It has been claimed that an insect called the froghopper (Philaenus spumarius) is the best jumper in the animal kingdom. This insect can accelerate at 4100 m/s2 over a distance of 1.7 mm as it straightens its specially designed "jumping legs."
(a) Assuming a uniform acceleration, what is the velocity of the insect after it has accelerated through this short distance? (m/s)
(b) how long did it take to reach that velocity? (ms)
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Verified answer
vf^2=vi^2+2adf
vi=0
vf^2=2(4100)(1.7)
vf^2=118.07 m/s
df=vit+.5at^2
1.7=2050t^2
t=.02880s or 28.8 ms
a= 4100ms-2
s=1.7mm=0.0017m
u=0
v^2=u^2+2as
v^2= 2*4100*0.0017
v^2= 13.94
v= 3.73ms-1
v=u+at
t= (v-u)/a
=3.73/4100
= 9.11*10^-4 s