The distance moved by runner a is calculated using the formula s = ut +[tex]\frac{1}{2} at^2[/tex]
For runner a , u = 3 m/s, a = 0, s = 25 + x, where x is the distance traveled by runner b before they catch up.
So, 25 + x = 3*t
The distance moved by runner b is calculated using the formula s = ut +[tex]\frac{1}{2} at^2[/tex]
For runner b , u = 0 m/s, a = 0.1[tex]m/s^2[/tex], s = x
So, x = [tex]\frac{1}{2} *0.1*t^2=0.1t^2[/tex]
Substituting in the previous equation, [tex]25+0.05t^2=3t[/tex]
[tex]t^2-60t+500 = 0\\ \\ (t-50)(t-10)=0\\ \\ t=50 seconds, t=10 seconds[/tex]
So they will catch up after 10 seconds.
