Certain insects can achieve seemingly impossible accelerations while jumping. The click beetle accelerates at an astonishing 400

Question

3 years ago

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Certain insects can achieve seemingly impossible accelerations while jumping. The click beetle accelerates at an astonishing 400

g over a distance of 0.58 cm as it rapidly bends its thorax, making the "click" that gives it its name.
A. Assuming the beetle jumps straight up, at what speed does it leave the ground?
B. How much time is required for the beetle to reach this speed?
C. Ignoring air resistance, how high would it go?

Physics

1 answer:

olasank [31] · Answer 1 · 2022-04-19T01:46:01+03:00

A) Speed at which the insect leaves the ground: 6.7 m/s

B) Time needed: 1.7 ms

C) Maximum height reached: 2.3 m

Explanation:

A)

The motion of the click beetle is a uniformly accelerated motion, so we can find the speed at which it leaves the ground by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

During the phase of acceleration, we have:

u = 0 (it starts from rest)

$a=400g=400(9.8)=3920 m/s^2$ (acceleration)

s = 0.58 cm = 0.0058 m is the distance covered

Solving for v, we find the velocity at which the insect leaves the ground:

$v=\sqrt{2as}=\sqrt{2(3920)(0.0058)}=6.7 m/s$

B)

The time needed for the beetle to reach this speed can be found by using another suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this case, we have

u = 0

v = 6.7 m/s (found in part a)

$a=3920 m/s^2$

Therefore, the time needed is

$t=\frac{v-u}{a}=\frac{6.7-0}{3920}=1.7\cdot 10^{-3} s = 1.7 ms$

C)

In order to solve this part, we can apply the law of conservation of energy: in fact, the total mechanical energy of the beetle is conserved, therefore the initial kinetic energy is entirely converted into potential energy as the insect reaches the maximum height. So we can write

$K_i = U_f = \frac{1}{2}mv^2=mgh$