How Is Force Calculated When a Student Hits a Dashboard?

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Homework Help Overview

The problem involves a scenario where a student, not wearing a seatbelt, is brought to a stop after colliding with a dashboard in a car that is decelerating. The student has a mass of 73 kg and the car's initial speed is 100 km/h, which is reduced to 0 km/h in 2 seconds. The student travels a distance of 0.650 m before hitting the dashboard, and it takes 0.25 seconds to come to a stop after the impact. The objective is to find the force exerted by the dashboard on the student.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the acceleration of the student as he approaches the dashboard and question whether it can be assumed to match the car's deceleration. There are attempts to clarify the time it takes for the student to hit the dashboard and whether that time is necessary for calculations. Some participants explore the relationship between the student's velocity and the car's velocity during the collision.

Discussion Status

There are various approaches being explored, with some participants suggesting different methods for calculating the force and questioning the assumptions made about the velocities involved. Guidance has been offered regarding the change in momentum and the time intervals relevant to the collision, but no consensus has been reached on the correct approach or final calculations.

Contextual Notes

Participants note the complexity of the problem, including the need to consider relative velocities and the effects of deceleration on both the car and the student. There are mentions of potential discrepancies in calculations and the importance of maintaining consistent frames of reference.

  • #31
Oriako said:
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) = do + Vi x t + (1/2) a x t2

do 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.
 
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  • #32
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.
 

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