An electron with an initial velocity v_{0}=1.5 \times 10^{5} m/s enters a region of length L=1.00 cm where it is electrically accelerated as shown in the figure. It emerges with v=5.70 \times 10^{6} m/s. What is its acceleration, assumed constant?

#### Solution:

### As the question states that the electron is in constant acceleration, we can use the following equation to solve our problem.

### v^{2}=v_0^2+2a(x-x_{0})

##### (This is a fundamental equation that can be found in many physics books that describe the motion of a particle with constant acceleration.)

### Let us now isolate a in our equation. Thus we have:

### a=\frac{v^{2}-v_0^2}{2x}

### Let us now substitute the values given in our problem.

### a=\frac{(5.7\times 10^{6} m/s)^{2}-(1.5\times 10^{5} m/s)^2}{2(0.010 m)}

##### (note that we converted 1.00cm to m)

### a=1.62\times 10^{15} m/s^2

###### This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 23.

In above equation where you isolate the acceleration you write 2x. where is the initial position? Are you considering initial position 0?

Yes, try to see if you can always place the coordinate system (the starting location) at zero. Makes the math a lot simpler.