SOMEONE ASKED ๐
When jumping, a flea rapidly extends its legs, reaching a takeoff speed of 1.0 m/s over a distance of 0.50mm .?
HERE THE ANSWERS ๐

Assuming lift off as soon as the legs are fully extended over the distance S = .5E3 m and assuming the acceleration A is constant, then the Vavg = V/2 = .5 m/s when V = 1 mps take off speed.
So from S = Vavg T; we have T = S/Vavg = .5E3/.5 = 1E3 seconds or 1 millisecond. ANS.

For 0.5mm = 5e4m, the flea is accelerating at a constant rate. At d = 5e4 V(t) = 1m/s
We want to know when the flea is at d = 5e4 and V(t) = 1
V(t) = a*t = 1
5e4 = 1/2 at^2 => 10e4 = at^2 but we know that at = 1
10e4 = 0.001 = att = 1t = t so t = 0.001 = 1millisecond = 1ms

If we assume the takeoff speed is constant, then v = d/t.
1 = 0.0005 / t
t = 0.0005 s
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