Basic Acceleration and Distance Problem

In summary, the electron in the CRT accelerates from 14200 m/s to 3.73x10^6 m/s over 1.93 cm. Using the kinematic equation v^2 = v_0^2 + 2 a * Delta x, the acceleration is approximately 360432600000000 m/s^2 and the time it takes to travel this distance is 1.0345^-8 s. However, there may be some discrepancies due to rounding errors.
  • #1
balllla
12
0
1. An electron in a crt accelerates from 14200 m/s to 3.73x10^6 m/s over 1.93 cm. how long does the electron take to travel this distance? and What is it's acceleration (in m/s^2)

The supposedly "easiest" question on my physics worksheet has me stumped. I know that I have to convert so it would be 1420000 cm/s and 373000000 cm/s , but I'm am stuck because I do not know the time or acceleration? I just don't know what equation to use?! I appreciate anyones help.

Thank You.
 
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  • #2
Hi balllla, Welcome to PF.
From the set of kinematic equations, find out the equation, which relates initial velocity, final velocity, displacement and acceleration.
 
  • #3
LaTeX Code: v^2 = v_0^2 + 2 a \\Delta x

Would I use this equation which would find out the acceleration to later find out the displacement.?
 
  • #4
I would convert the 1.93 cm to 0.0193 m instead.
 
  • #5
ok but what actually am I trying to find. Am I trying to find time and are 14200m/s and 3.73 velocity?
 
  • #6
would this work

3.73x10^6 ^2= 14200m/s^2 + 2a(.0193m)

? what would that give me, my acceleration? but I still don't know how to do the first part of the problem.

many thanks.
 
  • #7
balllla said:
LaTeX Code: v^2 = v_0^2 + 2 a \\Delta x

Would I use this equation which would find out the acceleration to later find out the displacement.?

I believe you meant "time" instead of "displacement".
Put the velocities into m/s instead of cm/s and put the distance into meters.

Okay. Tell us which of those given numbers you think are the Vi, Vf, X, a, and t. Just label and list them. Then finding an equation should be simple.

You are trying to find the acceleration and the time.
 
  • #8
vi=14200 m/s
vf=3.73x10^6 m/s
x=.0193m
t=?
a=?
 
  • #9
balllla said:
would this work

3.73x10^6 ^2= 14200m/s^2 + 2a(.0193m)

? what would that give me, my acceleration? but I still don't know how to do the first part of the problem.

many thanks.
You are correct. It is a large number. Proceed.
 
  • #10
rl.bhat said:
You are correct. It is a large number. Proceed.

Are you sure? Would the v^2 on the left side of the equation be the final velocity or the average velocity? I'm not trying to confuse anyone, just asking.
 
  • #11
OK. My acceleration is approx. 360432600000000 m/s^2. When doing the algebra I get an answer of .00000001 seconds for time, but my online computer software is saying it is wrong, I know that my acceleration is right but how is my time wrong?!
 
  • #12
I am getting 1.0345^-8 s.
 

1. What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity, meaning it has both magnitude and direction.

2. How is acceleration calculated?

Acceleration is calculated by dividing the change in velocity by the time it takes for that change to occur. The formula for acceleration is a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What is the difference between average and instantaneous acceleration?

Average acceleration is the average rate of change of an object's velocity over a given period of time, while instantaneous acceleration is the rate of change of an object's velocity at a specific moment in time. Average acceleration is calculated by dividing the total change in velocity by the total change in time, while instantaneous acceleration is calculated by taking the derivative of the object's velocity function at a specific time.

4. How does distance relate to acceleration?

Distance is the total length an object has traveled, while acceleration is the rate at which an object is changing its velocity. The distance an object travels is directly related to its acceleration, as a greater acceleration will result in a greater change in velocity and therefore a greater distance traveled over a given period of time.

5. What are some real-life examples of basic acceleration and distance problems?

Some examples of basic acceleration and distance problems include a car accelerating from a stop sign, a roller coaster going down a steep drop, or a person jumping off a diving board into a pool. All of these situations involve an object changing its velocity and traveling a certain distance as a result of acceleration.

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