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The given problem describes the motion of an object with the position function \( s(t) = 1 - 13t + 7t^2 \), where \( s(t) \) is the distance from the origin at time \( t \).

### 1. Find the velocity \( v(t) \) of the object at any time \( t \):

To find the velocity, \( v(t) \), we need to take the derivative of the position function \( s(t) \) with respect to time \( t \). 

\[
v(t) = \frac{d}{dt} s(t)
\]

Given \( s(t) = 1 - 13t + 7t^2 \), differentiate term by term:

\[
v(t) = \frac{d}{dt}(1) - \frac{d}{dt}(13t) + \frac{d}{dt}(7t^2)
\]
\[
v(t) = 0 - 13 + 14t
\]
Thus, the velocity function is:

\[
v(t) = 14t - 13
\]

### 2. When is the object at rest?

The object is at rest when its velocity is zero. To find when the object is at rest, set \( v(t) = 0 \):

\[
14t - 13 = 0
\]
\[
14t = 13
\]
\[
t = \frac{13}{14}
\]

Thus, the object is at rest at \( t = \frac{13}{14} \) seconds.

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Would you like more details on these steps or have any other questions?

Here are 5 related questions to explore further:
1. What is the acceleration of the object at any time \( t \)?
2. What is the object's position at \( t = 0 \) and \( t = 1 \)?
3. How far does the object travel by the time it comes to rest?
4. What is the total displacement between \( t = 0 \) and \( t = 2 \)?
5. What type of motion (speeding up or slowing down) is the object experiencing before it comes to rest?

**Tip:** The velocity is the derivative of position, and acceleration is the derivative of velocity, which helps you analyze how an object moves over time.

The function s(t) = 1 - 13t + 7t^2 describes the distance s from the origin at time t of an object in rectilinear motion. Find the velocity v of the object at any time t. When is the object at rest?

Learn how to find the velocity function and determine when the object is at rest, given the position function s(t) = 1 - 13t + 7t^2. This calculus problem involves finding derivatives and solving for when the object’s velocity is zero.

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