truck physics question?

Question

The driver of an empty speeding truck slams on the brakes and skids to a stop in a distance Dm = 254.84 m. The truck's initial velocity is v0 = 50 m/s and its total mass is m = 3137.8 kg. The coefficient of kinetic friction between the tires and the road is uk = 0.5. If the truck were carrying an amount of cargo that doubled its total mass, what would its stopping distance have been? (Assume the road is straight and horizontal).

Answer 1

Since the stopping distance is related to the force as
m*g*u*d=.5*m*v^2
note that d is actually independent of mass.
d=v^2/(2*g*u)
j

Answer 2

The force stopping the truck is Friction and you find friction by the normal force times uk. (.5 x 9.8 x 3137 x 2) = 30750.44 N) then you find the accelleration with F = ma
(30750.44/(3137.8x2)) = 4.9 m/s then you find the distance with v(f)^2 = v(0)^2 + 2ad (0 = 50^2 + 2x4.9xd) so d = 255.10 m. This should be the same as the original distance because the masses from F = ma and the normal force in cancel. It is slightly different because I rounded some numbers, but you get the idea.

Answer 3

The regulation of inertia states that gadgets tend to maintain shifting uniformly till that stream is replaced by utilising an exterior rigidity. As such, the passengers of the truck mattress keep shifting fairly forward because of the fact the rigidity exerted upon them by utilising the mattress of the truck does no longer sufficiently regulate their direction jointly as the truck it extremely is self is changing direction, inflicting them to circulate relative to the truck. You journey this particularly in the cab of a motor vehicle. Feeling as while you're being pushed to the exterior of the turn although, doorways and seat belts keep you from being tossed out.