Froghopper insects have a typical mass of around 11.1 mg 11.1 mg and can jump to a height of 55.3 cm. 55.3 cm. The takeoff velocity is achieved as the insect flexes its legs over a distance of approximately 2.00 mm. 2.00 mm. Assume that the jump is vertical and that the froghopper undergoes constant acceleration while its feet are in contact with the ground. Ignore air resistance. What is the acceleration of the insect during the time of the jump (before it leaves the ground)?

Acceleration is 26114.500m/s^2.

Explanation:

To answer this question Mathematically let's look at two equations of motion.

(1) Vf=vi=at

(2) y(t)=y0+v0t=(1/2)at^2

we can solve it in two steps

first we will basically solve for vi which requires us to solve for t as well.

second we will solve for acceleration which will also require us to solve for t as well.

in both cases we will use these two equtions of motion.

step (1) solving for vf and t.

frog jums 55.3cm , wth initial velocity = 0 that is two piecies of information in eq(1).  and (2),

that gives us t = 3.3seconds and vf = 32.32ms.

step (2).

solving for accelerations . similarly we now that from attains vf in some time while jumping 2.00mm. that two piecies of informations is enought and substituting in (1) and (2) gives.

a =  26114.500m/s^2  and t = 0.0012376seconds.

i don't understand. explain it clearly

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm