A driver traveling 50 mph rounds a bend and sees a car stopped 40 yards (36.576 m) ahead. Assumethe following: the road is flat
1 answer:
Answer:
a)there is collision
b)V=7.79m/s
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)
{Vf^{2}-Vo^2}/{2.a} =X (2)
X=Xo+ VoT+0.5at^{2} (3)
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
To solve this problem, we use equation number 2 to calculate the distance traveled after the end of a movement with constant speed with a time of 0.2s, if it is less than 36.576m there is no collision.
Xt=X1+X2 (5)
X1= distance traveled with constant speed
X1=distance traveled with constant acceleration
for X1
X1=Vt
where
V=50mph=22.35m/s
t=0.2
X1=0.2x22.35m/s
X1=4.47m
for X2 using ecuation number 2
Vo=22.35m/s
Vf=0m/s
a=-6m/s^2
{Vf^{2}-Vo^2}/{2.a} =X
{0^{2}-22.35^2}/{(2).(6)} =X
x2=41.626m
Xt=x1+x2
Xt=41.626+4.47=49m
as the distance traveled is 49m the cars crash
b) to calculate the final speed we use the ecuation number tow(2)
X=36.576m
Vo=22.35m/s
a=-6m/s^2
{Vf^{2}-Vo^2}/{2.a} =X
solving for vf
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