An electron in a cathode-ray tube is traveling horizontally at 2.10×10^9 cm/s when deflection plates give it an upward acceleration of 5.30×10^17 cm/s^2 .

B.) What is its vertical displacement during this time?

My work http://i.imgur.com/AqKlTuX.jpg

I am asking this because the answer my teacher gave is 2.3cm

Thanks in advance


closed as too localized by Qmechanic Feb 12 '13 at 0:17

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  • $\begingroup$ For starters, y = y_0 + ut + 0.5at^2, where as you have omitted this 0.5 $\endgroup$ – Kenshin Feb 7 '13 at 5:24
  • $\begingroup$ And secondly you performed a calculator error as well. See my answer below. $\endgroup$ – Kenshin Feb 7 '13 at 5:28

$y = y_0 + ut + 0.5at^2$

Since $y_0$ and $u$ are $0$, we have $y = 0.5~at^2$.

In your calculation $t = 2.95 \times 10^{-9}$, which is correct.

So, $y = 0.5\times5.30×10^{17}\times (2.95 \times 10^{-9})^2$

Therefore, $y = 2.3~\text{cm}$

  • 1
    $\begingroup$ No problem user1530249! $\endgroup$ – Kenshin Feb 7 '13 at 5:31

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