).

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