Question
Two stones are thrown from the same point at the same time one vertically upwards with a speed of 40ms^-1 and the other downwards with a speed of 40ms^-1. How far are the stones after 2 seconds
Answers
GPT-4o mini
To find the positions of the two stones after 2 seconds, we can use the following kinematic equation:
\[
s = ut + \frac{1}{2}at^2
\]
where:
- \(s\) is the displacement,
- \(u\) is the initial velocity,
- \(a\) is the acceleration (for free-fall, this is approximated as \(-9.81 \, \text{m/s}^2\)),
- \(t\) is the time in seconds.
### For the stone thrown upwards:
- Initial velocity \(u = 40 \, \text{m/s}\)
- Acceleration \(a = -9.81 \, \text{m/s}^2\)
- Time \(t = 2 \, \text{s}\)
Substituting the values into the formula:
\[
s_{\text{up}} = 40(2) + \frac{1}{2}(-9.81)(2^2)
\]
\[
s_{\text{up}} = 80 + \frac{1}{2}(-9.81)(4)
\]
\[
s_{\text{up}} = 80 - 19.62
\]
\[
s_{\text{up}} = 60.38 \, \text{meters upwards}
\]
### For the stone thrown downwards:
- Initial velocity \(u = -40 \, \text{m/s}\) (since it is thrown downwards, we take it as negative)
- Acceleration \(a = -9.81 \, \text{m/s}^2\)
- Time \(t = 2 \, \text{s}\)
Substituting the values into the formula:
\[
s_{\text{down}} = -40(2) + \frac{1}{2}(-9.81)(2^2)
\]
\[
s_{\text{down}} = -80 + \frac{1}{2}(-9.81)(4)
\]
\[
s_{\text{down}} = -80 - 19.62
\]
\[
s_{\text{down}} = -99.62 \, \text{meters downwards}
\]
### Distance between the two stones after 2 seconds:
Now, to find the distance between the two stones, we need to take the absolute difference of their positions.
The upward stone is at \(60.38 \, \text{m}\) and the downward stone is at \(-99.62 \, \text{m}\).
To find the distance between them:
\[
\text{Distance} = |s_{\text{up}} - s_{\text{down}}| = |60.38 - (-99.62)|
\]
\[
\text{Distance} = |60.38 + 99.62| = |160|
\]
\[
\text{Distance} = 160 \, \text{meters}
\]
Therefore, after 2 seconds, the stones are **160 meters apart**.
\[
s = ut + \frac{1}{2}at^2
\]
where:
- \(s\) is the displacement,
- \(u\) is the initial velocity,
- \(a\) is the acceleration (for free-fall, this is approximated as \(-9.81 \, \text{m/s}^2\)),
- \(t\) is the time in seconds.
### For the stone thrown upwards:
- Initial velocity \(u = 40 \, \text{m/s}\)
- Acceleration \(a = -9.81 \, \text{m/s}^2\)
- Time \(t = 2 \, \text{s}\)
Substituting the values into the formula:
\[
s_{\text{up}} = 40(2) + \frac{1}{2}(-9.81)(2^2)
\]
\[
s_{\text{up}} = 80 + \frac{1}{2}(-9.81)(4)
\]
\[
s_{\text{up}} = 80 - 19.62
\]
\[
s_{\text{up}} = 60.38 \, \text{meters upwards}
\]
### For the stone thrown downwards:
- Initial velocity \(u = -40 \, \text{m/s}\) (since it is thrown downwards, we take it as negative)
- Acceleration \(a = -9.81 \, \text{m/s}^2\)
- Time \(t = 2 \, \text{s}\)
Substituting the values into the formula:
\[
s_{\text{down}} = -40(2) + \frac{1}{2}(-9.81)(2^2)
\]
\[
s_{\text{down}} = -80 + \frac{1}{2}(-9.81)(4)
\]
\[
s_{\text{down}} = -80 - 19.62
\]
\[
s_{\text{down}} = -99.62 \, \text{meters downwards}
\]
### Distance between the two stones after 2 seconds:
Now, to find the distance between the two stones, we need to take the absolute difference of their positions.
The upward stone is at \(60.38 \, \text{m}\) and the downward stone is at \(-99.62 \, \text{m}\).
To find the distance between them:
\[
\text{Distance} = |s_{\text{up}} - s_{\text{down}}| = |60.38 - (-99.62)|
\]
\[
\text{Distance} = |60.38 + 99.62| = |160|
\]
\[
\text{Distance} = 160 \, \text{meters}
\]
Therefore, after 2 seconds, the stones are **160 meters apart**.