An electron in a cathode-ray-tube (CRT) accelerates uniformly from 8.00 times 10

Question

An electron in a cathode-ray-tube (CRT) accelerates uniformly from 8.00 times 10^4 m/s to 6.00 times 10^6 m/s over 1.00 cm. (a) In what time interval does the electron travel this 1.00 cm? s (b) What is its acceleration? m/s^2 A parcel of air moving in a straight tube with a constant acceleration of -3.20 m/s^2 and has a velocity of 12.0 m/s at 10:05:00 a.m. (a) What is its velocity at 10:05:01 a.m.? m/s (b) What is its velocity at 10:05:04 a.m.? m/s (c) What is its velocity at 10:04:59 a.m.? m/s (d) Describe the shape of a graph of velocity versus time for this parcel of air.

Explanation / Answer

Hi,

10. In this case we have the initial and final velocity, along with the distance travelled. Besides we know that the rate of change in the velocity is constant, therefore:

v2 = vo2+ 2ax :::::::: a = (v2 - vo2)/2x = [ (6*106 m/s)2 - (8*106 m/s)2 ]/[2*1*10-2 m] = 1.8*1015 m/s2

v = vo + at ::::::: t = (v-vo)/a = (6*106 m/s - 8*104 m/s) / ( 1.8*1015 m/s2) = 3.3*10-9 s

11. In this case we have a constant acceleration and the velocity at an instant of time, depending of the other instants of time, it can be taken as an initial velocity or a final velocity, therefore:

(a) v = vo + at ; t = 1 s ; v = 12 m/s + (-3.2 m/s2)(1 s) = 8.8 m/s

(b) v = vo + at ; t = 4 s ; v = 12 m/s + (-3.2 m/s2)(4 s) = -0.8 m/s

(c) vo = v - at ; t = 1 s ; vo = 12 m/s - (-3.2 m/s2)(1 s) = 15.2 m/s

(d) It should be a straight line with a negative slope, so it should go down and its cuts with the axis (either x or y) depends on the scale of time used.

I hope it helps.

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