Colon Greene
04/27/2023 · Senior High School
Question 10 (Mandatory) ( 1 point) A car is driving along the highway behind a slower vehicle when it pulls out to pass. If the car's acceleration is uniform at \( 2.0 \mathrm{~m} / \mathrm{s} / \mathrm{s} \) for 4.0 s and it reaches a speed of 28 \( \mathrm{~m} / \mathrm{s} \), what was its speed when it first pulled out to pass the slower vehicle? \( 22 \mathrm{~m} / \mathrm{s} \) \( 24 \mathrm{~m} / \mathrm{s} \) \( 20 \mathrm{~m} / \mathrm{s} \) \( 18 \mathrm{~m} / \mathrm{s} \)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To find the initial speed of the car when it first pulled out to pass, we can use the formula for acceleration:
\[
v_f = v_i + a \cdot t
\]
where:
- \( v_f \) is the final speed (28 m/s),
- \( v_i \) is the initial speed (which we want to find),
- \( a \) is the acceleration (2.0 m/s²),
- \( t \) is the time (4.0 s).
Rearranging the formula to solve for \( v_i \):
\[
v_i = v_f - a \cdot t
\]
Now, substituting the known values:
\[
v_i = 28 \, \text{m/s} - (2.0 \, \text{m/s}^2 \cdot 4.0 \, \text{s})
\]
Calculating the acceleration term:
\[
2.0 \, \text{m/s}^2 \cdot 4.0 \, \text{s} = 8.0 \, \text{m/s}
\]
Now substituting back into the equation for \( v_i \):
\[
v_i = 28 \, \text{m/s} - 8.0 \, \text{m/s} = 20 \, \text{m/s}
\]
Thus, the speed of the car when it first pulled out to pass the slower vehicle was \( 20 \, \text{m/s} \).
The correct answer is:
**\( 20 \mathrm{~m/s} \)**
Quick Answer
The initial speed of the car was \( 20 \mathrm{~m/s} \).
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