MATH SOLVE

5 months ago

Q:
# When jumping, a flea rapidly extends its legs, reaching a takeoff speed of 1.0 m/s over a distance of 0.50 mm . part a what is the flea's acceleration as it extends its legs?

Accepted Solution

A:

The flea's acceleration is 1,000 m/s².Further explanationThe problem we overlook this time is about uniformly accelerated motion.The list of variables to be considered is as follows.[tex]\boxed{u \ or \ v_i = initial \ velocity}[/tex][tex]\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}[/tex][tex]\boxed{a = acceleration \ (constant)}[/tex][tex]\boxed{d = distance \ travelled}[/tex]Given that a take-off speed of 1.0 m/s over a distance of 0.50 mm. The initial speed was zero from rest. It was asked about flea's acceleration when she extends her legs. The formula we follow for this problem are as follows:[tex]\boxed{ \ v^2 = u^2 + 2ad \ }[/tex]a = acceleration (in m/s²)u = initial velocity = 0 m/sv = take off speed = 1.0 m/sd = distance travelled = 0.50 mmConverted 0.50 mm to meters: [tex]\boxed{0.50 \ mm = 0.50 \times 10^{-3} \ m = 5.0 \times 10^{-4} \ m}[/tex]Steps:We set the formula so that the acceleration becomes the subject.[tex]\boxed{ \ v^2 = u^2 + 2ad \ }[/tex][tex]\boxed{ \ v^2 - u^2 = 2ad \ }[/tex][tex]\boxed{ \ 2ad = v^2 - u^2 \ }[/tex][tex]\boxed{ \ a = \frac{v^2 - u^2}{2d} \ }[/tex]Substitute all data into equations.[tex]\boxed{ \ a = \frac{(1.0)^2 - (0)^2}{2(5.0 \times 10^{-4})} \ }[/tex][tex]\boxed{ \ a = \frac{1}{10 \times 10^{-4}} \ }[/tex][tex]\boxed{ \ a = \frac{1}{1 \times 10^{-3}} \ }[/tex][tex]\boxed{ \ a = 1 \times 10^{3}\ = 1,000 \ m/s^2 \ }[/tex]We get the flea's acceleration, i.e., 1,000 m/s².Learn moreAbout scientific notation the mass of substance in cubic : flea, when jumping, rapidly extends its legs, take-off speed, initial velocity, distance, acceleration, uniformly accelerated motion, convert