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:
[tex]v^2-u^2=2as[/tex]
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)
[tex]a=400g=400(9.8)=3920 m/s^2[/tex] (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:
[tex]v=\sqrt{2as}=\sqrt{2(3920)(0.0058)}=6.7 m/s[/tex]
B)
The time needed for the beetle to reach this speed can be found by using another suvat equation:
[tex]v=u+at[/tex]
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)
[tex]a=3920 m/s^2[/tex]
Therefore, the time needed is
[tex]t=\frac{v-u}{a}=\frac{6.7-0}{3920}=1.7\cdot 10^{-3} s = 1.7 ms[/tex]
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
[tex]K_i = U_f = \frac{1}{2}mv^2=mgh[/tex]
where
m is the mass of the beetle
h is the maximum height reached
g is the acceleration due to gravity
v is the initial speed
Re-arranging the equation,
[tex]h=\frac{v^2}{2g}[/tex]
And substituting
v = 6.7 m/s
[tex]g=9.8 m/s^2[/tex]
We find
[tex]h=\frac{6.7^2}{2(9.8)}=2.3 m[/tex]
Learn more about accelerated motion:
brainly.com/question/9527152
brainly.com/question/11181826
brainly.com/question/2506873
brainly.com/question/2562700
#LearnwithBrainly
