(a) What is the acceleration as it moves the first 0.21 cm? 1 m/s 2 (b) How long
ID: 2251264 • Letter: #
Question
(a) What is the acceleration as it moves the first 0.21 cm?1 m/s2
(b) How long does it take to move the 2.1-cm distance in the nose?
2 ms
(c) Sketch a graph of vx(t). vx(t). A typical sneeze has a maximum speed of 44 m/s. Suppose the material emitted in the sneeze begins inside the nose at rest, 2.1 cm from the nostrils. It has a constant acceleration for the first 0.21 cm and then moves at constant velocity for the remainder of the distance. What is the acceleration as it moves the first 0.21 cm? How long does it take to move the 2.1-cm distance in the nose? Sketch a graph of vx(t).
Explanation / Answer
a.) Apply the fourth SUVAT equation:
v^2 = u^2 + 2as
44^2 = 0^2 + 2*a*0.0021 (0.21 cm converted into metres)
1936 = 0 + 0.0042a
0.0042a = 1936
a =460952
The acceleration is460952 m/s/s over the first 0.21 cm.
b.) Apply the first SUVAT equation to calculate the time to travel the first 0.25 cm:
v = u + at
44 = 0 + 460952t
t =9.54*10^-5 seconds
This is the time fore the first 0.25 seconds so now apply the third SUVAt equation:
s = ((u + v)/2)*t
0.0189 = ((44 + 44)/2) * t
t = 4.29*10^-4 seconds
Now add the times together:
0.000525
So the total time is 5.249 * (10^-4)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.