Stopping distance of 2 Vehicles

[PhysicsAdvice]

A truck with a mass of 4500kg and a car with a mass 1500kg both apply the brakes at the same time, if the coefficient of friction is 0.15, and both vehicles are traveling at 20m/s, how far will it take each vehicle to stop?

Ff=0.15Fn
W=FappD
Ek2-Ek1=W
Fn=mg
Ek=1/2mv^2

for the truck:

Fn=4500kg(-9.81m/s^2)= -44145N Ek2-Ek=0-(1/2(4500kg)(20m/2)^2)
=-900000j
Ff=0.15(-44145N)
=-6621.75N

-900000j=-6622.75N x D=135.8952097m

For Car: Fn= -24525N Ek2-Ek1=-500000j

Ff= 0.15(-24525N)=-3678.25N

-500000j=-3678.25N x D=135.9342078m

Should the distances really be the same?

 

[cupid.callin]

yes it would be same

here frictional force on anybody is : μmg
so deceleration due to friction is: μmg/m = μg
as deceleration is same so stopping distance will also be same

 

[PhysicsAdvice]

should the time it takes also be equal since t=v2-v1/a and a=μmg/m?

 

[cupid.callin]

yes it will be same

 

[rwisz]

Based on the calculations, the stopping distance for the truck is slightly longer than the stopping distance for the car. This is due to the fact that the truck has a larger mass and therefore, a greater force of friction acting against it. However, the difference in stopping distance is very small and may not be noticeable in real-world scenarios. It is important to note that these calculations are based on ideal conditions and may vary in real-life situations. Other factors such as road conditions, tire grip, and braking systems can also impact the stopping distance of a vehicle.