Explanation:
Given that,
Initial speed of cells is 0 as they were at rest
The acceleration of the cell, [tex]a=5.3\times 10^{7}\ m/s^2[/tex]
Time, t = 700 ns
We need to find the maximum speed reached by the cells and the distance traveled during the acceleration. Let v is the final speed. So,
[tex]v=u+at\\\\\because u =0\\\\v=at\\\\v=5.3\times 10^7\times 700\times 10^{-9}\\\\v=37.1\ m/s[/tex]
Let d is the distance traveled. Using equation of motion as follows :
[tex]d=\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 5.3\times 10^7\times (700\times 10^{-9})^2\\\\d=1.29\times 10^{-5}\ m[/tex]
Hence, this is the required solution.
