A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots" of the car's position at equal 0.5 s time intervals. (Assume the positive direction is to the right. Indicate the direction with the sign of your answer.) 1=0s 1=0.5 s 1= 1.0 s 1= 1.5 s 1=2.0 s 0m 0,1 m 0.4 m 0.9 m 1,6 m (a) What is the car's average velocity in the interval between t = 0,5 s to t= 1.0 s? 0.6 m/s (b) Using data from t = 0.5 s to t = 1.5 s, what is the car's acceleration at t = 1.0 s? 1.4 Recall that the definition of average acceleration is the change in velocity divided by the change in time. What is the initial velocity? What is the final velocity? Be careful about the change in time as well-if you measure velocities at "midpoints" within two separate intervals, what is the separation between those midpoints? m/s? (c) Is the car's speed increasing or decreastng with time? O increasing O decreasing O constant O not enough information

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### Analyzing Motion of a Remote Controlled Toy Car

A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots" of the car's position at equal 0.5 s time intervals. (Assume the positive direction is to the right. Indicate the direction with the sign of your answer.)

**Snapshot Positions:**
- At \( t = 0 \) s: Position = \( 0 \) m
- At \( t = 0.5 \) s: Position = \( 0.1 \) m
- At \( t = 1.0 \) s: Position = \( 0.4 \) m
- At \( t = 1.5 \) s: Position = \( 0.9 \) m
- At \( t = 2.0 \) s: Position = \( 1.6 \) m
   
### Questions and Solutions

#### (a) What is the car's average velocity in the interval between \( t = 0.5 \) s to \( t = 1.0 \) s?
**Answer:** \( 0.6 \) m/s ✔

**Explanation:**
To find the average velocity, we use:
\[ \text{Average velocity} = \frac{\text{Change in position}}{\text{Change in time}} \]
From \( t = 0.5 \) s to \( t = 1.0 \) s:
\[ \text{Change in position} = 0.4 \, \text{m} - 0.1 \, \text{m} = 0.3 \, \text{m} \]
\[ \text{Change in time} = 1.0 \, \text{s} - 0.5 \, \text{s} = 0.5 \, \text{s} \]
\[ \text{Average velocity} = \frac{0.3 \, \text{m}}{0.5 \, \text{s}} = 0.6 \, \text{m/s} \]

#### (b) Using data from \( t = 0.5 \) s to \( t = 1.5 \) s, what is the car's acceleration at \( t = 1.0 \) s?
**Answer:** \(
Transcribed Image Text:### Analyzing Motion of a Remote Controlled Toy Car A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots" of the car's position at equal 0.5 s time intervals. (Assume the positive direction is to the right. Indicate the direction with the sign of your answer.) **Snapshot Positions:** - At \( t = 0 \) s: Position = \( 0 \) m - At \( t = 0.5 \) s: Position = \( 0.1 \) m - At \( t = 1.0 \) s: Position = \( 0.4 \) m - At \( t = 1.5 \) s: Position = \( 0.9 \) m - At \( t = 2.0 \) s: Position = \( 1.6 \) m ### Questions and Solutions #### (a) What is the car's average velocity in the interval between \( t = 0.5 \) s to \( t = 1.0 \) s? **Answer:** \( 0.6 \) m/s ✔ **Explanation:** To find the average velocity, we use: \[ \text{Average velocity} = \frac{\text{Change in position}}{\text{Change in time}} \] From \( t = 0.5 \) s to \( t = 1.0 \) s: \[ \text{Change in position} = 0.4 \, \text{m} - 0.1 \, \text{m} = 0.3 \, \text{m} \] \[ \text{Change in time} = 1.0 \, \text{s} - 0.5 \, \text{s} = 0.5 \, \text{s} \] \[ \text{Average velocity} = \frac{0.3 \, \text{m}}{0.5 \, \text{s}} = 0.6 \, \text{m/s} \] #### (b) Using data from \( t = 0.5 \) s to \( t = 1.5 \) s, what is the car's acceleration at \( t = 1.0 \) s? **Answer:** \(
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