You need to think of an energy solution. At 10 mph you have stored up
one-half-mass-times-velocity-squared (.5MV^2) dynamic energy. In the crash
this energy is used up in deforming materials, making sound, producing heat
and the post impact velocity. There is also the force-times-distance (Fd)
energy which is what you are referring to.
Lets make it easy, assume that you and the bike are a rigid body, the impact
brings you to a complete stop and ignore all losses except (Fd). Mass is
weight/gravity and 1 mph = 1.47 mph. Then:
..5MV^2 = Fd
..5(150/32.2)(10)^2(1.467)^2 = Fd
Or:
Fd = 503 ft-lb
It takes a 503 lb force to stop in one foot (ouch!)
Braking at .6g, F = .6(150) = 90 lbs and d = 10 ft
In an impact you can use a conservation of momentum solution as well.