How Is Force Calculated When a Student Hits a Dashboard?

[Oriako]

Homework Statement

A car is brought from 100 km/h to 0 km/h in 2 seconds. A student with a mass of 73kg(not wearing a seat belt) flies through the air toward the dashboard which is 0.650m away. He smashes into the dashboard and it takes 0.25 seconds to slow him to the velocity of the car at this time. Find the force exerted by the dashboard on the student.

Homework Equations

Kinematic Equations

F x t = m x (Vf - Vi)

F = m x a

The Attempt at a Solution

I've tried many different ways but don't know how to find the time it takes for the person to hit the dashboard as it is the 0.25s to stop completely once hitting the dashboard.

I got an acceleration of -13.89 m/s of the car but am unsure of what to say about the student.

THANKS!

 

[Oriako]

Is it possible that the passenger has an acceleration of 13.89 m/s as he approaches the dashboard, as from the reference of the interior of the car he goes from a Vi of 0 to a Vf you need to find. Delta t is also a mystery, but if acceleration could be assumed to be 13.89 m/s as the car is moving away from him then I can find the Vf?

 

[Booke]

I'm a bit rusty but here's my try since no one's post yet.
Your first equation should be all you need. F=dp/dt
dp=0-73*100km/h*.278(m/s)/(km/h) dt=.25s

The time it takes the person to hit the dashboard shouldn't be needed.

 

[Oriako]

Are you sure about that?

The change in momentum is certainly m x -Vi, but how can you say that happens over the course of 0.25s? Once he hits the dashboard, then it takes him 0.25s to slow down to whatever speed the car is while in the middle of negative acceleration. Or can you just say that since he moves at a constant velocity while in the space between his seat and dashboard? An explanation would be helpful.

 

[Booke]

As the car is brought to a stop, there is no force acting on the person until he/she hits the dashboard, if no force acts on him/her, F=ma, there could be no acceleration on the person. The person's change in momentum only occurs once he/she hits the dashboard so time we are concern with starts after that.

 

[Oriako]

Wow, I've been stressing over this question for hours doing complicated systems of equations and (0.65m + x) variables, etc.

I can't believe it is honestly that simple... I just question it because my teacher said otherwise, but you never know. He could easily be wrong.

 

[Oriako]

Wait, can Vf actually be 0? Because since we used 27.78 m/s (relative to the ground) in our calculation of Vi, we must be consistent and use our Vf relative to the ground.

The car is still moving when HE stops moving from the internal perspective. You would need to find the velocity the car/passenger system is moving at the time he has a velocity of zero (relative to the interior).

0.25 seconds to slow him to the velocity of the car at this time.

 

[gneill]

Look at it from the point of view of the frame of reference of the student. He continues to travel inertially until he hits the dashboard. From his point of view the dashboard starts at the given distance away, and suddenly begins to accelerate towards him. You have already calculated the car's acceleration (deceleration), so it should be a simple matter to calculate both the time of 'first contact' and the velocity of the student/dashboard 0.25 seconds after that.

 

[Oriako]

On gneill's suggestion I ended up getting a force of 1211.79N. Can anyone verify?

Also, does this mean that Booke was wrong?

 

[gneill]

Perhaps you should show your work.

 

[SammyS]

On gneill's suggestion I ended up getting a force of 1211.79N. Can anyone verify?

Also, does this mean that Booke was wrong?

I got a different answer. I do believe Booke was wromg.

Using gneill's suggestion: How long does it take an object to cover a distance of 0.65 m if it starts from rest an accelerates at 13.89 m/s² ?

d = (1/2)a(t²) .

 

[Oriako]

I did:
Rearranged Vf^2 = Vi^2 + 2adeltaD
Before the collision
Vf = sqrt(2(a)(d))
Vf = sqrt(2(13.89m/s)(0.65m))
Vf = sqrt((1.24m)(13.89m/s))
Vf = 4.15m/s

During collision, so Vf in the previous is Vi here:
Fnet x t = m(Vf-Vi)
Fnet x t = (73kg)((0m/s)-(4.15m/s))
Fnet = (302.95 kg x m/s)/(0.25s)
Fnet = (1211.79N)

Amirite? D:

 

[gneill]

I did:
Rearranged Vf^2 = Vi^2 + 2adeltaD
Before the collision
Vf = sqrt(2(a)(d))
Vf = sqrt(2(13.89m/s)(0.65m))
Vf = sqrt((1.24m)(13.89m/s))
Vf = 4.15m/s

Okay, that's the relative velocity between the dash and student when they first meet.

