A passenger car and a heavy truck are driving side by side along the straight test road with
equal speed of
v0 =100km/h . The mass of the truck is assumed to be 20 times that of the
passenger car. The test road surface as well as the tire surfaces may be assumed to be
qualitatively equivalent for both vehicles. At a certain instant both vehicles will start brake
with wheels locked. The passenger car will stop after sliding 100 meter distance. What will
be the braking distance of the truck?

Question

A passenger car and a heavy truck are driving side by side along the straight test road with
equal speed of 
v0 =100km/h . The mass of the truck is assumed to be 20 times that of the
passenger car. The test road surface as well as the tire surfaces may be assumed to be
qualitatively equivalent for both vehicles. At a certain instant both vehicles will start brake
with wheels locked. The passenger car will stop after sliding 100 meter distance. What will
be the braking distance of the truck?

anonymous · Answer

100 m. 
(v_i)^2=(v_f)^2+2ad
v_i for both 100
v_f for both 0
what is a? a=F/m
F=muk*m...mass cancels when finding a so a is same in both cases.

anonymous · Answer

Hi, what is muk, and after mass cancels out how can i get the distance of the truck.

anonymous · Answer

muk:$$\mu k$$ the coefficient of friction. the values are not needed since it states in the problem that both vehicles experience the same road conditions. it's just important to know in cases like these, mass would cancel so the mass is not relevant.

anonymous · Answer

i am trying it bit still a bit stuck, coefficient for friction is 1 right, if you can solve it will help me with other questions like this in my coming exam

anonymous · Answer

if you'd like, you can plug in any number because it's not given in the problem, if that will help you solve the problem. you have all the info you need for the car so use kinematics equation to solve for a.
100km/h=27.8m/s
0^2=(27.8m)^2+2a(100m)

anonymous · Answer

Thats where i was stuck for now. after i get a for p car then

anonymous · Answer

a=-3.86m/s/s. now you want to solve for coefficient of friction.$$a=(\mu kmg)/m$$
plug in numbers...we'll use 5kg for car just to see what happens
3.86=(muk*5kg*9.8m/s/s)/5kg
you can cancel 5kg so divide both side by 9.8 to get muk
muk:.394

anonymous · Answer

ok, now we have this coefficient of friction, plug it into kinematics equation to find distance for the truck. truck is 20 times the mass so it will be 100kg.
first find a
a=(.394*100kg*9.8m/s/s)/100kg=3.86m/s/s...looks familiar?
and plug in to find distance...
0^2=(27.8m/s)^2+2(-3.86)x
x=??

anonymous · Answer

i can now understand. big ups to you. thanks now let me try another similar problem.

anonymous · Answer

np, glad you understand