Finding constant acceleration without time

[Uriah Graves]

Homework Statement

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An X-ray tube gives electrons constant acceleration over a distance of 14 cm . If their final speed is 1.0×107 m/s , what are the electrons' acceleration?
Express your answer to two significant figures and include the appropriate units.

Homework Equations

My prof. has provided me with a series of kinematic equations to use to solve for constant acceleration. None of them allow me to solve this without "t" time. This problem does not give me time. For this reason, I resorted to the internet to get help.

I have noticed many people use a = [Vf2 - Vi2] / [2d] for a very similar question, but I am unsure where that formula comes from, as I thought acceleration was a difference of velocity/time, not a difference of velocity/2*distance.

I have solved this equation with Vf^2 = Vi^2 + 2ad. I am not sure where this formula comes from either, but using it worked.

The Attempt at a Solution

3.6*10^14 m/s^2

 

[haruspex]

My prof. has provided me with a series of kinematic equations to use to solve for constant acceleration. None of them allow me to solve this without "t" time.

There are five standard equations for constant acceleration. Using u and v for initial and final velocities and a, s (your d), and t as the others, they are known as the SUVAT equations. (Google that.) Each equation omits one of the five, so yes, there is one without time. These are worth learning.

Vf^2 = Vi^2 + 2ad

That is indeed the equation, and it is the same as the other you quoted, just rearranged. (Vf2 meaning V_f².)

 

[Uriah Graves]

I will Google that! This is exactly why I am here! Thank you for taking the time to reply to my post! I genuinely appreciate it!

 

[jbriggs444]

I have solved this equation with Vf^2 = Vi^2 + 2ad. I am not sure where this formula comes from either, but using it worked.

One way of looking at this equation is as a description of work done by a constant force. Multiply every term by m. Now you have $$mv_f^2 = mv_i^2 + 2mad$$Now, if we we divide every term by two and substitute in for $F=ma$ we wind up with$$\frac{1}{2}mv_f^2=\frac{1}{2}mv_i^2 + Fd$$Which is simply a statement that the final energy is equal to the initial energy plus work done.

[I went through first year physics without ever memorizing this particular SUVAT formula. I re-cast every relevant problem in terms of energy and work and solved them that way]