Question

An electron in the cathode ray tube of a tele-
vision set enters a region where it acceler-
ates uniformly from a speed of 53800 m/s to
a speed of 4.58 × 106 m/s in a distance of
2.82 cm.
What is its acceleration?
Answer in units of m/s2

Answer 1

Acceleration is the change in velocity over time:

Acceleration = (Final velocity - initial velocity)/time

Since there are no directions involved, speed is used instead of velocity.

So you know the initial speed:

Initial speed = 53800 m/s

And you know the final speed:

Final speed = 4.58 x 10^6 m/s

So, we can now do the speed part of the formula:

4.58 x 10^6 - 53800 = 4.53 x 10^6 m/s

This is the change that occurred in the distance 2.82 cm. So to find out how long that change took to occur we look at the following: (I will convert 2.82 cm to m = .0282 m)

0.0282 m * (1/4.53 x 10^6) s/m = 6.23 x 10^-9 s

So, now we have the initial speed, final speed, and the time so acceleration is:

Acceleration = (4.53 x 10^6 m/s)/(6.23 x 10^-9 s) = 7.27 x 10^14 (m/s)/s