Acceleration Question.Please Check My Answer

[cbrowne]

Homework Statement

An electron in a cathode ray tube of a TV set enters a region where it accelerates
uniformly from a speed of 3.5 × 104 m/s to a speed of 1.4 × 106 m/s in a distance
of 2.0 cm. (a) What is the acceleration of the electron in this region? (b) How
long is the electron in the region where it accelerates?

Homework Equations

The Attempt at a Solution

A) Vi= 3.5 x 104 m/s Vf= 1.4x 10⁶ m/s
d= 0.02 m

Vf²= Vi²+ 2ad

(1.4x10^6)² = (3.5x 10^4)² + 2a(0.02)

(1.958775 x 10^12) = 2a(0.02)

(1.958775x10^12) = 0.04a
0.04 0.04


4.8969375x 10¹³= a

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b) Vf= Vi+ at

(1.4x10^6)= (3.5x10^4)+ ( 4.8969375x10^13)t

136500 = 4.8969375x10^13t
4.8969375x10^13t 4.8969375x10^13t

= 0.000000003 --> t = 3.0x 10^-9

 

[jmarcian]

seems you used the correct equations, everything looks good to me!

 

[alphysicist]

*******************************************

b) Vf= Vi+ at

(1.4x10^6)= (3.5x10^4)+ ( 4.8969375x10^13)t

136500 = 4.8969375x10^13t

I believe you have a calculator error here; it looks like you dropped a zero on the left hand side. (1.4 million minus 35 thousand does not give a little over a hundred thousand.)

If you correct that, you should get the right answer for the time.

4.8969375x10^13t 4.8969375x10^13t

= 0.000000003 --> t = 3.0x 10^-9

 

[cbrowne]

THANK SOO MUCH FOR NOTICING THAT MISTAKE! I end up getting 2.8 x10^-8 for my answer. Could you please verify this

 

[alphysicist]

THANK SOO MUCH FOR NOTICING THAT MISTAKE! I end up getting 2.8 x10^-8 for my answer. Could you please verify this

That looks right to me.