Stopping distance of 2 Vehicles

AI Thread Summary
The discussion analyzes the stopping distances of a truck and a car, both weighing 4500kg and 1500kg respectively, when braking from a speed of 20m/s with a friction coefficient of 0.15. Calculations show that the stopping distances for both vehicles are approximately 135.9 meters. The frictional force is calculated using the formula Ff=μFn, leading to similar deceleration rates for both vehicles. Consequently, the time taken to stop is also equal for both vehicles due to the same deceleration. The conclusion confirms that both the stopping distance and time will be the same for the truck and car under these conditions.
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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?
 
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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
 
should the time it takes also be equal since t=v2-v1/a and a=μmg/m?
 
yes it will be same
 
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