Question
a car which weight is 1000N is travelling at 90kmn/hr and come to rest in a distance of 50m after applying the brakes. assume uniform acceleration, calculate how large its stopping force was exerted by the friction between the wheel and the road.
Answers
Damon
90,000 meters / 3600 seconds = 25 m/s=Vi
v = Vi + a t
0 = 25 + a t
so a t = -25 and t = -25/a
d = Vi t + .5 a t^2
50 = 25 (-25/a) + .5 a (625/a^2)
50 = -625/a + .5 (625/a)
50= -625/2a
100 a = -625
a = -6.25
then
F = m a
v = Vi + a t
0 = 25 + a t
so a t = -25 and t = -25/a
d = Vi t + .5 a t^2
50 = 25 (-25/a) + .5 a (625/a^2)
50 = -625/a + .5 (625/a)
50= -625/2a
100 a = -625
a = -6.25
then
F = m a