Relative Velocity of a Passenger Hitting the Dashboard in a Car Crash

[brochesspro]

Homework Statement

Given below.

Relevant Equations

"SUVAT" Equations
Equation for "Relative Velocity"

Where exactly have I gone wrong? I think it is the part where I assume that the person gains the deceleration of the car, but I have no other way to proceed in this case. Also please only use the equations that I have posted below, and it would help if you would not use the equation for "Relative Velocity" but if it is not possible any other way then please help yourself. But do not go beyond that scope. I thank you for your guidance.

 

[haruspex]

Homework Statement:: Given below.
Relevant Equations:: "SUVAT" Equations
Equation for "Relative Velocity"

View attachment 292263
Where exactly have I gone wrong? I think it is the part where I assume that the person gains the deceleration of the car, but I have no other way to proceed in this case. Also please only use the equations that I have posted below, and it would help if you would not use the equation for "Relative Velocity" but if it is not possible any other way then please help yourself. But do not go beyond that scope. I thank you for your guidance.
View attachment 292261
View attachment 292262

In your $v^2-v_0^2=$ equation, those are initial and final velocities of something that has an acceleration a and travels a distance s over the interval of interest. What is the something you are considering here? Does it satisfy those conditions?

It really is much simpler using relative velocities, specifically, the frame of reference of the passenger. From her perspective, the car accelerates towards her and covers 0.6m.

 

[brochesspro]

I think the something is the passenger. The passenger gains the acceleration of the car when it collides. But I am sure that there is something wrong with this approach because of the concept of inertia. The passenger will retain the initial velocity of the car just before the moment it collided and as it collides. I will do another approach I thought of and in the mean time, please correct me if anything is wrong.

 

[brochesspro]

 

[PeroK]

That doesn't look right. You've assumed that the passenger travels $0.6m$, but the car is also moving after the collision. So, $0.6m$ would only be correct if the car stopped immediately with almost infinite deceleration.

 

[brochesspro]

That doesn't look right. You've assumed that the passenger travels $0.6m$, but the car is also moving after the collision. So, $0.6m$ would only be correct if the car stopped immediately with almost infinite deceleration.

So what do you think I should do? I am confused what I should do in each frame of reference.

 

[PeroK]

So what do you think I should do?

There are two approaches.

1) Set up separate equations for the position and velocity of the passenger and the position and velocity of the car. Calculate when the positions coincide and calculate the relative speed from that.

2) Be a bit cleverer and analyse the problem from a reference frame moving at the initial velocity of the car before the collision.

 

[brochesspro]

I shall try it.

 

[brochesspro]

 

[PeroK]

That looks correct for the time until the passenger hits the dashboard.

 

[brochesspro]

I see. Thank you.

 

[PeroK]

I see. Thank you.

Can you finish it off from there?

 

[brochesspro]

 

[brochesspro]

So this is the full solution, right?

 

[brochesspro]

Never mind, thank you very much.

 

[PeroK]

So this is the full solution, right?

That's more than enough! You got there in the end.

Note that if we take the second approach above, in a frame moving at initial speed of the car the passenger remains at rest and the dashboard starts at rest and accelerates towards the passenger at $a$. It's speed after the displacement $s = 0.6m$ is:$$v^2 = 2as = 240 m^2/s^2$$Hence $$v = \sqrt{240} m/s$$As you got after considerably more calculations!

 

[brochesspro]

That's more than enough! You got there in the end.

Note that if we take the second approach above, in a frame moving at initial speed of the car the passenger remains at rest and the dashboard starts at rest and accelerates towards the passenger at $a$. It's speed after the displacement $s = 0.6m$ is:$$v^2 = 2as = 240 m^2/s^2$$Hence $$v = \sqrt{240} m/s$$As you got after considerably more calculations!

There is also that the vehicle comes to rest before the crash between the person and the dashboard. I did the mathematics below, although I had to see the solution in the end. The final conclusion is below, the solution is not complete with just the previous image. But thanks a lot anyways.

 

[PeroK]

There is also that the vehicle comes to rest before the crash between the person and the dashboard. I did the mathematics below, although I had to see the solution in the end. The final conclusion is below, the solution is not complete with just the previous image. But thanks a lot anyways.
View attachment 292288

That's a good point. I assumed the question wasn't that tricky!