Vo = 50mi/h = 22.22 m/s.
Wt. = 3000Lbs = 1362 kg.
d = 200Ft = 60.61 m.
a = -Vo^2/2d = -(22.22^2)/121.2 =
-4.07 m/s.
Fb = M*a = 1362 * (-4.07) = -5543
N. = Braking force.
V = Vo + a*t = 0.
t = -Vo/a = -22.22/-4.07 = 5.45 s. = Time required to stop.
A CAR IS MOVING INITIALLY AT A SPEED OF 50MI/H WEIGHTING 3000LB IS BROUGT TO A STOP IN A DISTANCE OF 200FT FIND THE BRAKING FORCE.AND FIND TIME REQUIRED TO STOP ASSUMING THE SAME BRAKING FORCE..AND FIND THE DISTANCE AND TIME REQUIRED TO STOP IF THE CAR WERE GOING 25MI/H INITIALLY?
2 answers
m=1362kg,v=22.22m/s,d=61m;
we know that a moving object has a kinetic energy=1/2mv^2.So the braking action would be the negative of the work done against Kinetic energy,
W(braking)=-K.E i.e 0.5*1362*22.2*22.2,
we get,W=-335624.04.
now,F=W/stopping dist.
i.e F=335624.04/61
This would be -5542.95 approx
and for time you can use impulse equation,F*t=p;so t=30263.64/5543=5.45s..
for the seconf just replace the value of v;
we know that a moving object has a kinetic energy=1/2mv^2.So the braking action would be the negative of the work done against Kinetic energy,
W(braking)=-K.E i.e 0.5*1362*22.2*22.2,
we get,W=-335624.04.
now,F=W/stopping dist.
i.e F=335624.04/61
This would be -5542.95 approx
and for time you can use impulse equation,F*t=p;so t=30263.64/5543=5.45s..
for the seconf just replace the value of v;
2 answers