Question
A 4000 kg truck traveling at constant velocity and a braking force of 20000 N is suddenly applied. It takes 6 seconds for the truck to stop.
a) what is the deceleration of the truck?
b) what is the velocity of the truck before the brakes were applied
c) what distance did the truck travel from the time the brakes were applied to when it stopped.
a) what is the deceleration of the truck?
b) what is the velocity of the truck before the brakes were applied
c) what distance did the truck travel from the time the brakes were applied to when it stopped.
Answers
bobpursley
F=ma
a=Fmax/m
if it takes six seconds...
Vf=Vi - F/m
solve for Vi
distance?
Vf^2=Vi^2-2ad solve for d.
a=Fmax/m
if it takes six seconds...
Vf=Vi - F/m
solve for Vi
distance?
Vf^2=Vi^2-2ad solve for d.
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