If a child ran into the road 65 to 70 feet ahead of your vehicle, what is the highest speed from which you could stop with good brakes before hitting him?

Let v =  the highest speed of the vehicle.
Let d = the stopping distance.

The formula for stopping distance is
d = v²/(2μg)
where
μ = 0.8, the static coefficient of friction for good brakes (normal conditions)
g = 32.2 ft/s², acceleration due to gravity.

In terms of v,
v = √(2μgd)

Note that 88 ft/s = 60 mph.

Consider d = 65 ft.
v = √2*0.8*32.2*65) = 57.87 ft/s
or
v = 57.87*(60/88) = 39.5 mph

Consider d = 70 ft.
v = √(2*0.8*32.2*70) = 60.05 ft/s
or
v = 60.05*(60/88) = 40.95 mph

The lower of these two speeds should be the highest allowable speed in order to avoid hitting the child.

The highest speed is 39.5 mph or 57.9 ft/s.

This is the answer: body 1 + engine 2