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The problem describes a proton (with mass \( m = 1.67 \times 10^{-27} \) kg) being accelerated along a straight line with an acceleration \( a = 6.70 \times 10^{10} \, \text{m/s}^2 \). The proton initially has a speed \( v_i = 1.70 \times 10^6 \, \text{m/s} \), and it travels a distance of \( d = 2.40 \, \text{cm} = 2.40 \times 10^{-2} \, \text{m} \).

We need to find:
(a) The final speed \( v_f \) of the proton.
(b) The increase in its kinetic energy.

### Part (a): Final Speed

We can use the following kinematic equation to find the final speed:

\[
v_f^2 = v_i^2 + 2ad
\]

Substituting the known values:

\[
v_f^2 = (1.70 \times 10^6)^2 + 2 \times (6.70 \times 10^{10}) \times (2.40 \times 10^{-2})
\]

First, calculate each term:
\[
v_i^2 = (1.70 \times 10^6)^2 = 2.89 \times 10^{12} \, \text{m}^2/\text{s}^2
\]
\[
2ad = 2 \times 6.70 \times 10^{10} \times 2.40 \times 10^{-2} = 3.22 \times 10^9 \, \text{m}^2/\text{s}^2
\]

Now, summing these:

\[
v_f^2 = 2.89 \times 10^{12} + 3.22 \times 10^9 = 2.89322 \times 10^{12}
\]

Taking the square root:

\[
v_f = \sqrt{2.89322 \times 10^{12}} \approx 1.70 \times 10^6 \, \text{m/s}
\]

The final speed is approximately the same as the initial speed, meaning the acceleration had a very small effect over this short distance.

### Part (b): Increase in Kinetic Energy

The kinetic energy is given by:

\[
KE = \frac{1}{2} m v^2
\]

We want to find the increase in kinetic energy, which is:

\[
\Delta KE = \frac{1}{2} m (v_f^2 - v_i^2)
\]

Substitute the known values:

\[
\Delta KE = \frac{1}{2} \times (1.67 \times 10^{-27}) \times (2.89322 \times 10^{12} - 2.89 \times 10^{12})
\]

\[
\Delta KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times 3.22 \times 10^9
\]

\[
\Delta KE = 2.69 \times 10^{-18} \, \text{J}
\]

So, the increase in kinetic energy is approximately \( 2.69 \times 10^{-18} \, \text{J} \).

---

Let me know if you'd like further details on these steps! Here are some related questions:

1. How is the kinematic equation used in this context?
2. How does acceleration affect a particle over a short distance?
3. What is the significance of the proton's initial and final speed being similar?
4. How is the change in kinetic energy calculated for accelerating particles?
5. What factors limit the increase in kinetic energy in this scenario?

**Tip**: When dealing with very small or large numbers, scientific notation helps maintain clarity and precision.

A proton (mass m = 1.67 × 10^−27 kg) is being accelerated along a straight line at 6.70 × 10^10 m/s² in a machine. If the proton has an initial speed of 1.70 × 10^6 m/s and travels 2.40 cm, what is (a) its speed and (b) the increase in its kinetic energy?

Learn how to calculate the final speed and the increase in kinetic energy of a proton accelerated at 6.70 × 10^10 m/s², starting with an initial speed of 1.70 × 10^6 m/s over a 2.40 cm distance. Explore key kinematic and energy formulas in this detailed physics problem solution.

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