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
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
$-32 \mathrm{~m}$
Solution
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
From 0 to 2 seconds: The area is a triangle with a base of 2 seconds and a height of -4 m/s.
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times (-4) = -4 \, \text{m}
\]
From 2 to 4 seconds: The area is a triangle with a base of 2 seconds and a height of 4 m/s.
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4 \, \text{m}
\]
From 4 to 6 seconds: The area is a triangle with a base of 2 seconds and a height of 4 m/s.
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4 \, \text{m}
\]
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:
\[
\boxed{0 \, \text{m}}
\]