For the jumping flea that reaches a takeoff speed of 1.0 m/s over a distance of 0.50 mm, we have:
a) The flea's acceleration as it extends its legs is 1000 m/s².
b) It takes 1.0x10⁻³ s for the flea to leave the ground after it begins pushing off.
a) The flea's acceleration as it extends its legs is the following:
[tex] v_{f}^{2} = v_{i}^{2} + 2ad [/tex]
Where:
[tex] v_{f}[/tex]: is the final speed = 1.0 m/s
[tex] v_{i}[/tex]: is the initial speed = 0
a: is the acceleration =?
d: is the distance = 0.50 mm
The acceleration is:
[tex] a = \frac{v_{f}^{2}}{2d} = \frac{(1.0 m/s)^{2}}{2*0.50 \cdot 10^{-3} m} = 1000 m/s^{2} [/tex]
Hence, the flea's acceleration is 1000 m/s².
b) The time can be calculated with the next equation:
[tex] v_{f} = v_{i} + at [/tex]
Where:
t: is the time =?
Then, the time is:
[tex] t = \frac{v_{f}}{a} = \frac{1.0 m/s}{1000 m/s^{2}} = 1.0 \cdot 10^{-3} s [/tex]
Therefore, it takes 1.0 ms for the flea to leave the ground.
You can find another example of acceleration calculation here: https://brainly.com/question/559318?referrer=searchResults
I hope it helps you!
