Acceleration, distance, velocity, time?

Question

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 4200 m/s^2 over a distance of 1.6 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, and how long did it take to reach that velocity?

(b) How high would the insect jump if air resistance could be ignored? Note that the actual height obtained is about 0.7 m, so air resistance is important here.

Answer 1

we know a = 4200 and we know x (displacement) = 1.6E-3 and we know initial velocity 0 (why?).  so we can apply

vf^2 =vi^2 +2ax

vf = sqrt(2ax)

For the time, again we know v =0 , so we can use

x = vit + 1/2at^2 or

x = 1/2at^2  or t = sqrt(2x/a)  plug and chug

b)  We assume the insect jumps vertically, we take as its initial velocity in freefall as the aswer in a sqrt(2ax) upwards.  While in free fall, it has an acceleration of g downwards.  Since these are in opposite directions, take acc = -g.  Finally at the top of its jump its velocity is ....zero (why?) so...

vf^2 = vi^2 + 2(-g)y

solve for y

y = (vf^2 - vi^2)/2y

but vi = sqrt 2ax = sqrt 2*4200*1.6E-3

plug and chug.  good luck!