Answer:
The constant acceleration, [tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]
Solution:
As per the question:
[tex]v_{1} = 3 x 10^5 m/s[/tex]
[tex]v_{2} = 6.40 x 10^6 m/s[/tex]
length of the re region, l = 1 cm = 0.01 m
Now,
Using the third equation of motion:
[tex]v_{2}^2 = v_{1}^2 + 2a_{c}l[/tex]
[tex]a_{c} = \frac{v_{2}^2 - v_{1}^2}{2l}[/tex]
[tex]a_{c} = \frac{(6.40 x 10^6)^2 - (3 x 10^5)^2}{2\times 0.01}[/tex]
[tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]
