Question
An object starts from rest with an acceleration of 10 m/s2. How far does it go in 0.5 s?
*
1 point
1.25 m
2.5 m
0 m
*
1 point
1.25 m
2.5 m
0 m
Answers
GPT-4o mini
To calculate the distance traveled by an object starting from rest under constant acceleration, you can use the following kinematic equation:
\[
d = v_i t + \frac{1}{2} a t^2
\]
where:
- \(d\) is the distance,
- \(v_i\) is the initial velocity (0 m/s since it starts from rest),
- \(t\) is the time (0.5 s),
- \(a\) is the acceleration (10 m/s²).
Since the object starts from rest, \(v_i = 0\). Thus, the equation simplifies to:
\[
d = \frac{1}{2} a t^2
\]
Substituting the values into the equation:
\[
d = \frac{1}{2} \times 10 \, \text{m/s}^2 \times (0.5 \, \text{s})^2
\]
\[
d = \frac{1}{2} \times 10 \times 0.25
\]
\[
d = \frac{1}{2} \times 2.5
\]
\[
d = 1.25 \, \text{m}
\]
So, the correct answer is **1.25 m**.
\[
d = v_i t + \frac{1}{2} a t^2
\]
where:
- \(d\) is the distance,
- \(v_i\) is the initial velocity (0 m/s since it starts from rest),
- \(t\) is the time (0.5 s),
- \(a\) is the acceleration (10 m/s²).
Since the object starts from rest, \(v_i = 0\). Thus, the equation simplifies to:
\[
d = \frac{1}{2} a t^2
\]
Substituting the values into the equation:
\[
d = \frac{1}{2} \times 10 \, \text{m/s}^2 \times (0.5 \, \text{s})^2
\]
\[
d = \frac{1}{2} \times 10 \times 0.25
\]
\[
d = \frac{1}{2} \times 2.5
\]
\[
d = 1.25 \, \text{m}
\]
So, the correct answer is **1.25 m**.