I am not sure i understood the question completely...if he pushes against the wall he acquires a velcity v0 in x direction

If the professor slides only in the horizontal direction...

 

Fcos 45 =ma

f cos 45 = m dv/dt

or dv =f cos 45/ m dt

dv = 1/2 m (b- ct) dt

integrating on both sides

change in velocity

v = 1/m2 (bt-ct²/2)

hence v_t=v₀+ 1/m2 (bt-ct²/2)