1. Jan 6, 2014

### george ozua

Two cars move at a speed of 16 meters per second. Each of them has a passenger that is not wearing the seat-belt. Both of these cars decelerate abruptly. Car number one completely stops in 0.25 seconds. Car number two completely stops in two seconds. We know that in both cases, the passengers will continue moving after the car stops (due to their inertia).

After the cars stop, passenger one and passenger two will be launched with the same speed?
In other words, the time that the cars use to stop provokes different speeds in their occupants when launched due to inertia?

I believe that time plays a role in determining the speed of the passengers when cars one and two stop. I am confused because the law of inertia tells that bodies continue moving at their same speed. But how can we explain that the passengers of a car that decelerates in 0.1 seconds end up crashing with the wind shield and that the passengers of a car that decelerates in ten seconds barely move in relation to the car?
Thanks.

2. Jan 6, 2014

### lendav_rott

If a car stops so suddenly, it means they hit a tank or a rather sturdy tree.
Newton says that a body will attempt to maintain its velocity and direction. If your car moves 16m/s per second and gradually slows down, it means the difference in velocities as it goes down step by step is not enough to generate a force great enough to send the two people airborne.

As you are driving your body also obtains a kinetic energy of mv²/2 , where m is your mass and v is the car's velocity. Imagine if the car stopped in 0.25 seconds, to us it is quicker than in a blink of an eye, so it might as well be instantenous. Suddenly, the car stopped, but you still try to move at the velocity of v. What happens to all that kinetic energy? Does it just dissipate into thin air? No, it doesn't. The car's kinetic energy will most likely be transferred to mechanical energy and heat, but the person's body will want to keep moving in the same direction and at the same velocity - inertia.

Now, imagine you know ahead of time you have to slow down, as you decelerate slowly, your body will also, but you can still feel the forward thrust if you hit the brakes too hard. If you decelerate slowly enough, the difference in your initial kinetic energy and your kinetic energy after each passing time interval will not be great enough to actually lift your body off the seat.

3. Jan 6, 2014

### CWatters

No not "after the cars stop". If there is no friction the passengers start moving forward in their seat the moment the brakes are applied, not once the cars stop.

Friction between the passenger and the seat will play a role if the car stops slowly. The force propelling them forward in their seat will be ma (where "m" is their mass and "a" the rate of deceleration).

The friction force is mgμ (where m is their mass, g is the acceleration due to gravity and μ is the coefficient of friction)

So to remain in their seat..

ma < mgμ

mass cancels leaving

a < gμ

I've no idea what the coefficient of friction is for a car seat but a similar homework question google found had μ = 0.5.

So for the passenger to remain in their seat the deceleration must be less than about 0.5 g

If the car stops from 16m/s in 2 seconds and we assume constant deceleration then the SUVAT equation can be used to calculate the actual deceleration...

v = u +at

v=0
u=16
t=2

a = -u/t
= 16/2
= 8m/s/s

That is much greater than 0.5g so it's very unlikely the passenger will remain in their seat even if the car takes 2 seconds to stop.

Obviously the passenger might help friction by bracing with their arms and legs but I don't recommend trying it at home.

Last edited: Jan 6, 2014
4. Jan 6, 2014

### CWatters

PS If you ignore friction then both passengers will start sliding towards the windscreen the instant the brakes are applied. Both will hit the windscreen but the question is how hard...

From the passengers perspective it appears as if the windscreen is accelerating towards them. How hard you hit it depends on how hard the car decelerates...

Lets say the distance between passenger and screen starts off as 1m.

Above I calculated that the 2 second car decelerates at 8m/s/s. So using...

v2 = u2 +2as

we can work out how fast the passenger hits the screen..

u = 0 (the initial relative velocity is zero as both are going same speed)
a = 8 m/s/s
s = 1m

v = sqrt(2as)
= squrt(2*8*1)
v = 4m/s

So without friction the passenger hits the screen at 4m/s in the car that stops in 2 seconds.

I'll let you work out the deceleration and impact velocity in the other car.