Vi = 72,000 meters/3600 seconds
= 20 m/s
v = Vi + a t
what is a ?
a = F/m
F = -m g (.5)
so
a = -.5 g = -9.81/2 = -4.9 close enough
v = 20 - 4.9 t
when v = 0
0 = 20 - 4.9 t
so t is about 4 seconds
average speed during stop = 20/2 = 10 m/s
so
d = 10 m/s * 4 s = 40 meters
( nearly half a football field )
Consider a car moving along a straight horizontal road with a speed of 72km/h.If the coefficient of static friction between the tyres and the road is 0.5,the shortest distance in which the car can be stopped is(taking g=10m/s^2)
(a)30m (b)40m (c)72m (d)20m
2 answers
U - 72km/h
=72×5/18 = 20m/s
V=0
A=-g
=-0.5×10=5m/s^2
V^2=u^2+2as
0=400+2×(-5)×s
0=400-10s
10s=400
S=400/10=40m
=72×5/18 = 20m/s
V=0
A=-g
=-0.5×10=5m/s^2
V^2=u^2+2as
0=400+2×(-5)×s
0=400-10s
10s=400
S=400/10=40m
2 answers