Question

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 and straight, average human reaction time is 0.2 seconds, and theaverage braking acceleration of a US car is 6 m/s​2​. [1 mile=1609.344 meters ]a.Is there a collision?b.If not, how close do they come? If so, at what speed?

Answer 1

a)there is collision

b)V=7.79m/s

Step-by-step 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

`Vf=\sqrt[2]{2ax+Vo^(2) }`

`Vf=\sqrt[2]{2(-6)(36.576)+22.352^(2) }=7.79m/s`