During collision, so Vf in the previous is Vi here:
Fnet x t = m(Vf-Vi)
Fnet x t = (73kg)((0m/s)-(4.15m/s))
Fnet = (302.95 kg x m/s)/(0.25s)
Fnet = (1211.79N)

Amirite? D:

I'm not sure that it's quite that simple to calculate Fnet, because the dash continues to decelerate during the time when the student is matching speeds with it.

What I chose to do was find the speed of the car 0.25 seconds after the first contact of student with dashboard. Apparently the student and dashboard are "in agreement" at that time. So the student went from his initial 100 kph to whatever that car speed is in the 0.25 seconds allotted. Two speeds, one time interval, voila.

 

[Oriako]

Right.

So then Fnet = 1014N? Did you get that?

I found Vf for the dashboard collision to be 0.68 m/s

 

[gneill]

No, those aren't the numbers I'm seeing. I have the car still traveling at about 20 m/s when the student matches velocity.

What is your value for the time from the moment the crash begins to the first contact of the student and the dashboard?

 

[Oriako]

I found the velocity of the passenger as he hits the dashboard to be 4.15 m/s.

Then I calculated using a = (Vf - Vi)/t, what Vf was.
Vf = (0.25s)(-13.89m/s^2) + (4.15m/s)
Vf = 0.68 m/s

Should that acceleration be positive? I'm fairly sure it shouldnt.

Then Fnet = m(0.68m/s - 4.15m/s) / (0.25s)

Why do I need the time it takes for him to hit the dashboard? I figured that out using

the relative velocity between the dash and student when they first meet.

btw, thank you so much for all the help.

 

[SammyS]

In that last .25 seconds, the car changes velocity by another (.25)·(13.89)=3.47 m/s,
so the student changes velocity by 7.72m/s in .25 seconds

 

[Oriako]

I see how the car changes another 3.47m/s... but isn't the 4.15m/s relative to the interior and not the ground? How do you relate the two? I don't see how it ends up being 7.72m/s. Why did you add them together?

 

[Oriako]

I'm fairly certain that I accounted for the deceleration of the car when I calculated:

Vf = (0.25s)(-13.89m/s^2) + (4.15m/s)
Vf = 0.68 m/s

 

[gneill]

I see how the car changes another 3.47m/s... but isn't the 4.15m/s relative to the interior and not the ground? How do you relate the two? I don't see how it ends up being 7.72m/s. Why did you add them together?

Only the ground frame is inertial after the student meets the dashboard. Everything else will be experiencing accelerations. The car interior is accelerating (or decelerating) with respect to the ground frame, as is the student at another rate. The force that the student feels depends how his inertial mass is accelerated overall.

 

[Oriako]

Oh my goodness. That makes sense now. But, how do I calculate that? I don't see where the velocity change of 7.72 m/s comes from...

 

[SammyS]

From the time the brakes are applied, until the the time the student's velocity matches the dashboard's velocity, the car has changed velocity by about 7.72m/s; that takes about 0.556 seconds. The student begins with the same velocity as the car and again matches the velocity of the car; so he/she has the same velocity change, but it occurs over only .25 seconds. Figure out the student's average acceleration, then multiply by mass.

 

[gneill]

Also, going back over earlier calculations, I think you might want to revisit the numbers for the initial velocity difference between man and machine. You've been using 4.15m/s but it should be slightly different. Your method of calculation was fine, and the numbers you showed look fine too, but the result somehow drifted. Should be more like 4.25m/s.

Gotta go catch some shuteye now (Zzzzz). Back in the a.m.

 

[Oriako]

My teacher said that the deceleration of the car does not matter during the 0.25s. The force would just be whatever his initial velocity was (4.25m/s) and his final velocity (relative to the interior of the car) is surely 0 since his initial velocity of 4.25m/s is relative to the interior of the car.

Therefore:
F x t = m (Vf - Vi)
F x t = (73kg)(0 - 4.25m/s)
F = (73kg)(-4.25m/s) / (0.25s)
F = 1241N

...Can someone please disprove me? And just do out the work for the whole question and explain why.

 

[gneill]

My teacher said that the deceleration of the car does not matter during the 0.25s. The force would just be whatever his initial velocity was (4.25m/s) and his final velocity (relative to the interior of the car) is surely 0 since his initial velocity of 4.25m/s is relative to the interior of the car.

