A bush baby, an African primate, is capable of leaping vertically to the remarkable height of 2.3 m. To jump this high, the bush baby accelerates over a distance of 0.14m while extending the legs. The acceleration during the jump is approximately constant.
1)What is the acceleration during the pushing-off phase, in m/s2?
2)What is the acceleration during the pushing-off phase, in g's?
I honestly just don't know where to start. Ive tried several things but cant seem to get the right answer
To find the acceleration during the pushing-off phase, we can use the kinematic equation:
vf^2 = vi^2 + 2as
where:
vf = final velocity (0 m/s, since the bush baby stops at the highest point)
vi = initial velocity (which is also 0 m/s since the bush baby starts from rest)
a = acceleration
s = distance covered during the pushing-off phase
We are given:
s = 0.14 m
Rearranging the equation, we have:
a = (vf^2 - vi^2) / (2s)
Since vf = 0 m/s and vi = 0 m/s, the equation becomes:
a = (0 - 0) / (2s)
a = 0 / (2s)
a = 0 m/s^2
Therefore, the acceleration during the pushing-off phase is 0 m/s^2.
To calculate the acceleration in g's, we need to know the acceleration due to gravity, which is approximately 9.8 m/s^2.
We can convert the acceleration to g's using the following equation:
acceleration in g's = acceleration in m/s^2 / acceleration due to gravity
acceleration in g's = 0 m/s^2 / 9.8 m/s^2
acceleration in g's ≈ 0
Therefore, the acceleration during the pushing-off phase is approximately 0 g's.
To solve this problem, we can use the kinematic equation for uniformly accelerated motion:
\[d = \frac{1}{2}at^2\]
Where:
- \(d\) is the distance traveled during the jump (0.14m).
- \(a\) is the acceleration during the jump.
- \(t\) is the time taken for the jump.
Since we want to find the acceleration, we need to rearrange the equation. Solving for \(a\), we get:
\[a = \frac{2d}{t^2}\]
To find the time taken for the jump, we need to use the equation for vertical motion:
\[d = \frac{1}{2}gt^2\]
Where:
- \(g\) is the acceleration due to gravity (approximately 9.8 m/s²).
Rearranging this equation to solve for \(t^2\), we get:
\[t^2 = \frac{2d}{g}\]
Now we have the value for \(t^2\), we can substitute it back into the equation for acceleration:
\[a = \frac{2d}{(2d/g)} = g\]
1) The acceleration during the pushing-off phase is equal to the acceleration due to gravity, approximately 9.8 m/s².
2) To find the acceleration in g's, we divide the acceleration by the acceleration due to gravity:
\[a_{\text{g's}} = \frac{a}{g} = \frac{9.8 \, \text{m/s}^2}{9.8 \, \text{m/s}^2} = 1 \, \text{g}\]
Therefore, the acceleration during the pushing-off phase is equal to 1g.