An electron, starting from rest and moving with a constant acceleration, travels 2.00 cm in 5.00 ms. What is the magnitude of this acceleration ?

Answer

To solve the problem of finding the magnitude of the acceleration of an electron that travels 2.00 cm in 5.00 ms starting from rest, we can use the equations of motion. Here are the steps to arrive at the solution:

### Step 1: Convert units to SI
We need to convert the distance and time into SI units:
- Distance: \( 2.00 \, \text{cm} = 2.00 \times 10^{-2} \, \text{m} \)
- Time: \( 5.00 \, \text{ms} = 5.00 \times 10^{-3} \, \text{s} \)

### Step 2: Identify known values
From the problem, we have:
- Initial velocity \( u = 0 \, \text{m/s} \) (since it starts from rest)
- Distance \( s = 2.00 \times 10^{-2} \, \text{m} \)
- Time \( t = 5.00 \times 10^{-3} \, \text{s} \)
- Acceleration \( a \) is what we need to find.

### Step 3: Use the equation of motion
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values into the equation:
\[ 2.00 \times 10^{-2} = 0 \cdot (5.00 \times 10^{-3}) + \frac{1}{2} a (5.00 \times 10^{-3})^2 \]
This simplifies to:
\[ 2.00 \times 10^{-2} = \frac{1}{2} a (25.00 \times 10^{-6}) \]

### Step 4: Solve for acceleration \( a \)
Rearranging the equation to isolate \( a \):
\[ 2.00 \times 10^{-2} = \frac{1}{2} a (25.00 \times 10^{-6}) \]
Multiplying both sides by 2:
\[ 4.00 \times 10^{-2} = a (25.00 \times 10^{-6}) \]
Now, divide both sides by \( 25.00 \times 10^{-6} \):
\[ a = \frac{4.00 \times 10^{-2}}{25.00 \times 10^{-6}} \]
Calculating this gives:
\[ a = \frac{4.00}{25.00} \times 10^{4} = 0.16 \times 10^{4} \, \text{m/s}^2 \]
Thus:
\[ a = 1.6 \times 10^{3} \, \text{m/s}^2 \]

### Final Answer
The magnitude of the acceleration of the electron is \( 1.6 \times 10^{3} \, \text{m/s}^2 \).

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