Therefore:
F x t = m (Vf - Vi)
F x t = (73kg)(0 - 4.25m/s)
F = (73kg)(-4.25m/s) / (0.25s)
F = 1241N

...Can someone please disprove me? And just do out the work for the whole question and explain why.

Well, I humbly beg to differ, but would welcome being corrected. To me that's a bit like saying the swing of the bat doesn't affect the force that the baseball experiences, it's all in the pitch. The car frame of reference is not an inertial one while the crash is ongoing. A pseudo-force representing the acceleration would have to be invoked in order to make calculations in that frame of reference work.

First contact of student and dashboard occurs at time t1 = 0.306s into the crash.

At that time the student is traveling at v1 = 100 kph (forward).

0.25 seconds later (t2 = t1 + 0.25s) the student has matched velocities with the car. The car's speed at that time is (given the previously calculated value for the car's acceleration, a, over the 2s crash),

v2 = 100 kph - a*t2 = 72.20 kph (forward)

The student's change of velocity with respect to an inertial frame of reference (in this case either the ground or his/her initial uniform 100 kph state) is ∆v = (v2 - v1) = -7.721 m/s, or -27.80 kph for those who like consistent units used throughout.

That is the change in velocity that the student undergoes as viewed from any inertial reference frame (in Classical Newtonian mechanics). The force that this change in velocity represents is:

f = M*∆v/dt = 2250N

where dt is the 0.25 second interval and M is the 73kg mass of the student.

 

[Oriako]

Thank you very much for the explanation, but would you be able to explain how you solved t1 as being 0.306s into the crash?!

 

[gneill]

Thank you very much for the explanation, but would you be able to explain how you solved t1 as being 0.306s into the crash?!

The car is decelerating at a = 13.889 m/s². The initial separation is d_o = 0.650m. From the point of view of the student, the dashboard suddenly starts accelerating towards him/her at rate a. So the expression for the separation is

d = d_o - (1/2) a t²

Solve for t when d = 0.

 

[Oriako]

How is Vi x t = 0.65m? I'm sorry, I really apologize for not understanding as this is probably really easy. But isn't his Vi just 0 so that entire term becomes zero? And why are you solving for when d = 0? I've done so many of these questions before and I've never really seen anything quite like this...

 

[gneill]

How is Vi x t = 0.65m? I'm sorry, I really apologize for not understanding as this is probably really easy. But isn't his Vi just 0 so that entire term becomes zero? And why are you solving for when d = 0? I've done so many of these questions before and I've never really seen anything quite like this...

It's purely an acceleration question. Since both objects begin at the same velocity (100 kph), we jump to their frame of reference where they are mutually at rest to begin with, and separated by 0.650m.

When the crash begins, the student remains at rest in this frame of reference (traveling inertially). The dash, however, begins to accelerate towards the student. The distance covered by an object accelerating at a over time t is (1/2) a t². The student and dash meet when the dash has covered the 0.650m separation.

 

[Oriako]

Apparently I'm retarded but I just don't understand how you used this formula the way you did.

d = Vi x t + (1/2) a x t^2

As far as I know, this formula is for calculating the distance that an object travels over a specified time interval when you are given how long it takes, the initial velocity, and the acceleration.

Why do you sub in Vi x t as being the distance from the dashboard? Sure, the acceleration is -13.89 m/s^2, but isn't the d on the left side of the equation 0.65m?

 

[gneill]

Apparently I'm retarded but I just don't understand how you used this formula the way you did.

d = Vi x t + (1/2) a x t^2

As far as I know, this formula is for calculating the distance that an object travels over a specified time interval when you are given how long it takes, the initial velocity, and the acceleration.

Why do you sub in Vi x t as being the distance from the dashboard? Sure, the acceleration is -13.89 m/s^2, but isn't the d on the left side of the equation 0.65m?

The complete general expression is

d(t) = d_o + Vi x t + (1/2) a x t²

d_o is the initial displacement.

d(t) is the separation at time t.

In this case, in the frame of reference of the moving car and student at the instant the crash begins, Vi = 0; they are moving at the same rate in the ground frame of reference, so in the student/car frame their relative velocity Vi is zero. When you then put in the negative value for the acceleration, you obtain the formula I gave.

 

[Oriako]

Wow. Why do they remove that d sub-zero term on my formula sheet then!? WTF. It should be: ∆d = Vi x t + (1/2) a x t^2

Thank you so much for all the help you've given me. This problem has been bothering me for days.