Relative distances (motion problem) --

[christian0710]

Homework Statement

In a collision, an automobile initially traveling at 50 km/h decelerates at a constant rate of 200 m/s^2. A passenger not wearing a seat belt crashes against the dashboard. Before the collision, the distance between the passenger and the dashboard was 0.60 m. With what speed, relative to the automobile, does the passenger crash into the dashboard? Assume that the passenger has no deceleration before contact with the dashboard.

Homework Equations

+ [/B]
The equations for the cars decceleration, velocity and distance traveled are:

a=-a=200m/s^2
v=--a*t +V0
X(t) = -a/2 * t^2 +V0*t 3.The attempt at a solution
we change velocity into m/s
50km/h = 50*1000/60^2 = (125/9 ) m/s

The time at which the car comes to a halt is when velocity = zero
0= a*t –V0
t=V0/a = (125/9)/200 = 5/72sec

The distance the car traveled in that time is
X(5/72) = -200/2 * (125/9)^2 +(125/9)*(5/72) = 0.48 meters

The speed of the passenger must be
0,06m/5/72sec = 8.64m/s

The speed of the car must be
0.48/(5/27) = 6.9m/s

The difference between 8.64 and 6.9 ( I assume is what is meant by relative distance? ) can impossible be 15.6 m/s which is the answer.

Help Is a ppreciated.

The Attempt at a Solution

 

[NascentOxygen]

The solution hinges on calculating details about when the passenger's face "overtakes" his car, because until that happens there is no damage incurred to his head. The time the car takes to come to a standstill has no real relevance, the damage has been done long before the car halts, usually.

The question is asking for relative velocity, not relative distance.

 

[gneill]

The problem doesn't state that the car remains at a halt when its speed reaches zero. If you find that the car reaches zero speed before the passenger hits the dashboard then I suppose you'll have to make an assumption about any continuing motion of the car after that instant.