about 20 meters/second initial speed
mass not relevant
speed at end = original speed + a t
a t = 0 - 20
t = -20/a (Note a is negative of course)
distance = Vi t + 0.5 a t^2
40 = 20 (-20/a) +(0.5) (a) (-20/a)^2
40 = -400/a + 200/a = -200/a
a = -200/40 = -5 m/s^32
t = -20/-5 = 4 seconds
check my arithmetic
A car of mass 1000kg travelling @72km/hr is brought to rest in a distance of40m find average stoping for nd time taken
4 answers
Vo = 72km/h = 72000m/3600s = 20 m/s.
V^2 = Vo^2 + 2a*d = 0.
20^2 + 2a*40 = 0.
a = -5 m/s^2.
V = Vo + a*t = 0.
20 + (-5)t = 0,
t = 4 s. = stopping time.
V^2 = Vo^2 + 2a*d = 0.
20^2 + 2a*40 = 0.
a = -5 m/s^2.
V = Vo + a*t = 0.
20 + (-5)t = 0,
t = 4 s. = stopping time.
I don't understand were you get 40
What what is the formular for retardation
5 answers