An object's position is given by x = 09 m/s, 9 m/s -3 m/s, 3 m/s -6 m/s, 6 m/s O-10 m/s, 10 m/s 5 - 1². Att = 3 s, its velocity and speed are

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I'm unsure where to start in this proble. By comparing equations, does that mean Vo=5? How can we tell what acceleation is? Does the power rule allow us to make 1/2t^2 just 1t? I've attached images below

### 2.2 Speed and Velocity

An object's position is given by the equation:

\[ x = 5 - \frac{1}{2}t^2 \]

At \( t = 3 \) seconds, its velocity and speed are:

- \( 9 \, \text{m/s}, \ 9 \, \text{m/s} \)
- \(-3 \, \text{m/s}, \ 3 \, \text{m/s} \)
- \(-6 \, \text{m/s}, \ 6 \, \text{m/s} \)
- \(-10 \, \text{m/s}, \ 10 \, \text{m/s} \)

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Transcribed Image Text:### 2.2 Speed and Velocity An object's position is given by the equation: \[ x = 5 - \frac{1}{2}t^2 \] At \( t = 3 \) seconds, its velocity and speed are: - \( 9 \, \text{m/s}, \ 9 \, \text{m/s} \) - \(-3 \, \text{m/s}, \ 3 \, \text{m/s} \) - \(-6 \, \text{m/s}, \ 6 \, \text{m/s} \) - \(-10 \, \text{m/s}, \ 10 \, \text{m/s} \) Click "Save for Later" to continue later.
### Physics Problem: Object's Position and Velocity

**Problem #9: An object's position is given by the equation \( x = 5 - \frac{1}{2}t^2 \). At \( t = 3 \) seconds, find the velocity (V).**

#### Steps:

1. **Compare the given equation with the standard equation:**
   \[
   x = V_ot + \frac{1}{2}at^2
   \]
   Given: 
   \[
   x = 5 - \frac{1}{2}t^2
   \]

2. **Determine Initial Conditions:**
   \[
   V_0 = ?
   \]
   
#### Known:
   \[
   t = 3 \text{s}
   \]

3. **Calculate the Velocity (V):**
   \[
   V = \frac{\Delta D}{\Delta t}
   \]
   \[
   \text{Does this mean } V_0 = 5?
   \]
   \[
   \text{How do we find } \Delta D?
   \]

This set of equations and logical deductions will help to solve for the unknown initial velocity \((V_0)\) and understand the object's motion characteristics under given conditions. 

#### Explanation:

- The equation \( x = 5 - \frac{1}{2}t^2 \) is compared with the kinematic equation \( x = V_ot + \frac{1}{2}at^2 \) to deduce the parameters such as initial velocity (\(V_0\)) and acceleration.
- The problem involves differentiating and understanding each component to eventually determine the velocity at \( t = 3 \) seconds.

Make sure to apply calculus for finding derivative for velocity based on position-time function if needed and identify how changes in each parameter affect the overall motion of the object.
Transcribed Image Text:### Physics Problem: Object's Position and Velocity **Problem #9: An object's position is given by the equation \( x = 5 - \frac{1}{2}t^2 \). At \( t = 3 \) seconds, find the velocity (V).** #### Steps: 1. **Compare the given equation with the standard equation:** \[ x = V_ot + \frac{1}{2}at^2 \] Given: \[ x = 5 - \frac{1}{2}t^2 \] 2. **Determine Initial Conditions:** \[ V_0 = ? \] #### Known: \[ t = 3 \text{s} \] 3. **Calculate the Velocity (V):** \[ V = \frac{\Delta D}{\Delta t} \] \[ \text{Does this mean } V_0 = 5? \] \[ \text{How do we find } \Delta D? \] This set of equations and logical deductions will help to solve for the unknown initial velocity \((V_0)\) and understand the object's motion characteristics under given conditions. #### Explanation: - The equation \( x = 5 - \frac{1}{2}t^2 \) is compared with the kinematic equation \( x = V_ot + \frac{1}{2}at^2 \) to deduce the parameters such as initial velocity (\(V_0\)) and acceleration. - The problem involves differentiating and understanding each component to eventually determine the velocity at \( t = 3 \) seconds. Make sure to apply calculus for finding derivative for velocity based on position-time function if needed and identify how changes in each parameter affect the overall motion of the object.
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