a) Vo = 50 km/h = 13.89 m/s
T = "Crumpling time" = X/(Vo/2)
= 2*0.50 m/13.89 m/s = 0.072 s
(Vo/2 is the average speed of the car during crumpling)
Acceleration = Vo/T = Vo^2/(2X)
= 193 m/s^2
b) It is not reasonable to assume that the crumple distance will remain 50 cm at the higher speed. More information is needed.
PLEASE ANSWER ASAP!!!!!!!!!! step by step
a) find the acceleration experienced by the driver of a car traveling at 50km/h if the car hits a pole and crumples 50cm.
b) repeat part a) for a car traveling 100km/h
3 answers
can you explain it a bit clearer
v=50 km/h =50000/3600=13.89 m/s
For accelerated motion
s=vₒt+at²/2 … (1)
v=vₒ+at………(2).
Final v=0 => t= -vₒ/a.
Substitute “t” in (1)
a= - vₒ²/2•s= - (13.89)²/2•0.5=
= - 195.9 m/s²
For accelerated motion
s=vₒt+at²/2 … (1)
v=vₒ+at………(2).
Final v=0 => t= -vₒ/a.
Substitute “t” in (1)
a= - vₒ²/2•s= - (13.89)²/2•0.5=
= - 195.9 m/s²
2 answers