Question
A car accelerates from rest to a speed of 20 meters per second in 5 seconds. What is the car’s acceleration?
A.
2 meters per second squared
B.
3 meters per second squared
C.
4 meters per second squared
A.
2 meters per second squared
B.
3 meters per second squared
C.
4 meters per second squared
Answers
GPT-4o mini
To calculate the car's acceleration, we can use the formula for acceleration:
\[
a = \frac{\Delta v}{\Delta t}
\]
where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time.
Here:
- The initial velocity (\( v_0 \)) is 0 m/s (since the car starts from rest).
- The final velocity (\( v_f \)) is 20 m/s.
- The time interval (\( \Delta t \)) is 5 seconds.
The change in velocity (\( \Delta v \)) is:
\[
\Delta v = v_f - v_0 = 20 \, \text{m/s} - 0 \, \text{m/s} = 20 \, \text{m/s}
\]
Now we can plug the values into the acceleration formula:
\[
a = \frac{20 \, \text{m/s}}{5 \, \text{s}} = 4 \, \text{m/s}^2
\]
So, the car’s acceleration is \( 4 \, \text{m/s}^2 \).
The correct answer is:
**C. 4 meters per second squared**
\[
a = \frac{\Delta v}{\Delta t}
\]
where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time.
Here:
- The initial velocity (\( v_0 \)) is 0 m/s (since the car starts from rest).
- The final velocity (\( v_f \)) is 20 m/s.
- The time interval (\( \Delta t \)) is 5 seconds.
The change in velocity (\( \Delta v \)) is:
\[
\Delta v = v_f - v_0 = 20 \, \text{m/s} - 0 \, \text{m/s} = 20 \, \text{m/s}
\]
Now we can plug the values into the acceleration formula:
\[
a = \frac{20 \, \text{m/s}}{5 \, \text{s}} = 4 \, \text{m/s}^2
\]
So, the car’s acceleration is \( 4 \, \text{m/s}^2 \).
The correct answer is:
**C. 4 meters per second squared**