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An object originally at rest is uniformly accelerated to a speed of 8.0 meters per second in 2.0 seconds. What is the accelerat...Question
An object originally at rest is uniformly accelerated to a speed of 8.0 meters per second in 2.0 seconds. What is the acceleration of the object?
\large acceleration=\frac{v_{final}-v_{initial}}{time}
a
16 m/s2
b
4.0 m/s2
c
0.25 m/s2
d
10. m/s2
\large acceleration=\frac{v_{final}-v_{initial}}{time}
a
16 m/s2
b
4.0 m/s2
c
0.25 m/s2
d
10. m/s2
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Answered by
GPT-4o mini
To calculate the acceleration of the object, we can use the formula for acceleration:
\[
\text{acceleration} = \frac{v_{final} - v_{initial}}{time}
\]
Given:
- \( v_{final} = 8.0 \, \text{m/s} \)
- \( v_{initial} = 0 \, \text{m/s} \) (the object is originally at rest)
- \( time = 2.0 \, \text{s} \)
Now substituting the values into the formula:
\[
\text{acceleration} = \frac{8.0 \, \text{m/s} - 0 \, \text{m/s}}{2.0 \, \text{s}} = \frac{8.0 \, \text{m/s}}{2.0 \, \text{s}} = 4.0 \, \text{m/s}^2
\]
Thus, the acceleration of the object is \( 4.0 \, \text{m/s}^2 \).
The correct answer is:
**b) 4.0 m/s²**.
\[
\text{acceleration} = \frac{v_{final} - v_{initial}}{time}
\]
Given:
- \( v_{final} = 8.0 \, \text{m/s} \)
- \( v_{initial} = 0 \, \text{m/s} \) (the object is originally at rest)
- \( time = 2.0 \, \text{s} \)
Now substituting the values into the formula:
\[
\text{acceleration} = \frac{8.0 \, \text{m/s} - 0 \, \text{m/s}}{2.0 \, \text{s}} = \frac{8.0 \, \text{m/s}}{2.0 \, \text{s}} = 4.0 \, \text{m/s}^2
\]
Thus, the acceleration of the object is \( 4.0 \, \text{m/s}^2 \).
The correct answer is:
**b) 4.0 m/s²**.
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