Based on the calculations, the distance traveled by this electron is equal to 0.232 m or 23.2 cm.
Given the following data:
- Initial velocity, u = 0 m/s (since it's starting from rest).
- Acceleration, a = 5.33 × 10¹² m/s².
- Initial time, t₁ = 0.150 × 10⁻⁶ secs.
- Final time, t₂ = 0.200 × 10⁻⁶ secs.
- Deceleration, d = -2.67 × 10¹³ m/s².
How to calculate the distance traveled?
In order to calculate the distance traveled by this electron, we would determine the change in displacement and initial velocity for each path of its motion.
For the initial displacement, we would apply the second equation of kinematic motion as follows:
S₁ = ut₁ + ½at₁²
S₁ = 0(0.150 × 10⁻⁶) + ½ × 5.33 × 10¹² × (0.150 × 10⁻⁶)²
S₁ = 0.06 meter.
For the initial velocity, we would apply the first equation of kinematic motion as follows:
V₁ = u + at₁
V₁ = 0 + (5.33 × 10¹² × 0.150 × 10⁻⁶)
V₁ = 8.0 × 10⁵ m/s.
For the second displacement, we have:
S₂ = V₁t₂
S₂ = 8.0 × 10⁵ × 0.200 × 10⁻⁶
S₂ = 0.16 meter.
For the final displacement, we would apply the third equation of kinematic motion as follows:
[tex]S_3 = \frac{U^2\; - \;V_1^2}{2d} \\\\S_3 = \frac{0^2\; - \;(8.5 \times 10^5)^2}{2\times (-2.67 \times 10^{13})}[/tex]
S₃ = 0.012 meter.
Now, we can determine the total distance traveled by this electron:
Distance = 0.06 + 0.16 + 0.012
Distance = 0.232 m or 23.2 cm.
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