Question

1. an electron in a cathode-ray tube accelerates uniformly from ( 2.00 \times 10^4 ) m/s to ( 6.00 \times 10^6 ) m/s over 1.50 cm. (a) in what time interval does the electron travel this 1.50 cm? (b) what is its acceleration?

Explanation:

Step1: List given values

Initial velocity $v_0 = 2.00 \times 10^4 \, \text{m/s}$, final velocity $v = 6.00 \times 10^6 \, \text{m/s}$, displacement $\Delta x = 1.50 \, \text{cm} = 0.0150 \, \text{m}$

Step2: Find average velocity

$$v_{\text{avg}} = \frac{v_0 + v}{2} = \frac{2.00 \times 10^4 + 6.00 \times 10^6}{2} = 3.01 \times 10^6 \, \text{m/s}$$

Step3: Solve for time interval (part a)

Use $\Delta x = v_{\text{avg}} \cdot t$, rearrange for $t$:
$$t = \frac{\Delta x}{v_{\text{avg}}} = \frac{0.0150}{3.01 \times 10^6} \approx 4.98 \times 10^{-9} \, \text{s}$$

Step4: Solve for acceleration (part b)

Use kinematic equation $v^2 = v_0^2 + 2a\Delta x$, rearrange for $a$:
$$a = \frac{v^2 - v_0^2}{2\Delta x}$$
Substitute values:
$$a = \frac{(6.00 \times 10^6)^2 - (2.00 \times 10^4)^2}{2 \times 0.0150} = \frac{3.60 \times 10^{13} - 4.00 \times 10^8}{0.0300} \approx 1.20 \times 10^{15} \, \text{m/s}^2$$

Answer:

(a) $4.98 \times 10^{-9} \, \text{seconds}$
(b) $1.20 \times 10^{15} \, \text{m/s}^2$