An object is moving along a horizontal line such that it's velocity is V(t) : = What is the object's displacement between t = 0 and t = 5? O 1.558 O 1.944 2.558 O 2.944 3 (3t+4)*

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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### Problem Description

An object is moving along a horizontal line with its velocity given by the equation \( V(t) = \frac{3}{(3t + 4)} \).

### Question

What is the object's displacement between \( t = 0 \) and \( t = 5 \)?

- \( 1.558 \)
- \( 1.944 \)
- \( 2.558 \)
- \( 2.944 \)

### Explanation

To find the object's displacement over the given time interval, you need to integrate the velocity function \( V(t) \) with respect to time \( t \) from \( t = 0 \) to \( t = 5 \).

The displacement \( s \) can be found as follows:
\[ s = \int_0^5 V(t) \, dt = \int_0^5 \frac{3}{3t + 4} \, dt \]

Use integration techniques to solve this definite integral. The correct answer will correspond to one of the provided options.
Transcribed Image Text:### Problem Description An object is moving along a horizontal line with its velocity given by the equation \( V(t) = \frac{3}{(3t + 4)} \). ### Question What is the object's displacement between \( t = 0 \) and \( t = 5 \)? - \( 1.558 \) - \( 1.944 \) - \( 2.558 \) - \( 2.944 \) ### Explanation To find the object's displacement over the given time interval, you need to integrate the velocity function \( V(t) \) with respect to time \( t \) from \( t = 0 \) to \( t = 5 \). The displacement \( s \) can be found as follows: \[ s = \int_0^5 V(t) \, dt = \int_0^5 \frac{3}{3t + 4} \, dt \] Use integration techniques to solve this definite integral. The correct answer will correspond to one of the provided options.
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