Answer:
Distance traveled in third second = 25 meter.
Explanation:
We have equation of motion, [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
During the first second it travels 5.0 meters
In this case u = 0 m/s, t = 1 second and s = 5 meters
Substituting
[tex]5=0*1+\frac{1}{2} *a*1^2\\ \\ a=10m/s^2[/tex]
Now we need to find how much distance it travel during third second.
Distance traveled in third second = Distance covered in 3 seconds - Distance covered in 2 seconds
Distance covered in 3 seconds = [tex]0*3+\frac{1}{2} *10*3^2=45m[/tex]
Distance covered in 2 seconds = [tex]0*2+\frac{1}{2} *10*2^2=20m[/tex]
Distance traveled in third second = 45 - 20 = 25 meter
