Santiago Tran
02/14/2024 · Senior High School
Question 3 (Mandatory) (1 point) Over a period of 3.0 s a car's velocity changes from \( 18 \mathrm{~m} / \mathrm{s}[\mathrm{W}] \) to \( 12 \mathrm{~m} / \mathrm{s} \) [W]. What is the value of the car's acceleration during this time? \( 2.0 \mathrm{~m} / \mathrm{s} \) [E] \( 10 \mathrm{~m} / \mathrm{s} / \mathrm{s} \) [W] \( 2.0 \mathrm{~m} / \mathrm{s} / \mathrm{s} \) [W] \( 2.0 \mathrm{~m} / \mathrm{s} / \mathrm{s} \) [E]
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Step-by-step Solution
To find the acceleration of the car, we can use the formula for acceleration:
\[ \text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time}} \]
Given:
- Initial velocity (\( v_i \)) = \( 18 \, \text{m/s} \) [W]
- Final velocity (\( v_f \)) = \( 12 \, \text{m/s} \) [W]
- Time (\( t \)) = 3.0 s
We can calculate the change in velocity as:
\[ \text{Change in velocity} = v_f - v_i \]
Substitute the given values:
\[ \text{Change in velocity} = 12 \, \text{m/s} - 18 \, \text{m/s} \]
Now, we can calculate the acceleration using the formula:
\[ \text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time}} \]
Let's calculate the acceleration.
Calculate the value by following steps:
- step0: Calculate:
\(\frac{\left(12-18\right)}{3}\)
- step1: Remove the parentheses:
\(\frac{12-18}{3}\)
- step2: Subtract the numbers:
\(\frac{-6}{3}\)
- step3: Divide the terms:
\(-2\)
The acceleration of the car during this time is \( -2 \, \text{m/s}^2 \) [E].
Therefore, the correct answer is \( 2.0 \, \text{m/s}^2 \) [E].
Quick Answer
The acceleration of the car is \( 2.0 \, \text{m/s}^2 \) [E].
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