Questions: The graph below shows the velocity versus time of a toy car moving across the floor. Which of the following best approximates the displacement of the car between 0 and 8 seconds? A 0 m B 10 m C -22 m D -32 m

The graph below shows the velocity versus time of a toy car moving across the floor.

Which of the following best approximates the displacement of the car between 0 and 8 seconds?

A
0 m

B
10 m

C
-22 m

D
-32 m
Transcript text: The graph below shows the velocity versus time of a toy car moving across the floor. Which of the following best approximates the displacement of the car between 0 and 8 seconds? A 0 m B 10 m C $-22 \mathrm{~m}$ D $-32 \mathrm{~m}$
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Solution

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Solution Steps

Step 1: Understanding the Problem

The problem requires us to find the displacement of a toy car between 0 and 8 seconds using the given velocity vs. time graph.

Step 2: Analyzing the Graph

The graph shows the velocity of the car over time. The area under the velocity-time graph represents the displacement. We need to calculate the area under the curve from 0 to 8 seconds.

Step 3: Calculating the Area
  1. From 0 to 2 seconds: The area is a triangle with a base of 2 seconds and a height of -4 m/s. \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times (-4) = -4 \, \text{m} \]

  2. From 2 to 4 seconds: The area is a triangle with a base of 2 seconds and a height of 4 m/s. \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4 \, \text{m} \]

  3. From 4 to 6 seconds: The area is a triangle with a base of 2 seconds and a height of 4 m/s. \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4 \, \text{m} \]

  4. From 6 to 8 seconds: The area is a triangle with a base of 2 seconds and a height of -4 m/s. \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times (-4) = -4 \, \text{m} \]

Step 4: Summing the Areas

Sum the areas calculated in the previous step to find the total displacement. \[ \text{Total Displacement} = -4 + 4 + 4 - 4 = 0 \, \text{m} \]

Final Answer

The displacement of the car between 0 and 8 seconds is: \[ \boxed{0 \, \text{m}} \]

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