Guys I'm confused with this, different answers all over

>>7762981
apply newtons second law. that should be all you need to figure it out from here.

>>7762981

the distance would double (roughly) because the more mass means more inertia.

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Using basic mechanics you can get mass to divide out on both sides of the equation. This means that the mass doesn't matter, so the time will be the same.

Basically
Ffriction = Fnormal * mu
Fnormal = weight
substitute known quantities
prophet

>>7762997
This is true, but mathematically the inertia doesn't matter when calculating the stopping distance. I.E. mass (inertia) cancels.

>>7762981

>If you work it out mathematically it would be 1/2 the original stopping distance.

What? No it wouldn't. You fucked something up. If it's twice the mass it has twice the kinetic energy, since the frictional force of the brakes is assumed to be the same in both trucks, and force*distance is work, the work needed to cancel the KE will be twice the distance times the force (which remains the same)

>>7763030
It skids, you retard. Brakes don't do work.
Is reading a highschool problem such a challenge?

I guess the friction force doubles as the mass doubles, therefore they cancer each other out.

>>7762981
Should be the same, based on the high-school physics analysis. Frictional force doubles because the normal force doubles, but inertia doubles as well, so by F = ma the deceleration should be the same.

>>7762981
You should assume that the vehicle has the same speed in both cases, and that in both cases the brakes immediately stop the wheels from turning. Since momentum and friction force are both proportional to mass, changing the mass won't change the skidding distance.

>>7763030
>frictional force is the same

the frictional force between the locked tires and the road surface is dependent on mass, einstein

>>7762981
Engineer here.

In reality, the tires are fairly plastic and thus the coefficient of friction will also be dependent on mass, as well as the strength of the deceleration which just compounds the stopping force.

So a heavy truck does indeed stop faster.

Except in the case where the tires overheat or even fail catastrophically, at which point the coefficient of friction drops significantly and then the heavier truck takes much longer to come to a halt.

>>7762981
Kinetic energy of the "double truck" : m*v^2
Where m is the mass of the simple truck.
Let D be the stopping distance of the double truck, and d the stopping distance of the simple truck.
Ft is the tangential force of the wheels on the road for the simple truck, this force being proportional to the mass (Ft = Fn*f = g*m*f) the tangential force for the double truck is 2*Ft.
d*Ft = 0.5*m*v^2 (1)
D*2*Ft = m*v^2 (2)
D*Ft = 0.5*m*v^2
D*Ft = d*Ft (substitute with 1)
D = d

>>7763515
I simply used two
X = vi(t) + 1/2a(t^2)
X = vi(t) + 1/22a(t^2)=X = vi(t) + a(t^2)

Comparing them 2x=x the other distance is half of the other one.