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A typical sneeze has a maximum speed of 45 m/s. Suppose the material emitted in the...

A typical sneeze has a maximum speed of 45 m/s. Suppose the material emitted in the sneeze begins inside the nose at rest, 1.8 cm from the nostrils. It has a constant acceleration for the first 0.29 cm and then moves at constant velocity for the remainder of the distance. Enter your answer in scientific notation.

(a) What is the acceleration as it moves the first 0.29 cm?

(b) How long does it take to move the 1.8 cm distance in the nose?

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Answer #1

if any material insode nose initial at rest means u is zero

and if got accelarted in distance of 0.29 cm means it got maximum velocity till reach point at a distance of 0.29 cm

so by the v2 =u2 + 2aS

where v is fianl velocity u is initial velocity and S is distance travelled =.0029 m

v=45m/s

452= 02+ 2*a*0.0029

a=349137.931 m/s2

part b

total time taken = time to travel 0.29 cm(t1) + time to tarvel from 0.29 to 1.8 cm(t2)

v= u+ at

45=349137.931 t1

t1=0.12888ms

t2=(1.8-0.29)*10-2/45=0.3355ms

t=0.12888+ 0.3355=0.464435 ms

  

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