In a cathode ray tube in an old TV, electrons are...

In a cathode ray tube in an old TV, electrons are accelerated from rest with a constant acceleration of magnitude $7.03 \times 10^{13} \mathrm{m} / \mathrm{s}^{2}$ during the first $2.0 \mathrm{cm}$ of the tube's length; then they move at essentially constant velocity another $45 \mathrm{cm}$ before hitting the screen. (a) Find the speed of the electrons when they hit the screen. (b) How long does it take them to travel the length of the tube?

In a cathode ray tube in an old TV, electrons are accelerated from rest with a constant acceleration of magnitude $7.03 \times 10^{13} \mathrm{m} / \mathrm{s}^{2}$ during the first $2.0 \mathrm{cm}$ of the tube's length; then they move at essentially constant velocity another $45 \mathrm{cm}$ before hitting the screen.
(a) Find the speed of the electrons when they hit the screen. (b) How long does it take them to travel the length of the tube?

Key Concepts

Kinematics

Kinematics is the branch of mechanics that deals with the motion of objects without explicitly considering the forces causing the motion. It involves using equations that relate displacement, velocity, acceleration, and time, which are essential for analyzing motions involving acceleration, deceleration, and uniform motion phases.

Constant Acceleration

Constant acceleration refers to motion where the acceleration remains unchanged throughout the time interval considered. This concept allows the application of specific kinematic equations—such as v² = u² + 2as—to determine the final velocity of an object when it has been subject to continuous acceleration over a known distance from an initial state of rest.

Uniform (Constant) Motion

Uniform motion occurs when an object moves with a constant velocity, meaning there is no acceleration. In such cases, the time to cover a certain distance is directly calculated by dividing the distance by the constant speed. This concept simplifies the analysis of the motion once the acceleration phase has ended.

Motion Segmentation

Motion segmentation involves breaking down a complex motion into distinct phases, each with their own characteristics, such as an initial phase of constant acceleration followed by a phase of uniform motion. This approach enables the separate analysis of each phase using the appropriate kinematic relationships and then combining the results to understand the overall motion.

Transcript

00:01 We need to calculate how much speed the electrons pick up during the first two centimeters of the tube's length.

00:06 So to do this, we'll use our equation, final velocity squared, is equal to initial velocity squared plus 2a times the displacement of x.

00:20 We know that they start from rest, which means that v .i squared is zero.

00:25 So we're solving for vf squared.

00:28 That's going to be equal to zero plus two times the constant.

00:34 Or the magnitude of acceleration, which is 7 .03 times 10 to the 13th.

00:48 And i'm going to multiply x from two centimeters to meters, just because our previous numbers are all in meters, and we want to have like terms.

00:58 So two times 10 to the negative two, are the same as two dividing by 100, will give us meters.

01:05 So vf squared from here, if you solve, is going to be equal to 1 .68, times 10 to the 6 meters per second.

01:21 And since after the first two cent meters of the tube's length, they all move at a constant velocity that tells us that from here on out that this will be the final velocity when they reach the screen.

01:35 Now from here, we want to find out how long it takes the electrons to get there.

01:41 And so we need to find out the total time it takes to reach this velocity.

01:48 And then how long it takes them to travel the remaining 45 centimeters.

01:53 So for the first part, we get vf, the final velocity, see what's at vi, the initial velocity, plus a times the displacement of t...

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